This book will help readers understand the fundamentals of integration, and how to apply them in practical situations.

This book is for anyone who wants to learn more about integration theory and how to apply it in practical situations. Whether you are a student preparing for a calculus class, or an engineer working on a complex problem, this book has something for you.

Integration is one of the fundamental operations in calculus. It allows us to find the area under a curve, the volume of a solid of revolution, and many other applications. integration is basically finding the sum of an infinite number of infinitesimal pieces. It’s crazy to think about, but it turns out to be incredibly useful.

So integration is important because it helps us calculate all sorts of things that we wouldn’t be able to calculate without it. And that’s why it’s such an integral part of the calculus. Get it? Integral? Anyway, that’s why integration is important.

integration can be used to find the area under a curve. It is essentially the inverse of differentiation. Whereas differentiation allows you to find the slope of a curve at any given point, integration allows you to find the area under a curve between two points. Indefinite integration is integration without limits, which means that you are finding the area under the curve for all points between two given points.

As with differentiation, there are many rules and properties that you can use to simplify integration problems. However, it is still generally a complex process. Nevertheless, indefinite integration is a powerful tool that can be used to solve a variety of problems in mathematics and physics. And who knows, maybe one day you’ll be able to use it to find the area under your favorite curve!

There are two main types of integration: indefinite and definite. Indefinite integration is when we don’t have bounds on our integrals, while definite integration has specific boundaries. Both are important, and each has its own uses.

So whether you’re just getting started with integration or you’re a seasoned pro, I hope this article gives you a better understanding of what indefinite integration is and how it works.

Integration by substitution is a technique for finding the integral of a function by making an appropriate substitution. The idea is to find a new function, f(x), whose derivative is equal to the original function, g(x). Then, using the known formula for the integration of f(x), we can find the integral of g(x). This technique can be used to find integrals that would be difficult or impossible to calculate using other methods.

In addition, it can be used to simplify complex integrals. Integration by substitution is a powerful tool that can be used to solve a variety of problems.

If you’ve ever had to integrate a function, then you know that it can be a pain. However, there is a method that can make integration easier: Riemann sums. Riemann sums involve taking the area under the curve and dividing it into small intervals, each of which can be approximated by a rectangle. The heights of the rectangles are determined by the function values at the endpoint of each interval. Then, all of the rectangles are added together to approximate the area under the curve.

Although Riemann sums may seem like a lot of work, they can actually make integration easier because they provide a way to break up the area into manageable pieces. Plus, once you get the hang of it, integration using Riemann sums can be quite satisfying. So next time you’re stuck trying to integrate a function, give Riemann sums a try!

So what’s the big deal with area and Riemann sums? In a nutshell, they’re used to calculate the area under a curve. To do this, we divide the curve into a bunch of small rectangles, calculate the area of each rectangle, and then add them all up. The more rectangles we use, the more accurate our answer will be.

But what if we want to be really precise? That’s where integration comes in. Integration is just a fancy way of saying “area under the curve,” and it gives us a way to calculate the exact area of any shape, no matter how complex. So if you’re ever stuck trying to find the area of something, remember: integration is your friend!

More formally, integration is the process of finding a function that represents the rate of change of a quantity over time.

The definite integral is a tool that allows us to calculate the exact area under a curve. It is defined as the limit of the sum of an infinite number of small rectangles, each infinitesimally close to the curve. The height of each rectangle is equal to the value of the function at its x-coordinate, and the width of each rectangle is infinitesimally small. As the number of rectangles approaches infinity, the sum of their areas approaches the true value of the integral.

This definition may seem confusing at first, but it provides a precise way to calculate the area under any smooth curve. In practice, integration is used to solve problems in physics, engineering, and finance. It can be used to calculate things like displacement, velocity, and acceleration. It can also be used to find out how much money will be required to fund a project over time.

In short, integration is a powerful tool that can be used to solve a wide variety of problems.

The Fundamental Theorem of Calculus is a pretty amazing result. It states that integration and differentiation are inverse operations. In other words, if you know how to do one, you can do the other. And that’s not all! The theorem has two versions, each of which is pretty darn useful.

The first version says that if you have a function f(x), and you want to find its integral, all you need to do is find another function F(x) such that F’(x)=f(x). This is often called the differential form” of the theorem, and it’s very handy when you’re trying to find an antiderivative but you’re having trouble coming up with a closed-form solution.

The second version says that if you have a function f(x), and you want to find its derivative, all you need to do is find another function F(x) such that $$F(a)-F(b) = \int_a^b f(x) dx$$ for any values of a and b. This is sometimes called the integral form” of the theorem, and it’s very handy when you’re trying to find a derivative but you don’t have access to the original function.

So there you have it: the Fundamental Theorem of Calculus in all its glory. Whether you need to find an antiderivative or a derivative, this theorem has got your back!

The Fundamental Theorem of Calculus is a theorem that states that integration and differentiation are inverse operations. In other words, if a function is integrated, the resulting function will be the derivative of the original function.

The theorem has two versions: the first version states that integration is the antiderivative of a function, while the second version states that integration is the area under a curve. The theorem is named after Isaac Newton and Gottfried Leibniz, who are credited with its discovery. Today, the theorem is an important part of calculus and has many applications in mathematics and physics.

Without it, we would not be able to solve differential equations or to perform statistical inference. So if you ever wondered what calculus is good for, wonder no more: it’s good for pretty much everything!

Integration can be used to find the area of a region, the volume of a solid, the rate of change of a function, and much more. Integration is a powerful tool that can be used to solve many problems in mathematics and physics. However, it is not always easy to calculate an integral. In many cases, integration must be done numerically by approximating the area under the curve. This can be done using integration formulas or by using numerical integration methods such as trapezoidal rule or Simpson’s rule. Integration is an important topic in calculus and is essential for understanding many concepts in physics and engineering.

Teaching integration can be tough. Some students love the challenge of finding a function that represents the area under a curve, while others find it tedious. In this book, I’ve tried to find a balance between the two.

We’ll start with the basics of integration, and gradually work our way up to more difficult concepts. And along the way, we’ll apply what we’ve learned to practical situations.

In addition, the book will include worked examples and practice problems to help readers master the material. Whether you’re a student learning integration for the first time, or a seasoned mathematician looking for a refresher, this book will give you the tools you need to succeed.

So whether you’re a student who loves integration or one who just wants to get through it, this book is for you.

## 4.2 Antiderivatives

We go over antidifferentation by defining an antiderivative function and working out examples on finding antiderivatives. We also concentrate on the following problem: if $$F$$ is an antiderivative of the continuous function $$f,$$ then any other antiderivative of $$f$$ must have the form $$F(x)+C$$ where $$C$$ is some constant.
Thus showing, if a function $$F(x)$$ is an antiderivative of the function $$f(x)$$ then so is $$F(x)+C$$ where $$C$$ is called an arbitrary constant. The mean value theorem will be used to show that all derivatives of $$f(x)$$ are of the form $$F(x)+C$$ and that there are no others.

Definition 4.1 A function $$F$$ is called an antiderivative of a given function $$f$$ on an interval $$I$$ if $$F'(x)=f(x)$$ for all $$x$$ in $$I.$$

Theorem 4.1 If $$F$$ is an antiderivative of the continuous function $$f,$$ then any other antiderivative $$G$$, of $$f$$ must have the form $$G(x)=F(x)+C$$ where $$C$$ is some constant.

Proof. We show that if $$F$$ is differentiable in $$[a,b]$$ and $$F'(x)=0$$ for all $$x$$ in $$[a,b],$$ then $$F(x)=F(a)$$ for all $$x$$ in $$[a,b].$$ By the mean-value theorem applied to $$[a,x]$$ for any $$x$$ such that $$a<x\leq b,$$
$\frac{F(x)-F(a)}{x-a}=F'(c)=0$ where $$a<c<x.$$ Thus $$F(x)=F(a)=0,$$ so $$F(x)=F(a),$$ and $$F(x)$$ is constant in $$[a,b].$$ So in fact, for $$F'(x)=f(x)$$ with all $$x$$ in $$[a,b],$$ if we suppose $$G'(x)=f(x)$$ also for $$x$$ in $$[a,b]$$ then $\frac{d(G(x)-F(x))}{dx}=G'(x)-F'(x)=0.$ Thus $$G(x)-F(x)=G(a)-F(a).$$ If we let $$C=G(a)-F(a),$$ then $$G(x)=F(x)+C.$$

The notation $$\int f(x) \, dx=F(x)+C$$ where $$C$$ is an arbitrary constant means that $$F$$ is an antiderivative of $$f.$$ The function $$F$$ is called the indefinite integral of $$f$$ and satisfies the condition that $$F'(x)=f(x)$$ for all $$x$$ in the domain of $$f.$$ It is important to remember that $$F(x)+C$$ represents a family of functions.

## 4.3 Integral Notation

Example 4.1 Find the family of antiderivatives of the function $$f(x)=\sin x$$ and write an equation using the indefinite integral notation.

Solution. If $$F(x)=-\cos x,$$ then $$F'(x)=\sin x,$$ and so an antiderivative of sine is $$-\cos x.$$ Thus the general antiderivative is $$G(x)=-\cos x+C.$$ Therefore, $$\int \sin xdx=-\cos x+C$$ where $$C$$ is an arbitrary constant.

Example 4.2 Find the family of antiderivatives of the function $$f(x)=x^n$$, $$n\geq 0,$$ and write an equation using the indefinite integral notation.

Solution. Since $\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right)=\frac{(n+1)x^n}{n+1}=x^n$ the general antiderivative of $$f$$ is $F(x)=\frac{x^{n+1}}{n+1}+C,$ where $$C$$ is a constant, which is valid for $$n\geq 0$$ because $$f(x)=x^n$$ is defined on the interval $$(-\infty ,+\infty ).$$ Therefore, $\int x^n \, dx=\frac{x^{n+1}}{n+1}+C$ where $$C$$ is an arbitrary constant and $$n\geq 0.$$

Theorem 4.2 Suppose $$f$$ and $$g$$ are integrable functions and $$a$$, $$b$$, and $$c$$ are constants. Then

• $$\displaystyle \int c f(u)du=c\int f(u) \, du$$
• $$\displaystyle \int [f(u)+g(u)] \, du=\int f(u) \, du+\int g(u) \, du$$
• $$\displaystyle \int [f(u)-g(u)] \, du=\int f(u) \, du-\int g(u) \, du$$
• $$\displaystyle \int [a f(u)du+b g(u)] \, du=a\int f(u)du+b \int g(u) \, du$$

Theorem 4.3 Suppose $$C$$ is an arbitrary constant. Then
$\int x^n \, dx= \begin{cases} \displaystyle \frac{x^{n+1}}{n+1}+C & n\neq -1 \\ \ln |x|+C & n=-1. \end{cases}$

Theorem 4.4 Suppose $$u$$ is a differentiable function of $$x$$ and $$C$$ is an arbitrary constant.

• $$\displaystyle \int 0 \, du=0+C$$
• $$\displaystyle \int e^u \, du=e^u+C$$
• $$\displaystyle\int \sin u \, du=-\cos u+C$$
• $$\displaystyle\int \cos u \, du=\sin u+C$$
• $$\displaystyle\int \sec ^2 u \, du=\tan u+C$$
• $$\displaystyle\int \sec u \tan u \, du=\sec u+C$$
• $$\displaystyle\int \csc u \cot u \, du=-\csc u+C$$
• $$\displaystyle\int \csc ^2 u \, du=-\cot u+C$$
• $$\displaystyle\int \frac{1}{\sqrt{1-u^2}} \, du=\sin ^{-1}u+C$$
• $$\displaystyle\int \frac{1}{1+u^2} \, du=\tan ^{-1}u+C$$
• $$\displaystyle\int \frac{1}{|u|\sqrt{u^2-1}} \, du=\sec ^{-1}u+C$$

Example 4.3 Find the general antiderivative of the function $f(x)=\frac{x^2}{x^2+1}.$

Solution. Since $\frac{x^2}{x^2+1}=\frac{x^2+1-1}{x^2+1}=1-\frac{1}{x^2+1}$ we find \begin{align*} \int \frac{x^2}{x^2+1} \, dx & =\int \left(1-\frac{1}{x^2+1}\right) \, dx \\ & =\int dx-\int \frac{1}{x^2+1} \, dx =x-\tan ^{-1}x+C. \end{align*}

Example 4.4 Find the general antiderivative of the function $f(x)=\left(1+\frac{1}{x}\right)\left(1-\frac{4}{x^2}\right).$

Solution. We want to evaluate $$\int \left(1+\frac{1}{x}\right)\left(1-\frac{4}{x^2}\right)dx.$$ Since $\left(1+\frac{1}{x}\right)\left(1-\frac{4}{x^2}\right)=1+\frac{1}{x}-\frac{4}{x^2}-\frac{4}{x^3}$ we use the linearity rule and the power rule to find \begin{align*} \int \left(1+\frac{1}{x}\right)\left(1-\frac{4}{x^2}\right)dx & =\int \left(1+x^{-1}-4x^{-2}-4x^{-3}\right) \, dx \\ & =x+\ln x+\frac{4}{x}+\frac{2}{x^2}+C. \end{align*}

Example 4.5 Find $$f(x)$$ given $$f''(x)=x+\sqrt{x},$$ $$f(1)=1,$$ and $$f'(1)=2.$$

Solution. First we find $$f'$$ by $f'(x)=\int \left(x+\sqrt{x}\right) \, dx=\frac{x^2}{2}+\frac{2 x^{3/2}}{3}+C$ where $$C$$ is some constant which can be determined with $$f'(1)=2.$$ So $f'(1)=\frac{(1)^2}{2}+\frac{2 (1)^{3/2}}{3}+C=\frac{7}{6}+C=2 \qquad \Longrightarrow \qquad C=\frac{5}{6}.$ So in fact $f'(x)=\frac{x^2}{2}+\frac{2 x^{3/2}}{3}+\frac{5}{6}.$ Now to find $$f$$ we follow the same procedure $f(x)=\int \left(\frac{x^2}{2}+\frac{2 x^{3/2}}{3}+\frac{5}{6}\right) \, dx=\frac{x^3}{6}+\frac{4 x^{5/2}}{15}+\frac{5 x}{6}+K$ where $$K$$ is a constant which can be determined with $$f(1)=1.$$ So $f(1)=\frac{(1)^3}{6}+\frac{4 (1)^{5/2}}{15}+\frac{5 (1)}{6}+K=\frac{19}{15}+K=1\qquad \Longrightarrow \qquad K=-\frac{4}{15}.$ Therefore $f(x)=\frac{x^3}{6}+\frac{4 x^{5/2}}{15}+\frac{5 x}{6}-\frac{4}{15}$ as desired.

## 4.4 Finding Area

::: {#thm- } [Area Function] If $$f$$ is a continuous function such that $$f(x)\geq 0$$ for all $$x$$ on the closed interval $$[a,b],$$ then the area bounded by the curve $$y=f(x),$$ the $$x$$-axis, and the vertical lines $$x=a$$ and $$x=t,$$ viewed as a function of $$t,$$ is an antiderivative of $$f(t)$$ on $$[a,b].$$ :::

Example 4.6 Find the area under the parabola $$y=x^2$$ over the interval $$[0,1].$$

Solution. Since $$f(x)=x^2$$ is a continuous function with $$f(x)\geq 0$$ for all $$x$$. The area function is given by $A(t)=\int t^2 \, dt=\frac{1}{3}t^3+C$ and we can determine $$C$$ using $$A(0)=0$$ and so $A(0)=\frac{1}{3}(0)^3+C=0 \qquad \Longrightarrow \qquad C=0$ which means $$A(t)=\frac{1}{3}t^3.$$ Therefore the area under the curve from $$[0,1]$$ is $$A(1)=\frac{1}{3}(1)^3=\frac{1}{3}.$$

## 4.5 Applications of Integration

In the next example we find the demand function given the marginal revenue.

Example 4.7 A manufacturer estimates that the marginal revenue of a certain commodity is $$R'(x)=240+0.1x$$ when $$x$$ units are produced. Find the demand function $$p(x).$$

Solution. Since $R(x)=\int R'(x)dx=\int (240+0.1x) \, dx=240 x+0.05x^2+C$ and because $$R(x)=x p(x),$$ where $$p(x)$$ is the demand function, we must have $$R(0)=0$$ so that $$240(0)+0.05(0)+C+0$$ yielding $$C=0$$ and $p(x)=\frac{R(x)}{x}=\frac{240x+0.05x^2}{x}=240+0.05x.$

Example 4.8 A ball is thrown upward with a speed of 48 ft/s from the edge of a cliff 432 feet above the ground. Find its height above the ground $$t$$ seconds later. When does it reach its maximum height? When does it hit the ground?

Solution. The motion is vertical and we choose the positive direction to be upward. At time $$t$$ the distance above the ground is $$s(t)$$ and the velocity $$v(t)$$ is decreasing. Therefore the acceleration must be negative $$a(t)=\frac{dv}{dt}=-32.$$ Taking the antiderivative
$v(t)=\int a(t) \, dt=-32t+C.$ To determine $$C$$ we use the given information of $$v(0)=48.$$ Thus $$v(0)=-32(0)+C=48$$ and so $$C=48$$ and $$v(t)=-32t+48.$$ It follows the ball reaches its maximum height at $$v(t)=0$$ which means $$t=\frac{48}{32}s=1.5s.$$ Taking the antiderivative
$s(t)=v'(t)=\int (-32t+48) \, dt=-16 t^2+48 t+K$ and using $$s(0)=432$$ we find that $$s(0)=-16 (0)^2+48 (0)+K=432$$ and so $$K=432.$$ Therefore the height function is $s(t)=-16 t^2+48 t+432$ and so the ball hits the ground when $$s(t)=0$$ meaning $$t=\frac{3+3\sqrt{13}}{2}\approx 6.9 s$$ by using the quadratic formula.

Example 4.9 A company has found that the rate of change of its average cost for a product is $\bar{C}'(x)=\frac{1}{4}-\frac{100}{x^2}$ where $$x$$ is the number of units and cost is in dollars. The average cost of producing 20 units is 40,000 dollars.

• Find the average cost function for the product.
• Find the average cost of 100 units of the product.

Solution. To find $$\bar{C}(x)$$ we integrate, so $\bar{C}(x)=\int \left(\frac{1}{4}-\frac{100}{x^2}\right) \, dx=\frac{x}{4}+\frac{100}{x}+K$ to find the constant $$K$$ we use the given that the average cost of producing 20 units is 40,000. So we find the constant $$K$$ by $$\bar{C}(20)=\frac{20}{4}+\frac{100}{20}+K=10+K=40000$$ which means $$K=40000-10=39990.$$ (a) So the average cost function for the product is $$\bar{C}(x)=\frac{x}{4}+\frac{100}{x}+39990.$$ (b) The average cost of 100 units of the product is $\bar{C}(10)=\frac{10}{4}+\frac{100}{10}+39990=\frac{80005}{2}=40002.5$ dollars.

Example 4.10 An excellent film with a very small advertising budget must depend largely on world-of-mouth advertising. In this case, the rate at which weekly attendance might grow can be given by $\frac{dA}{dt}=\frac{-100}{(t+10)^2}+\frac{2000}{(t+10)^3}$ where $$t$$ is in the time in weeks since release and $$A$$ is attendance in millions.

• Find the function that describes weekly attendance at this film.
• Find the attendance at this film in the tenth week.

Solution. The function that describes weekly attendance at this film is found by integration \begin{align*} A(t)& =\int \frac{dA}{dt} \, dt =\int \left(\frac{-100}{(t+10)^2}+\frac{2000}{(t+10)^3}\right) \, dt \\ & =\int \left(\frac{-100}{(t+10)^2}+\frac{2000}{(t+10)^3}\right) \, dt =\frac{100}{t+10}-\frac{1000}{(t+10)^2}+K \end{align*} where $$t$$ is in the time in weeks since release and $$A$$ is attendance in millions. When $$t=0$$ the attendance was 0 and so we find $$K$$ by $A(0)=\frac{100}{0+10}-\frac{1000}{(0+10)^2}+K=K=0.$ Thus $A(t)=\frac{100}{t+10}-\frac{1000}{(t+10)^2}$ and so the attendance at this film in the tenth week is $A(10)=\frac{100}{10+10}-\frac{1000}{(10+10)^2}=\frac{5}{2}=2.5$ million people.

Example 4.11 Suppose the marginal cost for a product is $$\overline{M C}=60\sqrt{x+1}$$ and its fixed cost is 340.00. If the marginal revenue for the product is $$\overline{M R}=80x,$$ find the profit or loss from the production and sale of (a) 3 units (b) 8 units.

Solution. The marginal cost is $$\overline{M C}=60\sqrt{x+1}$$ and so the cost function is fond by integration $$C(x)=\int 60\sqrt{x+1}dx=40 (1+x)^{3/2}+K$$ where $$K$$ is a constant. Since $$C(0)=340$$ we find $$K=300$$ and so the cost function is $$C(x)=40 (1+x)^{3/2}+30.$$ The marginal revenue is $$\overline{MR}=80x$$ and so the revenue function is found by integration $$R(x)=\int 80 xdx=40 x^2.$$ Thus the profit function for this product is $P(x)=R(x)-C(x)=40 x^2-40 (1+x)^{3/2}-300.$ (a) The loss from the sale of 3 units is $$P(3)=40 (3)^2-40 (1+(3))^{3/2}-300=-260$$ dollars. (b) The profit from the sale of 8 units is $$P(8)=40 (8)^2-40 (1+(8))^{3/2}-300=1180$$ dollars.

## 4.6 Constant-Difference Theorem

The following theorem says that two functions with equal derivatives on an open interval differ by a constant on that interval.

::: {#thm- } Constant-Difference Theorem Let $$f$$ and $$g$$ be functions that are continuous on $$[a,b]$$ and differentiable on $$(a,b).$$ If $$f'(x)=g'(x)$$ for all $$x$$ in $$(a,b),$$ then $$f-g$$ is constant on $$(a,b);$$ that is, $$f(x)=g(x)+k$$ where $$k$$ is a constant. :::

Proof. Let $$F(x)=f(x)-g(x).$$ Then $$F'(x)=f'(x)-g'(x)=0$$ for all $$x$$ in $$(a,b).$$ Thus by the Zero Derivative Theorem, $$F(x)=k$$ for some constant $$k$$ and so $$f(x)=g(x)+k$$ as desired.

Example 4.12 Let $$g(x)=\sqrt{x^2+5}.$$ Find a function $$f$$ with $$f'(x)=g'(x)$$ and $$f(2)=1.$$

Solution. Let $$f(x)=\sqrt{x^2+5}+k$$ where $$k$$ is some constant to be determined. Then $$f'(x)=g'(x)$$ and to determine $$k$$ we use $$f(2)=1$$ to obtain $$f(2)=\sqrt{2^2+5}+k=3+k=1.$$ Therefore, $$f(x)=\sqrt{x^2+5}-2$$ is the function we desire.

Example 4.13 Show that $$f(x)=\frac{x+4}{5-x}$$ and $$g(x)=\frac{-9}{x-5}$$ differ by a constant. Are the conditions of the constant difference theorem satisfied? Does $$f'(x)=g'(x).$$

Solution. We simplify $$f'(x)=g'(x)=\frac{9}{(x-5)^2}$$ which is valid on any interval not containing $$x=5.$$ Thus on any interval not containing $$x=5,$$ the constant difference theorem applies. In fact, when $$x\neq 5$$, we have $f(x)-g(x)=\frac{x+4}{5-x}-\frac{-9}{x-5}=-1$

Example 4.14 Let $$f(x)=(x-2)^3$$ and $$g(x)=\left(x^2+12\right)(x-6).$$ Use $$f$$ and $$g$$ to demonstrate the Constant-Difference Theorem.

Solution. The functions $$f$$ and $$g$$ are polynomial functions so they are continuous and differentiable for all real numbers. Also, $f'(x)=3(x-2)^2=3x^2-12x+4=(2x)(x-6)+\left(x^2+12\right)=g'(x)$ for all real numbers. By the Constant Difference Theorem, we have $$f(x)=g(x)+k$$ for some real number $$k.$$

The following theorem is a partial converse to the statement that the derivative of a constant is 0.

::: {#thm- } [Zero-Derivative Theorem] Let $$f$$ be a function that is continuous on $$[a,b]$$ and differentiable on $$(a,b).$$ If $$f'(c)=0$$ for all $$c$$ in $$(a,b)$$ then $$f$$ is constant on $$[a,b].$$ :::

Proof. If $$x_1$$ and $$x_2$$ are different points in $$[a,b]$$ then by the Mean Value Theorem there exists a $$c$$ in $$\left(x_1,x_2\right)$$ such that $f'(c)=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}.$ By hypothesis $$f'(c)=0$$ and so $$f\left(x_2\right)-f\left(x_1\right)=0.$$ Since $$x_1$$ and $$x_2$$ were chosen arbitrarily, $$f$$ is a constant function on $$[a,b].$$

Example 4.15 Consider $$f(x)=\left\{ \begin{array}{ll} 1 & x\geq 0 \\ -1 & x<0 \end{array} \right..$$ Notice that $$f'(x)=0$$ for all $$x$$ in the domain, but $$f$$ is not a constant. Does this example contradict the Zero-Derivative Theorem?

Solution. No it does not, rather it shows that the assumptions of the zero-derivative theorem are necessary.

## 4.7 Exercises

Exercise 4.1 Find an antiderivative for the following sets of functions $$f(x)$$, $$g(x)$$, and $$h(x)$$. Write out an equation using an indefinite integral and each of these functions.

• $$f(x)=2x,$$ $$g(x)=x^2,$$ and $$h(x)=x^2-2x+1$$
• $$f(x)=-3x^{-4},$$ $$g(x)=x^{-4},$$ and $$h(x)=x^{-4}+2x+3$$
• $$f(x)=\frac{-2}{x^3},$$ $$g(x)=\frac{1}{2 x^3}$$ and $$h(x)=x^3-\frac{1}{x^3}$$
• $$f(x)=\frac{2}{3}x^{-1/3},$$ $$g(x)=\frac{1}{3}x^{-2/3}$$ and $$h(x)=\frac{-1}{3}x^{-4/3}$$
• $$f(x)=\frac{1}{3x},$$ $$g(x)=\frac{2}{5x},$$ and $$h(x)=1+\frac{4}{3x}-\frac{1}{x^2}$$
• $$f(x)=\sec ^2x,$$ $$g(x)=\frac{2}{3}\sec ^2\frac{x}{3}$$ and $$h(x)=-\sec ^2\frac{3x}{2}$$
• $$f(x)=e^{-2x}$$ $$g(x)=e^{4x/3},$$ and $$h(x)=e^{-x/5}.$$
• $$f(x)=3^x,$$ $$g(x)=2^{-x},$$ and $$h(x)=\left(\frac{5}{3}\right)^x.$$
• $$f(x)=x-\left(\frac{1}{2}\right)^x$$ $$g(x)=x^2+2^x,$$ and $$h(x)=\pi ^x-x^{-1}.$$

## 4.9 Understanding the Integration by Substitution Rule

Suppose it is known that in a certain country the life expectancy at birth of a female is changing at the rate of $g'(t)=\frac{5.45218}{(1+1.09t)^{0.9}}$ years per year. Here, $$t$$ is measured in years, with $$t=0$$ corresponding to the beginning of 1900. Can we find an expression $$g(t)$$ giving the life expectancy at birth (in years) of a female in that country if the life expectancy at the beginning of 1900 is 50.02 years. If so, for example what is the life expectancy at birth of a female born at the beginning of 2000 in that country?

We use integration to find $$g(t)$$ as follows, $g(t)=\int g'(t) \, dt = \int \frac{5.45218}{(1+1.09t)^{0.9}} \, dt$ by making a change of variables. Let $$u=1+0.09 t$$. Then $$du=1.09 \, dt$$ and so $g(t)=\frac{1}{1.09} \int \frac{5.45218}{u^{0.9}} \, du \approx 50.02 (1+1.09 t)^{0.1} +C.$ Since $$g(0)=50.02$$ we find $$C=0$$, and then $$g(t)=50.02(1+1.09 t)^{0.1}$$. The life expectancy at birth of a female in the year 2000 is $$g(100)=50.02 \left(110^{0.1}\right) \approx 80.04$$ years.

We turn our attention to the substitution rule which (as seen below) is basically the reverse of the chain rule. To see this let $$f$$, $$g$$, and $$u$$ be differentiable functions of $$x$$ such that $f(x)=g(u)\frac{du}{dx}$ Then $\int f(x) \, dx = \int g(u) \frac{du}{dx} \, dx = \int g(u) \, du =G(u)+C$ where $$G$$ is the antiderivative of $$g$$. Indeed, if $$G$$ is an antiderivative of $$g$$, then $$G'(u)=g(u)$$ and by the chain rule $f(x)=\frac{d}{dx}[G(u)]=G'(u)\frac{du}{dx} =g(u) \frac{du}{dx}.$ Now integrating both sides of this equation we obtain $\int f(x) \, dx = \int \left[ g(u) \frac{du}{dx} \right] \, dx =\int \left[ \frac{d}{dx} G(u) \right] \, dx = G(u)+C$ where $$C$$ is a constant.

## 4.10 The Integration by Substitution Rule

::: {#thm- } [Substitution Rule] If $$u=g(x)$$ is a differentiable function whose range is an interval $$I$$ and $$f$$ is continuous on $$I$$, then $\int f(g(x)) g'(x) \, dx =\int f(u) \, du.$ :::

Proof. By the chain rule, $$F(g(x))$$ is an antiderivative of $$f(g(x)) g'(x)$$ whenever $$F$$ is an antiderivative of $$f$$ since $\frac{d}{dx} F(g(x))=F(g(x)) g'(x) =f(g(x)) g'(x).$ If we make the substitution $$u=g(x)$$ then \begin{align*} \int f(g(x)) g'(x) \, dx & =\int \frac{d}{dx} F(g(x)) \, dx =F(g(x))+C =F(u) +C \\ & = \int F'(u)\, du = \int f(u) \, du. \end{align*}

The idea behind the substitution rule is to choose $$u=g(x)$$. Then find $$du=g'(x) \, dx$$. Next, make the substitution to obtain $$\int f(u) \, du$$. If it is possible to integrate the last integral involving $$u$$ only, then the change of variable, using the substitution rule can successful.

Example 4.16 Evaluate $$\displaystyle \int \frac{\cos \sqrt{\theta} }{\sqrt{\theta} \sin^2\sqrt{\theta}}\, d\theta$$.

Solution. Let $$u=\sin\sqrt{\theta}$$. Then $$du=\cos \sqrt{ \theta } \left(\frac{1}{2\sqrt{\theta}} \right) \, d\theta$$. By substitution \begin{align*} \int \frac{\cos \sqrt{\theta} }{\sqrt{\theta} \sin^2\sqrt{\theta}}\, d\theta =2\int \frac{1}{u^2} du =\frac{-2}{u}+C % \\ & \qquad \qquad =-\frac{2}{u}+C =-\frac{2}{\sin \sqrt{\theta}}+C =-2\csc \sqrt{\theta} +C \end{align*} where $$C$$ is a constant.

Example 4.17 Evaluate $$\displaystyle \int r^4 \left(7-\frac{r^5}{10}\right)^3\, dr$$.

Solution. Let $$u=7-\frac{r^5}{10}$$. Then $$du= \frac{-r^4}{2}\, dr$$. By substitution \begin{align*} \int r^4 \left(7-\frac{r^5}{10}\right)^3\, dr & = -2\int \left(7-\frac{r^5}{10}\right)^3\left(-\frac{1}{2}r^4\right) \, dr \\ & =-2\left( \frac{x^4}{4}\right)+C=-\frac{1}{2}\left(7-\frac{r^5}{10}\right)^4+C \end{align*} where $$C$$ is a constant.

Example 4.18 Evaluate $$\displaystyle \int \sqrt{\frac{x-1}{x^5}}\, dx$$.

Solution. Let $$u=1-\frac{1}{x}$$. Then $$du= \frac{1}{x^2}\, dx$$. By substitution \begin{align*} \int \sqrt{\frac{x-1}{x^5}}\, dx & = \int \sqrt{1-\frac{1}{x}} \left(\frac{1}{x^2}\right) \, dx \\ & =\int \sqrt{u} \, du = \frac{u^{3/2}}{3/2}+C =\frac{2}{3}\left(1-\frac{1}{x}\right)^{3/2}+C \end{align*} where $$C$$ is a constant.

Example 4.19 Evaluate $$\displaystyle \int e^{\sin^2 \theta} \sin 2\theta\, dx$$.

Solution. Let $$u=\sin^2\theta$$. Then $$du= 2\sin \theta \cos \theta \, d\theta =\sin 2\theta \, d\theta$$. By substitution \begin{align*} \int e^{\sin^2 \theta} \sin 2\theta\, dx =\int e^u \, du =e^u+C =e^{\sin^2 \theta}+C \end{align*} where $$C$$ is a constant.

Example 4.20 Evaluate $$\displaystyle \int \frac{1}{x\sqrt{x^4-1}}\, dx$$.

Solution. Let $$u=x^2$$. Then $$du= 2x \, dx$$. By substitution \begin{align*} \int \frac{1}{x\sqrt{x^4-1}}\, dx & =\int \frac{2x}{2x^2\sqrt{x^4-1}} \, dx =\int \frac{1}{2u\sqrt{u^2-1}} \, du \\ & =\frac{1}{2}\int\frac{1}{|u| \sqrt{u^2-1}} \, du =\frac{1}{2}\sec^{-1}u +C =\frac{1}{2}\sec^{-1}x^2 +C \end{align*} where $$C$$ is a constant.

Example 4.21 Evaluate $$\displaystyle \int \frac{ \displaystyle e^{\cos^{-1}x}}{\sqrt{1-x^2}} \, dx$$.

Solution. Let $$u=\cos^{-1}x$$. Then $$du= -\frac{1}{\sqrt{1-x^2}}\, dx$$. By substitution \begin{align*} \int \frac{e^{\cos^{-1}x}}{\sqrt{1-x^2}} \, dx =-\int e^u \, du +C =-e^{\cos^{-1}x}+C \end{align*} where $$C$$ is a constant.

Example 4.22 Evaluate $$\displaystyle \int \frac{1}{\sin^{-1}y \sqrt{1-y^2}}\, dy$$.

Solution. Let $$u=\sin^{-1}y$$. Then $$du= \frac{1}{\sqrt{1-y^2}}\, dy$$. By substitution \begin{align*} \int \frac{1}{\sin^{-1}y \sqrt{1-y^2}}\, dy =\int \frac{1}{u}\,du =\ln |u|+C =\ln |\sin^{-1} y|+C \end{align*} where $$C$$ is a constant.

Example 4.23 Evaluate $$\displaystyle \int \frac{\sin \sqrt{\theta}}{\sqrt{\theta \cos^3 \sqrt{\theta}}} \, d\theta$$.

Solution. Let $$u=\cos \sqrt{\theta}$$. Then $$du= -\sin \sqrt{\theta} \frac{1}{2\sqrt{\theta}}\, d\theta$$. By substitution \begin{align*} \int \frac{\sin \sqrt{\theta}}{\sqrt{\theta \cos^3 \sqrt{\theta}}} \, d\theta & = -\int \frac{-2\sin \sqrt{\theta}}{2\sqrt{\theta}\cos^{3/2}\sqrt{\theta}} \, d\theta \\ & =-2\int \frac{1}{u^{3/2}}\, du =\frac{4}{\sqrt{u}}+C =\frac{4}{\sqrt{\cos \sqrt{\theta}}}+C \end{align*} where $$C$$ is a constant.

Example 4.24 Solve the initial value problem: $y''(x)=4\sec^2 2x \tan 2x, \qquad y'(0)=4, \qquad y(0)=-1.$

Solution. First we find $y'=\int 4\sec^2 2x \tan 2x \, dx = \int 2u \, du =u^2+C=\tan^2 2x +C$ using $$u=\tan 2x$$ and $$du=2\sec^2 2x \, dx$$ and where $$C$$ can be determine using $$y'(0)=4$$. We find $$C=4$$ and so $$y'=\tan^2 2x+4$$. Using $$\tan^2 x=\sec^2 x-1$$, we find \begin{align*} y& =\int \left(\tan^2 2x +4\right)\, dx =\int \left(\sec^2 2x +3\right)\, dx \\ & =\int \sec^2 2x \, dx +\int 3\, dx = \frac{1}{2} \tan 2x +3x+C \end{align*} where $$C$$ can be determined by $$y(0)=1$$. We find $$C=-1$$ and therefore $y(x)=\frac{1}{2}\tan 2x+3x-1$ as desired.

Example 4.25 Evaluate $$\int 2 \sin x\cos x \, dx$$.

Solution. Since $$2\sin x \cos x=\sin 2x$$, we use $$u=2x$$. Then $$du=2dx$$ so $\int 2\sin x \cos x\, dx =\int \sin 2x \, dx =\int\frac{1}{2}\sin u \, du =-\frac{1}{2} \cos u +C =-\frac{1}{2} \cos 2x+C$ where $$C$$ is a constant. Alternatively, let $$v=\sin x$$. Then $$dv=\cos x\,dx$$ and so $\int 2\sin x \cos x\, dx =2\int v \, dv =v^2+K =\sin^2x +K$ where $$K$$ is a constant. Alternatively, let $$w=\cos x$$. Then $$dv=-\sin x\,dx$$ and so $\int 2\sin x \cos x\, dx =-2\int w \, dw =-w^2+L =-\cos^2x +L$ where $$L$$ is a constant.

Example 4.26 The slope at each point $$(x,y)$$ on the graph of $$y=F(x)$$ is given by $$x (x^2-1)^{1/3}$$ and the graph passes through the point $$(3,1)$$. Find $$F$$.

Solution. Since $$\frac{dy}{dx}=x(x^2-1)^{1/3}$$ we find $F(x)=\int x (x^2-1)^{1/3} \, dx =\frac{1}{2} \int u^{1/3} \, du =\frac{1}{2}\cdot \frac{3}{4} (x^2-1)^{1/3} +C$ Using, $$F(3)=\frac{3}{8} (2)^4+C=1$$ implies $$C=-5$$, it follows $$F(x)=\frac{3}{8} (x^2-1)^{4/3}-5$$.

## 4.11 An Application to Special Relativity

Example 4.27 According to Einstein’s special theory of relativity, the mass of a particle is given by $m=\frac{m_0}{\sqrt{\displaystyle 1-\frac{v^2}{c^2}}}$ where $$m_0$$ is the rest mass of the particle, $$v$$ is its velocity, and $$c$$ is the speed of light. Suppose that a particle starts from rest at $$t=0$$ and moves along a straight line under the action of a constant force $$F$$. Then, according to Newton’s second law of motion, the equation of motion is $F=m_0 \frac{d}{dt} \left(\frac{v}{\sqrt{1-\frac{v^2}{c^2}}}\right).$ Find the velocity and position functions of the particle. What happens to the velocity of the particle as time goes by?

Solution. From $$F=m_0 \frac{d}{dt} \left(\frac{v}{\sqrt{1-\frac{v^2}{c^2}}}\right)$$ we find $\frac{d}{dt} \left[\frac{v}{\sqrt{1-\frac{v^2}{c^2}}}\right]=\frac{F}{m_0}$ which implies $\frac{v}{\sqrt{1-\frac{v^2}{c^2}}}=\int \frac{F}{m_0} \, dt = \frac{F t}{m_0}+k$ where $$k$$ is a constant. But $$v(0)=0$$, so $$k=0$$. Therefore, $\frac{v}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{Ft}{m_0}$ and after solving for velocity $v=\frac{c F t}{\sqrt{m_0^2 c^2+F^2 t^2}}.$ Next we find the position function, $s=\int v(t) \, dt =\int \frac{c F t}{\sqrt{m_0^2 c^2+F^2 t^2}} \, dt$ Let $$u=m_0^2 c^2+F^2 t^2$$ then $$du = 2 F^2 t \, dt$$. Then by substitution, $s=\frac{c F}{2F^2} \int \frac{1}{u^{1/2}} \, du = \frac{c}{F} \left( m_0^2 c^2+F^2 t^2 \right)^{1/2} +C$ where $$C$$ is the constant of integration. Then $$s(0)=0$$ which shows $$C=-\frac{c^2 m_0}{F}$$, so $s(t)=\frac{c}{F}\left( m_0^2 c^2+F^2 t^2 \right)^{1/2} -\frac{c^2 m_0}{F}.$ Finally, $\lim_{t\to \infty } v(t)= \lim_{t\to \infty } \frac{c F t}{\sqrt{m_0^2+F^2 t^2}} = \lim_{t\to \infty } \frac{c F}{\sqrt{\frac{m_0^2 c^2}{t^2}+F^2}} =\frac{c F}{F}=c$ showing that its velocity approaches the speed of light.

## 4.12 Exercises

Exercise 4.2 Evaluate the following integrals.

• $$\int \frac{\ln x}{x} \, dx$$
• $$\int \sqrt{3x-5} \, dx$$
• $$\int (11-2x)^{-4/5}\, dx$$
• $$\int \csc^2 5x \, dx$$
• $$\int \cot \left[ \ln (x^2+1)\right] \frac{2x}{x^2+1}\, dx$$
• $$\int \frac{6x-9}{(x^2-3x+5)^3}\, dx$$
• $$\int \sin^3 x \cos x\, dx$$
• $$\int \frac{x^2}{x^3+1} \, dx$$
• $$\int \frac{4x}{2x+1} \, dx$$
• $$\int \frac{e^{\sqrt[3]{x}} }{x^{2/3}} \, dx$$
• $$\int x^3(x^2+4)^{1/2} \, dx$$
• $$\int \frac{\ln (x+1)}{x+1} \, dx$$
• $$\int \frac{1}{x^{2/3} (\sqrt[3]{x} +1)} \, dx$$
• $$\int \frac{e^{\sqrt{x}}}{ \sqrt{x} (e^{\sqrt{x}} +1) } \, dx$$

Exercise 4.3 The slope at each point $$(x,y)$$ on the graph of $$y=F(x)$$ is given by $$\frac{2x}{1-3x^2}$$. What is $$F(x)$$ if the graph passes through $$(0,5)$$?

Exercise 4.4 A particle moves along the $$t$$-axis in such a way that at time $$t$$, its velocity is $$v(t)=t^2(t^3-8)^{1/3}$$. At what time does the particle turn around? If the particle starts at a position which we denote as 1, where does it turn around?

Exercise 4.5 A rectangular storage tank has a square base 10 ft on a side. Water is flowing into the tank at the rate modeled by the function $R(t)=t(3t^2+1)^{-1/2}$ in units $$\text{ft}^3/\text{s}$$ at time $$t$$ seconds. If the tank is empty at time $$t=0$$, how mush water does it contain 4 sec later? What is the depth of the water at that time?

Exercise 4.6 Evaluate the following integrals.

• $$\int \frac{1}{(x+1)\sqrt{(x+1)^2}-9} \, dx$$
• $$\int \frac{1}{x\left[9+(\ln x)^2 \right] } \, dx$$
• $$\int \frac{\sqrt{a^2-x^2}}{x^4} \, dx$$
• $$\int \left[ (x^2-1)(x+1)\right]^{-2/3} \, dx$$

Exercise 4.7 Find the indefinite integral for the following then check you answer by differentiation.

• $$\int \left(3t^2+\frac{t}{2}\right) \, dt$$
• $$\int \left(1-x^2-3x^5\right)dx$$
• $$\int \left(x^{-1/3}\right) \, dx$$
• $$\int \left(\frac{\sqrt{x}}{2}+\frac{2}{\sqrt{x}}\right) \, dx$$
• $$\int 2x\left(1-x^{-3}\right)dx$$
• $$\int \left(\frac{4+\sqrt{t}}{t^3}\right) \, dt$$
• $$\int \left(7 \sin \frac{\theta }{3}\right) \, d\theta$$
• $$\int \left(-\frac{\sec ^2x}{3}\right) \, dx$$
• $$\int \left(e^{3x}+5e^{-x}\right) \, dx$$
• $$\int (1.3)^x \, dx$$
• $$\int \left(\frac{1+\cos 4t}{2}\right) \, dt$$
• $$\int \left(\frac{2}{\sqrt{1-y^2}}-\frac{1}{y^{1/4}}\right) \, dy$$
• $$\int \left(1+\tan ^2\theta \right) \, d\theta$$

Exercise 4.8 Verify the following formulas by differentiation where $$C$$ is an arbitrary constant.

• $$\int (7x-2)^3 \, dx=\frac{(7x-2)^4}{28}+C$$
• $$\int \csc ^2\left(\frac{x-1}{3}\right) \, dx=-3 \cot \left(\frac{x-1}{3}\right)+C$$
• $$\int x e^xdx=x e^x-e^x+C$$

Exercise 4.9 Determine which of the following formulas are right and which are wrong. Write an explanation for each. Assume $$C$$ is an arbitrary constant.

• $$\int \tan \theta \sec ^2\theta d\theta =\frac{\sec ^3\theta }{3}+C$$
• $$\int \tan \theta \sec ^2\theta d\theta =\frac{1}{2}\tan ^2 \theta +C$$
• $$\int \tan \theta \sec ^2\theta d\theta =\frac{1}{2}\sec ^2\theta +C$$
• $$\int \sqrt{2x+1} \, dx=\sqrt{x^2+x+C}.$$
• $$\int \sqrt{2x+1} \, dx=\sqrt{x^2+x}+C.$$
• $$\int \sqrt{2x+1} \, dx=\frac{1}{3}\left(\sqrt{2x+1}\right)^3+C.$$

Exercise 4.10 Solve the following initial value problems.

• $$\frac{dy}{dx}=\frac{1}{x^2}+x,$$ $$x>0;$$ $$y(2)=1.$$
• $$\frac{dy}{dx}=\frac{1}{2\sqrt{x}},$$ $$y(4)=0.$$
• $$\frac{d^2r}{dt^2}=\frac{2}{t^3},$$ $$\frac{dr}{dt}|_{t=1}=1,$$ and { }$$r(1)=1.$$
• $$\frac{d^3y}{dx^3}=6,$$ $$y\text{''}(0)=-8,$$ $$y'(0)=0,$$ and { }$$y(0)=5.$$

Exercise 4.11 Find a curve $$y=f(x)$$ with $$d^2y/dx^2=6x$$ and its graph passes through the point $$(0,1)$$ and has a horizontal tangent there. How many curves like this are there? How do you know?

Exercise 4.12 Given $$f(x)=\frac{d}{dx}\left(1-\sqrt{x}\right)$$ and $$g(x)=\frac{d}{dx}(x+2)$$, find each of the following.

• $$\int f(x) \, dx,$$
• $$\int g(x) \, dx,$$
• $$\int [-f(x)] \, dx,$$
• $$\int [-g(x)] \, dx,$$
• $$\int [f(x)+g(x)] \, dx,$$
• $$\int [f(x)-g(x)]dx.$$

## 4.14 Sigma Notation

In order to understand Riemann sums and integration theory correctly it is important to understand summations using sigma notation.

Definition 4.2 If $$a_m,$$ $$a_{m+1},$$ $$\ldots,$$ $$a_n$$ are real numbers such that $$m\leq n,$$ then the summation of these numbers written in sigma notation is $\sum _{i=m}^n a_i=a_m+a_{m+1}+\cdot \cdot \cdot +a_{n-1}+a_n$ and also using functional notation, $\sum _{i=m}^n f(i)=f(m)+f(m+1)+\cdot \cdot \cdot +f(n-1)+f(n)$ where $$f(i)=a_i.$$ The $$i$$ is called the index of summation, the $$a_i$$ are called the $$i$$th term of the sum, and the upper and lower bounds of the summation are $$n$$ and $$m,$$ respectively.

Example 4.28 Write the sum
$$\displaystyle \sum _{k=0}^4 \frac{2k-1}{2k+1}$$ in expanded form.

Solution. In expanded form the sum is \begin{align*} \sum _{k=0}^4 \frac{2k-1}{2k+1} & =\frac{2(0)-1}{2(0)+1}+\frac{2(1)-1}{2(1)+1}+\frac{2(2)-1}{2(2)+1}+\frac{2(3)-1}{2(3)+1}+\frac{2(4)-1}{2(4)+1} \\ & =-1+\frac{1}{3}+\frac{3}{5}+\frac{5}{7}+\frac{7}{9} =\frac{449}{315}. \end{align*}

Example 4.29 Write the sum
$$\displaystyle \left(\sum _{i=1}^6 \frac{i}{i+1}\right)-40$$ in expanded form.

Solution. In expanded form, the sum is \begin{align*} \left(\sum _{i=1}^6 \frac{i}{i+1}\right)-40 & =\frac{1}{1+1}+\frac{2}{2+1}+\frac{3}{3+1}+\frac{4}{4+1}+\frac{5}{5+1}+\frac{6}{6+1}-40 \\ & =\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\frac{5}{6}+\frac{6}{7}-40 =-\frac{4983}{140}. \end{align*}

Example 4.30 Write the sum $$\displaystyle \frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\cdot \cdot \cdot +\frac{1}{7225}$$ in sigma notation.

Solution. In sigma notation, the sum is $\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\cdot \cdot \cdot +\frac{1}{7225}=\sum _{k=1}^{85} \frac{1}{k^2}.$

Example 4.31 Write the sum
$$\displaystyle \frac{3}{7}+\frac{4}{8}+\frac{5}{9}+\frac{6}{10} +\cdot \cdot \cdot +\frac{23}{27}$$ in sigma notation.

Solution. In sigma notation the sum is $\sum _{k=7}^{27} \frac{k-4}{k}.$

Example 4.32 Write the sum $\left[1-\left(\frac{1}{4}\right)^2\right]+\left[1-\left(\frac{2}{4}\right)^2\right]+\left[1-\left(\frac{3}{4}\right)^2\right]+ \cdots +\left[1-\left(\frac{4}{4}\right)^2\right]$ in sigma notation.

Solution. In sigma notation the sum is $\sum _{k=1}^4 \left[1-\left(\frac{k}{4}\right)^2\right].$

Example 4.33 Write the sum $\left(\frac{1}{n}\right)\sqrt{1-\left(\frac{0}{n}\right)^2}+\left(\frac{1}{n}\right)\sqrt{1-\left(\frac{1}{n}\right)^2}+\cdots +\left(\frac{1}{n}\right)\sqrt{1-\left(\frac{n-1}{n}\right)^2}$ in sigma notation.

Solution. In sigma notation the sum is $\sum _{k=0}^{n-1} \left(\frac{1}{n}\right)\sqrt{1-\left(\frac{k}{n}\right)^2}$

## 4.15 Properties of Finite Sums

The linearity rule, subtotal rule, and the dominance rule are the basic properties of summations illustrated below. We also give summation formulas for $\sum _{k=1}^n k^i$ when $$i=1,2,\ldots,6$$ followed by some examples on taking limits of sums.

Theorem 4.5 If $$c$$ and $$d$$ are real numbers that do not depend on integers $$m$$ and $$n$$, then

• $$\displaystyle \sum _{k=1}^n c=n c$$
• $$\displaystyle \sum _{k=1}^n \left(a_k+b_k\right)=\sum _{k=1}^n a_k+\sum _{k=1}^n b_k$$
• $$\displaystyle \sum _{k=1}^n c a_k=c\left( \sum _{k=1}^n a_k \right)$$
• $$\displaystyle \sum _{k=1}^n \left(c a_k+d b_k\right)=c\left( \sum _{k=1}^n a_k \right)+d\left( \sum _{k=1}^n b_k \right)$$
• If $$1 < m < n,$$ then $$\displaystyle \sum _{k=1}^n a_k=\sum _{k=1}^m a_k+\sum _{k=m+1}^n a_k$$
• If $$a_k\leq b_k$$ for all $$k,$$ then $$\displaystyle \sum _{k=1}^n a_k\leq \sum _{k=1}^n b_k.$$

The following formulas will be necessary when determining limits of Riemann sums to find the area under a curve of a polynomial.

Theorem 4.6 The summation formulas for $$\sum _{k=1}^n k^i$$ when $$i=1,2,\ldots,7.$$

• $$\displaystyle \sum _{k=1}^n k=\frac{1}{2} n (n+1)$$
• $$\displaystyle \sum _{k=1}^n k^2=\frac{1}{6} n (n+1) (2 n+1)$$
• $$\displaystyle \sum _{k=1}^n k^3=\frac{1}{4} n^2 (n+1)^2$$
• $$\displaystyle \sum _{k=1}^n k^4=\frac{1}{30} n (n+1) (2 n+1) \left(3 n^2+3 n-1\right)$$
• $$\displaystyle \sum _{k=1}^n k^5=\frac{1}{12} n^2 (n+1)^2 \left(2 n^2+2 n-1\right)$$
• $$\displaystyle \sum _{k=1}^n k^6=\frac{1}{42} n (n+1) (2 n+1) \left(3 n^4+6 n^3-3 n+1\right)$$
• $$\displaystyle \sum _{k=1}^n k^7=\frac{1}{24} n^2 (n+1)^2 \left(3 n^4+6 n^3-n^2-4 n+2\right)$$

Example 4.34 Find the value of the sum $$\displaystyle \sum _{i=1}^{100} i \left(i^2+1\right).$$

Solution. The value of the sum is \begin{align*} & \sum _{i=1}^{100} i \left(i^2+1\right) =\sum _{i=1}^{100} \left(i^3+i\right) =\sum _{i=1}^{100} i^3+\sum _{i=1}^{100} i \\ & \qquad =\frac{1}{4} 100^2 (100+1)^2+\frac{1}{2} 100 (100+1) \\ & \qquad =25502500+5050 =25507550. \end{align*}

Example 4.35 Find the value of the sum $$\displaystyle \sum _{i=1}^4 \left(2^i+i^2\right).$$

Solution. The value of the sum is \begin{align*} \sum _{i=1}^4 \left(2^i+i^2\right) & =\left(2^1+1^2\right)+\left(2^2+2^2\right)+\left(2^3+3^2\right)+\left(2^4+4^2\right) \\ & =3+8+17+32 =60. \end{align*}

Example 4.36 Find the value of the sum $$\displaystyle \sum _{i=1}^n i^2 \left(i^2-i+1\right).$$

Solution. The value of the sum is \begin{align*} & \sum _{i=1}^n i^2 \left(i^2-i+1\right) =\sum _{i=1}^n \left(i^4-i^3+i^2\right) =\sum _{i=1}^n i^4-\sum _{i=1}^n i^3+\sum _{i=1}^n i^2 \\ & \qquad =\frac{1}{30} n (n+1) (2 n+1) \left(3 n^2+3 n-1\right)-\frac{1}{4} n^2 (n+1)^2+\frac{1}{6} n (n+1) (2 n+1) \\ & \qquad =\frac{1}{60} n (n+1) \left(12 n^3+3 n^2+7 n+8\right). \end{align*}

## 4.16 Definition of Riemann Sum

In this topic we illustrate how a Riemann sum can be used to approximate the area under a curve and in doing so, we anticipate the notion of definite integral. We will investigate the area under the curve $$y=2+6 x-5 x^2+x^3,$$ above the $$x$$-axis and between the vertical lines $$x=0$$ and $$x=5.$$

Definition 4.3 A Riemann sum for a function $$f$$ on the closed bounded interval $$[a,b]$$ is a sum of the form $\sum _{k=1}^n f\left(x_i{}^*\right)\Delta x_i$ where $$a=x_0<x_1<\cdot \cdot \cdot <x_n=b$$ and $$x_{i-1}\leq x_i{}^*\leq x_i$$ for $$i=1,\ldots,n$$. Let $$\Delta x_i=x_i-x_{i-1}$$ denote the width of each subinterval. The set $\mathcal{P}=\left\{x_0,x_1,\ldots.,x_n\right\}$ is called a partition of $$[a,b]$$, the largest of the $$\Delta x_i$$ is called the of $$\mathcal{P}$$, and the $$x_i{}^*$$ are called the for the Riemann sum.

Example 4.37 Given the function $$f(x)=x^3,$$ the closed bounded interval $$[0,1],$$ and the partition $$\left\{0,\frac{1}{2},\frac{3}{4},\frac{5}{6},1\right\}$$, compute a Riemann sum.

Solution. Organizing into a table we determine the values, $\begin{array}{c|c|c|c|c} i & \text{subinterval} & x_i{}^* & f\left(x_i{}^*\right) & \Delta x_i \\ \hline 1 & \left[0,\frac{1}{2}\right] & \frac{1}{3} & \left(\frac{1}{3}\right)^3=\frac{1}{27} & \frac{1}{2}-0=\frac{1}{2} \\ 2 & \left[\frac{1}{2},\frac{3}{4}\right] & \frac{2}{3} & \left(\frac{2}{3}\right)^3=\frac{8}{27} & \frac{3}{4}-\frac{1}{2}=\frac{1}{4} \\ 3 & \left[\frac{3}{4},\frac{5}{6}\right] & \frac{8}{10} & \left(\frac{8}{10}\right)^3=\frac{64}{125} & \frac{5}{6}-\frac{3}{4}=\frac{1}{12} \\ 4 & \left[\frac{5}{6},1\right] & \frac{9}{10} & \left(\frac{9}{10}\right)^3=\frac{729}{1000} & 1-\frac{5}{6}=\frac{1}{6} \end{array}$ So the Riemann sum for these $$x_i^*$$ is \begin{align*} \sum _{i=1}^4 f\left(x_i{}^*\right)\Delta x_i & =f\left(x_1{}^*\right)\Delta x_1+f\left(x_2{}^*\right)\Delta x_2+f\left(x_3{}^*\right)\Delta x_3+f\left(x_4{}^*\right)\Delta x_4 \\ & =\left(\frac{1}{27}\right)\left(\frac{1}{2}\right)+\left(\frac{8}{27}\right)\left(\frac{1}{4}\right)+\left(\frac{64}{125}\right)\left(\frac{1}{12}\right)+ \left(\frac{729}{1000}\right)\left(\frac{1}{6}\right) \\ & =\frac{2773}{10800}. \end{align*}

## 4.17 Estimating the Area Under a Curve

Example 4.38 Given the function $$f(x)=2+6 x-5 x^2+x^3,$$ the closed bounded interval $$[0,5],$$ and the partition $$\{0,1,2,3,4,5\},$$ compute a Riemann sum to approximate the area.

Solution. Organizing into a table we determine the values, $\begin{array}{c|c|c|c|c} i & \text{subinterval} & x_i{}^* & f\left(x_i{}^*\right) & \Delta x_i \\ \hline 1 & [0,1] & 0 & 2+6 (0)-5 (0)^2+(0)^3=2 & 1 \\ 2 & [1,2] & 1 & 2+6 (1)-5 (1)^2+(1)^3=4 & 1 \\ 3 & [2,3] & 2 & 2+6 (2)-5 (2)^2+(2)^3=2 & 1 \\ 4 & [3,4] & 3 & 2+6 (3)-5 (3)^2+(3)^3=2 & 1 \\ 5 & [4,5] & 4 & 2+6 (4)-5 (4)^2+(4)^3=10 & 1 \end{array}$ So the Riemann sum for these $$x_i{}^*$$ is $\sum _{i=1}^5 f\left(x_i{}^*\right)\Delta x_i =\text{(2)(1)+(4)(1)+(2)(1)+(2)(1)+(10)(1)} =20.$ Figure $$\ref{fig:graph32}$$ shows the Riemann sum as the approximate area under the given curve.

## 4.18 Refining Partitions

For our next example we will use a finer partition, say a partition using 14 subintervals between $$x=0$$ and $$x=5$$ (also with uniform width). We will still use left-endpoints for our subinterval representatives. Our second estimate for the area is $$28.1481$$.

Example 4.39 Given the function $$f(x)=2+6 x-5 x^2+x^3,$$ the closed bounded interval $$[0,5],$$ and the partition $\left\{0,\frac{1}{3},\frac{2}{3},1,\frac{4}{3},\frac{5}{3},2,\frac{7}{3},\frac{8}{3},3,\frac{10}{3},\frac{11}{3},4,\frac{13}{3},\frac{14}{3},5\right\},$ compute a Riemann sum to approximate the area.

Solution. Organizing into a table we determine the values, $\begin{array}{c|c|c|c|c} i & \text{subinterval} & x_i{}^* & f\left(x_i{}^*\right) & \Delta x_i \\ \hline 1 & \left[0,\frac{1}{3}\right] & 0 & 2+6 (0)-5 (0)^2+(0)^3=2 & \frac{1}{3} \\ 2 & \left[\frac{1}{3},\frac{2}{3}\right] & \frac{1}{3} & 2+6 \left(\frac{1}{3}\right)-5 \left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3=\frac{94}{27} & \frac{1}{3} \\ 3 & \left[\frac{2}{3},1\right] & \frac{2}{3} & 2+6 \left(\frac{2}{3}\right)-5 \left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\right)^3=\frac{110}{27} & \frac{1}{3} \\ 4 & \left[1,\frac{4}{3}\right] & 1 & 2+6 (1)-5 (1)^2+(1)^3=4 & \frac{1}{3} \\ 5 & \left[\frac{4}{3},\frac{5}{3}\right] & \frac{4}{3} & 2+6 \left(\frac{4}{3}\right)-5 \left(\frac{4}{3}\right)^2+\left(\frac{4}{3}\right)^3=\frac{94}{27} & \frac{1}{3} \\ 6 & \left[\frac{5}{3},2\right] & \frac{5}{3} & 2+6 \left(\frac{5}{3}\right)-5 \left(\frac{5}{3}\right)^2+\left(\frac{5}{3}\right)^3=\frac{74}{27} & \frac{1}{3} \\ 7 & \left[2,\frac{7}{3}\right] & 2 & 2+6 (2)-5 (2)^2+(2)^3=2 & \frac{1}{3} \\ 8 & \left[\frac{7}{3},\frac{8}{3}\right] & \frac{7}{3} & 2+6 \left(\frac{7}{3}\right)-5 \left(\frac{7}{3}\right)^2+\left(\frac{7}{3}\right)^3=\frac{40}{27} & \frac{1}{3} \\ 9 & \left[\frac{8}{3},3\right] & \frac{8}{3} & 2+6 \left(\frac{8}{3}\right)-5 \left(\frac{8}{3}\right)^2+\left(\frac{8}{3}\right)^3=\frac{38}{27} & \frac{1}{3} \\ 10 & \left[3,\frac{10}{3}\right] & 3 & 2+6 (3)-5 (3)^2+(3)^3=2 & \frac{1}{3} \\ 11 & \left[\frac{10}{3},\frac{11}{3}\right] & \frac{10}{3} & 2+6 \left(\frac{10}{3}\right)-5 \left(\frac{10}{3}\right)^2+\left(\frac{10}{3}\right)^3=\frac{94}{27} & \frac{1}{3} \\ 12 & \left[\frac{11}{3},4\right] & \frac{11}{3} & 2+6 \left(\frac{11}{3}\right)-5 \left(\frac{11}{3}\right)^2+\left(\frac{11}{3}\right)^3=\frac{164}{27} & \frac{1}{3} \\ 13 & \left[4,\frac{13}{3}\right] & 4 & 2+6 (4)-5 (4)^2+(4)^3=10 & \frac{1}{3} \\ 14 & \left[\frac{13}{3},\frac{14}{3}\right] & \frac{13}{3} & 2+6 \left(\frac{13}{3}\right)-5 \left(\frac{13}{3}\right)^2+\left(\frac{13}{3}\right)^3=\frac{418}{27} & \frac{1}{3} \\ 15 & \left[\frac{14}{3},5\right] & \frac{14}{3} & 2+6 \left(\frac{14}{3}\right)-5 \left(\frac{14}{3}\right)^2+\left(\frac{14}{3}\right)^3=\frac{614}{27} & \frac{1}{3} \end{array}$ So the Riemann sum $$\sum _{i=1}^{15} f\left(x_i{}^*\right)\Delta x_i$$ for these $$x_i{}^*$$ is $$\frac{760}{27}$$.

Example 4.40 Use a Riemann sum to approximate the area under the graph of $$f(x)=6x^2+2x+4$$ on $$[1,3]$$ with 8 subintervals.

Solution. As a partition we choose $$\mathcal{P}=\left\{1,\frac{5}{4},\frac{3}{2},\frac{7}{4},2,\frac{9}{4},\frac{5}{2},\frac{11}{4},3\right\}.$$ Organizing our computations and choices for our subinterval representations we have $\begin{array}{c|c|c|c|c|c} i & \text{subinterval} & x_i{}^* & f\left(x_i{}^*\right) & \Delta x_i & f\left(x_i{}^*\right)\Delta x_i \\ \hline 1 & \left[1,\frac{5}{4}\right] & 1 & 6(1)^2+2(1)+4=12 & \frac{1}{4} & 3 \\ 2 & \left[\frac{5}{4},\frac{3}{2}\right] & \frac{5}{4} & 6\left(\frac{5}{4}\right)^2+2\left(\frac{5}{4}\right)+4=\frac{127}{8} & \frac{1}{4} & \frac{127}{32} \\ 3 & \left[\frac{3}{2},\frac{7}{4}\right] & \frac{3}{2} & 6\left(\frac{3}{2}\right)^2+2\left(\frac{3}{2}\right)+4=\frac{41}{2} & \frac{1}{4} & \frac{41}{8} \\ 4 & \left[\frac{7}{4},2\right] & \frac{7}{4} & 6\left(\frac{7}{4}\right)^2+2\left(\frac{7}{4}\right)+4=\frac{207}{8} & \frac{1}{4} & \frac{207}{32} \\ 5 & \left[2,\frac{9}{4}\right] & 2 & 6(2)^2+2(2)+4=32 & \frac{1}{4} & 8 \\ 6 & \left[\frac{9}{4},\frac{5}{2}\right] & \frac{9}{4} & 6\left(\frac{9}{4}\right)^2+2\left(\frac{9}{4}\right)+4=\frac{311}{8} & \frac{1}{4} & \frac{311}{32} \\ 7 & \left[\frac{5}{2},\frac{11}{4}\right] & \frac{5}{2} & 6\left(\frac{5}{2}\right)^2+2\left(\frac{5}{2}\right)+4=\frac{93}{2} & \frac{1}{4} & \frac{93}{8} \\ 8 & \left[\frac{11}{4},3\right] & \frac{11}{4} & 6\left(\frac{11}{4}\right)^2+2\left(\frac{11}{4}\right)+4=\frac{439}{8} & \frac{1}{4} & \frac{439}{32} \end{array}$ So the Riemann sum for this partition $$\mathcal{P}$$ and these $$x_i^*$$ is $$\sum _{i=1}^8 f\left(x_i{}^*\right)\Delta x_i =493/8.$$ Thus, the approximate the area under the graph of $$f(x)=6x^2+2x+4$$ on $$[1,3]$$ with 8 subintervals is $$\frac{493}{8}$$ as shown in Figure $$\ref{fig:graph34}$$.

Example 4.41 Use a Riemann sum to approximate the area under the graph of $$f(x)=\sqrt{1+x^2}$$ on $$[0,1]$$ with 10 subintervals.

Solution. As a partition we choose $\mathcal{P}=\left\{0,\frac{1}{10},\frac{1}{5},\frac{3}{10},\frac{2}{5},\frac{1}{2},\frac{3}{5},\frac{7}{10},\frac{4}{5},\frac{9}{10},1\right\}.$ Organizing our computations and choices for our subinterval representations we have $\begin{array}{c|c|c|c|c|c} i & \text{subinterval} & x_i{}^* & f\left(x_i{}^*\right) & \Delta x_i & f\left(x_i{}^*\right)\Delta x_i \\ \hline 1 & \left[0,\frac{1}{10}\right] & \frac{1}{10} & \sqrt{1+\left(\frac{1}{10}\right)^2}=\frac{\sqrt{101}}{10} & \frac{1}{10} & \frac{\sqrt{101}}{100} \\ 2 & \left[\frac{1}{10},\frac{1}{5}\right] & \frac{1}{5} & \sqrt{1+\left(\frac{1}{5}\right)^2}=\frac{\sqrt{26}}{5} & \frac{1}{10} & \frac{\sqrt{\frac{13}{2}}}{25} \\ 3 & \left[\frac{1}{5},\frac{3}{10}\right] & \frac{3}{10} & \sqrt{1+\left(\frac{3}{10}\right)^2}=\frac{\sqrt{109}}{10} & \frac{1}{10} & \frac{\sqrt{109}}{100} \\ 4 & \left[\frac{3}{10},\frac{2}{5}\right] & \frac{2}{5} & \sqrt{1+\left(\frac{2}{5}\right)^2}=\frac{\sqrt{29}}{5} & \frac{1}{10} & \frac{\sqrt{29}}{50} \\ 5 & \left[\frac{2}{5},\frac{1}{2}\right] & \frac{1}{2} & \sqrt{1+\left(\frac{1}{2}\right)^2}=\frac{\sqrt{5}}{2} & \frac{1}{10} & \frac{1}{4 \sqrt{5}} \\ 6 & \left[\frac{1}{2},\frac{3}{5}\right] & \frac{3}{5} & \sqrt{1+\left(\frac{3}{5}\right)^2}=\frac{\sqrt{34}}{5} & \frac{1}{10} & \frac{\sqrt{\frac{17}{2}}}{25} \\ 7 & \left[\frac{3}{5},\frac{7}{10}\right] & \frac{7}{10} & \sqrt{1+\left(\frac{7}{10}\right)^2}=\frac{\sqrt{149}}{10} & \frac{1}{10} & \frac{\sqrt{149}}{100} \\ 8 & \left[\frac{7}{10},\frac{4}{5}\right] & \frac{4}{5} & \sqrt{1+\left(\frac{4}{5}\right)^2}=\frac{\sqrt{41}}{5} & \frac{1}{10} & \frac{\sqrt{41}}{50} \\ 9 & \left[\frac{4}{5},\frac{9}{10}\right] & \frac{9}{10} & \sqrt{1+\left(\frac{9}{10}\right)^2}=\frac{\sqrt{181}}{10} & \frac{1}{10} & \frac{\sqrt{181}}{100} \\ 10 & \left[\frac{9}{10},1\right] & 1 & \sqrt{1+(1)^2}=\sqrt{2} & \frac{1}{10} & \frac{1}{5 \sqrt{2}} \\ \hline & & & & \text{Total}\approx & 1.16909 \end{array}$ So the approximate the area under the graph of $$f(x)=\sqrt{1+x^2}$$ on $$[0,1]$$ using 10 subintervals is 1.16909 as shown in Figure $$\ref{fig:graph33}$$.

## 4.19 Area using Limits of Riemann Sums

It is important to be able to evaluate limits of sums so here are a couple of examples.

Theorem 4.7 Suppose $$f$$ is continuous and $$f(x)\geq 0$$ throughout the interval $$[a,b].$$ Then the area of the region under the curve $$y=f(x)$$ over this interval is $A=\lim _{\Delta x\to 0}\sum _{k=1}^n f(a+k \Delta x)\Delta x$ where $$\Delta x=(b-a)/n.$$

Example 4.42 Evaluate the limit $$\displaystyle \lim _{n\to +\infty }\sum _{k=1}^n \frac{k}{n^2}.$$

Solution. The value of the limit is \begin{align*} \lim _{n\to +\infty }\sum _{k=1}^n \frac{k}{n^2} & =\lim _{n\to +\infty }\frac{1}{n^2} \left(\sum _{k=1}^n k\right) \\ & =\lim _{n\to +\infty }\frac{1}{n^2} \left(\frac{1}{2} n (n+1)\right) =\lim _{n\to +\infty }\frac{n+1}{2 n} =\frac{1}{2}. \end{align*}

Example 4.43 Evaluate the limit $$\displaystyle \lim _{n\to +\infty }\sum _{k=1}^n \left(1+\frac{2k}{n}\right)^2\left(\frac{2}{n}\right).$$

Solution. The value of the limit is \begin{align*} & \lim _{n\to +\infty }\sum _{k=1}^n \left(1+\frac{2k}{n}\right)^2\left(\frac{2}{n}\right) =\lim _{n\to +\infty }\left(\frac{2}{n}\right)\sum _{k=1}^n \left(1+\frac{2k}{n}\right)^2 \\ & \qquad =\lim _{n\to +\infty }\left(\frac{2}{n}\right)\sum _{k=1}^n \left(\frac{4 k^2}{n^2}+\frac{4 k}{n}+1\right) =\lim _{n\to +\infty }\left(\frac{2}{n}\right)\left(\frac{13 n^2+12 n+2}{3 n}\right) \\ & \qquad =\lim _{n\to +\infty }\frac{2 \left(13 n^2+12 n+2\right)}{3 n^2} =\frac{26}{3}. \end{align*}

Example 4.44 Find the exact area under the curve $$y=x^2+4$$ on $$[2,10].$$

Solution. We will use the formula $A=\lim _{\Delta x\to 0}\sum _{k=1}^n f(a+k \Delta x)\Delta x$ with $$b=10$$ and $$a=2.$$ We see $$\Delta x=\frac{10-2}{n}=\frac{8}{ n};$$ and we notice that $$n\to +\infty$$ as $$\Delta x\to 0$$ and so the area $$A$$ is given by \begin{align*} A&=\lim _{\Delta x\to 0}\sum _{k=1}^n f(a+k \Delta x)\Delta x \\ & =\lim _{n\to \infty }\sum _{k=1}^n f\left(2+k \frac{8}{ n}\right)\frac{8}{ n} \\ & =\lim _{n\to \infty }\frac{8}{ n}\sum _{k=1}^n \left[\left(2+\frac{8 k}{ n}\right)^2+4\right] \\ & =\lim _{n\to \infty }\frac{8}{ n}\sum _{k=1}^n \left(8+4\left(\frac{8 k}{n}\right)+\frac{64 k^2}{n^2}\right) \\ & =\lim _{n\to \infty }\frac{8}{ n}\sum _{k=1}^n \left(8+4\left(\frac{8 k}{n}\right)+\frac{64 k^2}{n^2}\right) \\ & =\lim _{n\to \infty }\frac{8}{ n}\left[\sum _{k=1}^n 8+\frac{32}{n}\sum _{k=1}^n k+\frac{64 }{n^2}\sum _{k=1}^n k^2\right] \\ & =\lim _{n\to \infty }\frac{8}{ n}\left[8n+\frac{32}{n}\left(\frac{1}{2} n (n+1)\right)+\frac{64 }{n^2}\left(\frac{1}{6} n (n+1) (2 n+1)\right)\right] \\ & =\lim _{n\to \infty }\frac{64 \left(4+18 n+17 n^2\right)}{3 n^2} \qquad \text{(after algebraic simplification)} \\ & =\frac{64(17)}{3} \\ & =\frac{1088}{3}. \end{align*} Therefore, the exact area under the curve $$y=x^2+4$$ bounded by the lines $$y=0,$$ $$x=2,$$ and $$x=10$$ is $$\frac{1088}{3}.$$

Example 4.45 Find the exact area under the curve $$y=4x^3+3x^2$$ on $$[0,1].$$

Solution. We will use the formula $A=\lim _{\Delta x\to 0}\sum _{k=1}^n f(a+k \Delta x)\Delta x$ with $$b=1$$ and $$a=0.$$ We see $$\Delta x=\frac{1}{n};$$ and we notice that $$n\to +\infty$$ as $$\Delta x\to 0$$ and so the area $$A$$ is given by \begin{align*} A&=\lim _{\Delta x\to 0}\sum _{k=1}^n f(a+k \Delta x)\Delta x \\ & =\lim _{n\to \infty }\sum _{k=1}^n f\left(\frac{k}{n}\right)\frac{1}{n} \\ & =\lim _{n\to \infty }\frac{1}{ n}\sum _{k=1}^n \left[4\left(\frac{k}{n}\right)^3+3\left(\frac{k}{n}\right)^2\right] \\ & =\lim _{n\to \infty }\frac{1}{ n}\left[\frac{4}{n^3}\sum _{k=1}^n k^3+\frac{3}{n^2}\sum _{k=1}^n k^2\right] \\ & =\lim _{n\to \infty }\frac{1}{ n}\left[\frac{4}{n^3}\left(\frac{1}{4} n^2 (n+1)^2\right)+\frac{3}{n^2}\left(\frac{1}{6} n (n+1) (2 n+1)\right)\right] \\ & =\lim _{n\to \infty }\frac{3+7 n+4 n^2}{2 n^2} \qquad \text{(after algebraic simplification)} \\ & =2. \end{align*} Therefore, the exact area under the curve $$y=4x^3+3x^2$$ bounded by the lines $$y=0,$$ $$x=0,$$ and $$x=1$$ is $$2.$$

## 4.20 Exercises

Exercise 4.13 Use finite approximations to estimate the area under the graph of $$f(x)=x^3$$ between $$x=0$$ and $$x=1.$$

Exercise 4.14 Use finite approximations to estimate the area under the graph of $$f(x)=4-x^2$$ between $$x=-2$$ and $$x=2.$$

Exercise 4.15 Using rectangles whose height is given by the value of the function at the midpoint of the rectangle’s base (the midpoint rule) estimate the area under the graph of the function $$f(x)=x^2$$ between $$x=0$$ and $$x=1,$$ using first two and then four rectangles.

Exercise 4.16 Using rectangles whose height is given by the value of the function at the midpoint of the rectangle’s base (the midpoint rule) estimate the area under the graph of the function $$f(x)=4-x^2$$ between $$x=-2$$ and $$x=2,$$ using first two and then four rectangles.

Exercise 4.35 An object is shot straight upward from sea level with an initial velocity of $$400 \text{ft}/\sec .$$ (a) Assuming that gravity is the only force acting on the object, give an upper estimate for its velocity after 5 sec have elapsed. Use $$g=-32 \text{ft}\left/\sec ^2\right.$$ for the gravitational acceleration. (b) Find a lower estimate for the height attained after 5 sec. :::

Exercise 4.17 Use a finite sum to estimate the average of $$f(x)=\left(\frac{1}{2}\right)+\sin ^2\pi t$$ on the closed bounded interval $$[0,2]$$ by partitioning the interval into four subintervals of equal length and evaluating $$f$$ at the subinterval midpoints.

Exercise 4.18 Write the sum $$\sum _{k=1}^2 \frac{6k}{k+1}$$ without sigma notation and then evaluate the sum.

Exercise 4.19 Write the sum $$\sum _{k=1}^4 (-1)^k\cos k \pi$$ without sigma notation and then evaluate the sum.

Exercise 4.20 Rewrite the expression $$1-2+4-8+16-32$$ in sigma notation.

Exercise 4.21 Which formula is not equivalent to the other two?

• $$\sum _{k=2}^4 \frac{(-1)^{k-1}}{k-1}$$
• $$\sum _{k=0}^2 \frac{(-1)^k}{k+1}$$
• $$\sum _{k=-1}^1 \frac{(-1)^k}{k+2}$$

Exercise 4.22 Rewrite the sum using sigma notation $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}.$$

Exercise 4.23 Rewrite the sum using sigma notation $$\frac{-1}{5}+\frac{2}{5}-\frac{3}{5}+\frac{4}{5}-\frac{5}{5}.$$

Exercise 4.24 Suppose that $$\sum _{k=1}^n a_k=0$$ and $$\sum _{k=1}^n b_k=1.$$ Find the values of

• $$\sum _{k=1}^n 8 a_k$$
• $$\sum _{k=1}^n 250 b_k$$
• $$\sum _{k=1}^n \left( a_k+1\right)$$
• $$\sum _{k=1}^n \left(b_k-1\right)$$.

Exercise 4.25 Evaluate the following sums.

• $$\sum _{k=1}^7 (-2k)$$
• $$\sum _{k=1}^6 \left(k^2-5\right)$$
• $$\sum _{k=1}^5 \frac{k^3}{225}+\left( \sum _{k=1}^5 k \right)^3$$

Exercise 4.26 Graph the function $$f(x)=-x^2$$ over the interval $$[0,2].$$ Partition the interval into four subintervals of equal length. Then add to your sketch the rectangles associated with the Riemann sum $$\sum _{k=1}^n f\left(c_k\right)\Delta x_k ,$$ given that $$c_k$$ is the (a) left-hand endpoint, (b) right-hand endpoint, (c) the midpoint of the $$k$$-th subinterval . Make a separate sketch for each set of rectangles.

Exercise 4.27 Graph the function $$f(x)=\sin x\text{ }+1$$ over the interval $$[-\pi ,\pi ].$$ Partition the interval into four subintervals of equal length. Then add to your sketch the rectangles associated with the Riemann sum $$\sum _{k=1}^n f\left(c_k\right)\Delta x_k ,$$ given that $$c_k$$ is the (a) left-hand endpoint, (b) right-hand endpoint, (c) the midpoint of the $$k$$-th subinterval . Make a separate sketch for each set of rectangles.

## The Definition of the Definite Integral

Suppose a bounded function $$f$$ is given along with a closed interval $$[a,b]$$ on which $$f$$ is defined. First we will partition the interval $$[a,b]$$ into $$n$$ subintervals by choosing points $$\{x_0, x_1, \ldots, x_n\}$$ arranged in such a way that $a=x_0 < x_1 < x_2 < \cdots <x_{n-1}< x_n =b.$ Call this partition $$P$$. Now for $$i=1, 2, \ldots, n$$ the $$i^{\text{th}}$$ subinterval width is $$\Delta_i=x_i-x_{i-1}$$. The largest of these widths is called the norm of the partition $$P$$ and is denoted by $$\norm{P}$$; that is, $$\norm{P}=\max_{i=1,2,\ldots,n} \{\Delta_i\}$$. Choose a number (arbitrarily) from each subinterval, say $$x_i^*$$. For $$i=i,2,\ldots,n$$ the number $$x_i^*$$ chosen from the $$i^{\text{th}}$$ is called the $$i^{\text{th}}$$ subinterval representative of the partition $$P$$. Then the sum $\sum_{i=1}^n f(x_i^*) \Delta x_i$ is called the Riemann sum associated with $$f$$, the partition $$P$$, the chosen subinterval representatives.

Definition 4.4 Let $$f$$ be a bounded function defined on an interval $$[a,b]$$. Then the definite integral of $$f$$ over $$[a,b]$$ is defined as $\int_a^b f(x)\, dx := \lim_{\norm{P} \to 0} \sum_{k=1}^n f(c_k) \Delta x.$ if the limit exists for all partitions $$P$$ of $$[a,b]$$ and all choices of $$c_k$$ in $$[x_{k-1},x_k]$$.
When the above limit exists, we say the Riemann sums of $$f$$ on $$[a,b]$$ converge to the definite integral $$I=\int_a^b f(x) \, dx$$ and that $$f$$ is integrable over $$[a,b]$$.

For special cases we can define particular kinds of Riemann sums. For example, if we choose equally spaced subintervals using $$\Delta x_i= \Delta x =\frac{b-a}{n}$$ and $$x_i^*=a+k \Delta x$$ for $$i=1,2,\ldots,n$$ then the Riemann sum formed is $\sum_{i=1}^n f(a+k \Delta x) \Delta x$ and the partition $$P$$ used to form this Riemann sum is called a regular partition . For regular partitions, $$\norm{P}\to 0$$ if and only if $$n\to \infty$$ and so the definition of the definite integral sometimes is given as $\int_a^b f(x)\, dx := \lim_{n \to \infty } \sum_{k=1}^n f(c_k) \Delta x.$

Theorem 4.8 If a function $$f$$ is continuous on an interval $$[a,b]$$ then its definite integral over $$[a,b]$$ exists.

Example 4.46 Express $\lim_{\norm{P}\to 0 } \sum_{k=1}^n \sqrt{4-c_k^2} \Delta x_k$ where $$P$$ is a partition of $$[0,1]$$, as a definite integral and find its value using geometry.

Solution. See Figure $$\ref{fig:graph41}$$. Recall the area of a circle with radius 2 is $$4\pi$$. Let $$f(x)=\sqrt{4-x^2}$$. Since $$f$$ is continuous on $$[0,1]$$,
$\pi =\int_0^1 \sqrt{4-x^2}\, dx = \lim_{\norm{P}\to 0 } \sum_{k=1}^n \sqrt{4-c_k^2} \Delta x_k.$

Example 4.47 Express $\lim_{\norm{P}\to 0 } \sum_{k=1}^n (\tan c_k) \Delta x_k$ where $$P$$ is a partition of $$[0,\pi/4]$$, as a definite integral and find its approximate value using geometry.

Solution. Let $$f(x)=\tan x$$. Since $$f$$ is continuous on $$[0,\pi/4]$$,
$\frac{\pi}{8}\approx \int_0^{\pi/4} \tan x\, dx = \lim_{\norm{P}\to 0 } \sum_{k=1}^n (\tan c_k) \Delta x_k$ since the area under the curve is approximately equal to the area under the triangle with vertices $$(0,0)$$, $$(\pi/4,0)$$, and $$(\pi/4,1)$$.

## Properties of the Definite Integral

Being familiar with the basic properties of finite sums and limits many of the properties listed below are straightforward to understand.

Theorem 4.9 Suppose $$f$$ and $$g$$ are integrable functions on $$[a,b]$$. Then

• $$\displaystyle \int_a^b f(x) \, dx =-\int_b^a f(x) \, dx$$,
• $$\displaystyle \int_a^a f(x) \, dx =0$$,
• $$\displaystyle \int_a^b k f(x) \, dx = k \int_a^b f(x) \, dx$$,
• $$\displaystyle \int_a^b ( f(x) \pm g(x) ) \, dx = \int_a^b f(x) \, dx \pm \int_a^b g(x) \, dx$$, and
• $$\displaystyle \int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx$$.
• If $$f$$ has a maximum value $$M$$ and a minimum value $$m$$, then $m(b-a)\leq \int_a^b f(x) \, dx \leq M(b-a).$
• If $$f(x)\leq g(x)$$ on $$[a,b]$$, then $\int_a^b f(x)\, dx \leq \int_a^b g(x) \, dx.$

Example 4.48 Suppose $$\int_0^2 f(x) \, dx=3$$, $$\int_0^2 g(x) \, dx=-1$$, and $$\int_0^2 h(x) \, dx=3$$. Evaluate $\int_0^2 [2f(x)+5g(x)-6h(x)] \, dx$ and then find a constant $$s$$ such that $\int_0^2 [5 f(x)+s g(x)-7 h(x)] \, dx=0.$

Solution. Using the basic properties of the definite integral \begin{align*} \int_0^2 [2f(x)+5g(x)-7h(x)] \, dx & =2 \int_0^2 f(x)\, dx +5\int_0^2g(x)\, dx -7\int_0^2h(x) \, dx \\ & = 2(3)+5(-1)-7(3)=-20. \end{align*} Also \begin{align*} \int_0^2 [5 f(x)+s \, g(x)-6 h(x)] \, dx & = 5\int_0^2 f(x)\, dx +s \int_0^2 g(x)\, dx -7 \int_0^2h(x) \, dx \\ & =5(3) +s (-1) -6 (3) \, dx =-s-3=0 \end{align*} which implies $$s=-3$$.

Example 4.49 Sketch the graph of $$f$$ and use it to evaluate $$\int_{-1}^5 f(x) \, dx$$ given $f(x)= \begin{cases} 2 & \text{for } -1\leq x \leq 1 \\ 3-x & \text{for } 1\leq x \leq 4 \\ 2x-9 & \text{for } 4\leq x \leq 5. \end{cases}$

Solution. Figure $$\ref{fig:graph37}$$ is a sketch of the graph of $$f$$. Using the basic properties of the definite integral \begin{align*} \int_{-1}^5 f(x) \, dx & =\int_{-1}^1 f(x) \, dx +\int_{1}^4 f(x) \, dx+\int_{4}^5 f(x) \, dx \\ & =\int_{-1}^1 2 \, dx +\int_{1}^4 (3-x) \, dx+\int_{4}^5 (2x-9) \, dx \\ & = 2(2)+\frac{3}{2}-0=\frac{11}{2}. \end{align*}

Example 4.50 Suppose that $$f$$ and $$h$$ are integrable and that $$\int_1^9 f(x) \, dx=-1$$, $$\int_7^9 f(x) \, dx=5$$, and $$\int_7^9 h(x) \, dx=4$$. Find

• $$\int_1^9 -2f(x) \, dx$$
• $_7^9 [f(x)+h(x)] , dx$
• $_7^9 [2f(x)-3h(x)] , dx$
• $_1^7 f(x) , dx$
• $_9^1 f(x) , dx$
• $_9^7 [h(x)-f(x)] , dx$

Solution. Using the basic properties of the definite integral   - $$\int_1^9 -2f(x) \, dx =-2 \int_1^9 f(x) \, dx =-2 (-1)=2$$ - $$\int_7^9 [f(x)+h(x)] \, dx =\int_7^9 f(x)\, dx +\int_7^9 h(x) \, dx =5+4=9$$ - $$\int_7^9 [2f(x)-3h(x)] \, dx = 2 \int_7^9 f(x)\, dx -3\int_7^9h(x)] \, dx =2(5)-3(4)=-2$$ - $$\int_1^7 f(x) \, dx =\int_1^9 f(x) \, dx -\int_7^9 f(x) \, dx =-1-5=-6$$ - $$\int_9^1 f(x) \, dx =-\int_1^9 f(x) \, dx =-(-1)=1$$ - $$\int_9^7 [h(x)-f(x)] \, dx =\int_7^9 f(x)\, dx-\int_7^9 h(x)] \, dx =5-4=1$$

Example 4.51 What values of $$a$$ and $$b$$ minimize the value of $$\int_a^b(x^4-2x^2)\, dx$$?

Solution. By solving $$x^4-2x^2=0$$ we determine that $$x^4-2x^2\leq 0$$ on $$[-\sqrt{2},\sqrt{2}]$$. Thus we find that $$a=-\sqrt{2}$$ and $$b=\sqrt{2}$$ will minimize the value of $$\int_a^b(x^4-2x^2)\, dx$$. See Figure $$\ref{fig:graph38}$$.

Example 4.52

1. Find upper and lower bounds for $$\int_0^1 \frac{1}{1+x^2} \, dx$$.
2. Find upper and lower bounds for $$\int_0^{1/2} \frac{1}{1+x^2} \, dx$$ and $$\int_{1/2}^1 \frac{1}{1+x^2} \, dx$$. Use (b) to find another estimate of the integral in (a).

Solution. For (a) Since $$f(x)=\frac{1}{1+x^2}$$ is decreasing on $$[0,1]$$, it follows the maximum of $$f$$, $$M$$ occurs at 0 with $$M=1$$ and the minimum value of $$f$$, $$m$$ occurs at 1 with $$m=1/2$$. Therefore, $\frac{1}{2}(1-0) =\frac{1}{2} \leq \int_0^1 \frac{1}{1+x^2} \, dx \leq 1 (1-0) =1.$ For (b) with $$[0,1/2]$$ we find, $$M=1$$ and $$m=4/5$$ and so $\frac{4}{5} \left(\frac{1}{2}-0\right) =\frac{2}{5} \leq \int_0^{1/2} \frac{1}{1+x^2} \, dx \leq 1 (\frac{1}{2}-0) =\frac{1}{2}.$ For (b) with $$[0,1/2]$$ we find, $$M=4/5$$ and $$m=1/2$$ and so $\frac{1}{2} \left(1-\frac{1}{2}\right) =\frac{1}{4} \leq \int_{1/2}^1 \frac{1}{1+x^2} \, dx \leq \frac{4}{5} (1-\frac{1}{2}) =\frac{2}{5}$ Then for part (c), we have a better estimate on the bounds of the original integral $\frac{13}{20}=\frac{1}{4}+\frac{2}{5}\leq \int_0^{1/2} \frac{1}{1+x^2} \, dx + \int_{1/2}^1 \frac{1}{1+x^2} \, dx \leq \frac{1}{2}+\frac{2}{5}=\frac{9}{10}.$ namely, $\frac{13}{20}\leq \int_0^1 \frac{1}{1+x^2} \, dx \leq \frac{9}{10}.$

Example 4.53 Show that the value of $$\int_0^1 \sqrt{x+8} \, dx$$ lies between $$2\sqrt{2}\approx 2.8$$ and 3.

Solution. See Figure $$\ref{fig:graph40}$$. Since $$f(x)=\sqrt{x+8}$$ is increasing on $$[0,1]$$ it follows the maximum of $$f$$ is $$f(1)=\sqrt{1+8}=3$$ and the minimum of $$f$$ is $$f(0)=\sqrt{0+8}=2\sqrt{2}$$. Therefore, $2\sqrt{2} =2\sqrt{2} (1-0)\leq \int_0^1 \sqrt{x+8} \, dx\leq 3 (1-0)=3.$

## Displacement

Consider an object moving along a straight line. Assume the position of the object at time $$t$$ is given by the position function $$s(t)$$ and that its velocity at time $$t$$ is given by $$v(t)=s'(t)$$. If we happen to know that the object is always moving forward from $$t=a$$ to $$t=b$$, that is, $$v(t)> 0$$ on $$[a,b]$$, then the total distance travelled is $$s(b)-s(a)$$.

Theorem 4.10 The total distance travelled by an object with continuous velocity $$v(t)$$ along a straight line from $$t=a$$ to $$t=b$$ is $$S=\int_a^b |v(t)| \, dt.$$

Example 4.54 Suppose $$v(t)=\frac{1}{t+1}$$ is the velocity of an object moving along a straight line. Use the formula $$S_n=\sum_{k=1}^n | v(a+k \Delta t) | \Delta t$$ where $$\Delta t =\frac{b-a}{n}$$ to estimate (using right endpoints) the total distance travelled by the object during the time interval $$[0,1]$$.

Solution. Let $$a=0$$ with $$\Delta t=\frac{1}{4}$$. Then $v(a+k \Delta t)=v \left(\frac{k}{4}\right) = \frac{1}{1+\frac{k}{4}} =\frac{4}{k+4}$ Thus the total distance travelled by the object during the time interval $$[0,1]$$ is $S_4=\sum_{k=1}^4 \frac{4}{k+4}\left(\frac{1}{4}\right)\approx 0.635.$

## Area Under a Curve

Theorem 4.11 Suppose $$f$$ is a continuous function and $$f(x)\geq 0$$ on the closed interval $$[a,b]$$. Then the area under the curve $$y=f(x)$$ on $$[a,b]$$ is given by the definite integral of $$f$$ on $$[a,b]$$.

Example 4.55 Use the definition of the definite integral to find the area of the region between the curve $$y=3x^2$$ and the $$x$$-axis on the interval $$[0,b]$$.

Solution. Let $$\Delta x=\frac{b-0}{n}=\frac{b}{a}$$ and let $$x_0=0$$, $$x_1=\Delta x$$, $$x_2=2 \Delta x$$, , $$x_{n-1}=\Delta x$$, $$x_n=n \Delta x=b$$. Let the subinterval representatives, the $$c_k's$$ be the right endpoints of the subintervals, so $$c_1=x_1$$, $$c_2=x_2$$, and so on. The rectangles defined have areas: \begin{align*} & f(c_1)\Delta x = f(\Delta x)\Delta x =3(\Delta x)^2\Delta x=3 (\Delta x)^3 \\ & f(c_2)\Delta x = f(2\Delta x)\Delta x =3(2\Delta x)^2\Delta x=3 (2)^2 (\Delta x)^3 \\ & f(c_3)\Delta x = f(3\Delta x)\Delta x =3(3\Delta x)^2\Delta x=3 (3)^2 (\Delta x)^3 \\ & \vdots \\ & f(c_n)\Delta x = f(n\Delta x)\Delta x =3(n\Delta x)^2\Delta x=3 (n)^2 (\Delta x)^3 \end{align*} Then forming the Riemann sum \begin{align*} S_n & =\sum_{k=1}^n f(c_k) \Delta x = \sum_{k=1}^n 3 k^2(\Delta x) ^3 \\ & =3(\Delta x)^3 \sum_{k=1}^n k^2 =3\left(\frac{b^3}{n^3}\right) \left(\frac{n(n+1)(2n+1)}{6}\right) =\frac{b^3}{2}\left(2+\frac{3}{n}+\frac{1}{n^2}\right). \end{align*} Therefore, applying the definition of the definite integral $\int_0^b 3x^2 \, dx =\lim_{n\to \infty } \frac{b^3}{2}\left(2+\frac{3}{n}+\frac{1}{n^2}\right) =b^3.$

## Average Value

Let $$f$$ be a continuous function defined on $$[a,b]$$. We will ask the question: what are the average values that $$f$$ takes on over the interval $$[a,b]$$? To answer this, we will sample a finite number of values of $$f$$. For example, say we sample $$n$$ values of $$f$$ by dividing $$[a,b]$$ into $$n$$ subintervals of equal width $$\Delta x=\frac{b-a}{n}$$ and evaluating $$f$$ at a point $$c_k$$ in each subinterval. The average of the $$n$$ sampled values is \begin{align*} \frac{f(c_1)+f(c_2)+\cdots +f(c_n)}{n} & = \frac{1}{n} \sum_{k=1}^n f(c_k) \\ & = \frac{\Delta x}{b-a} \sum_{k=1}^n f(c_k) = \frac{1}{b-a} \sum_{k=1}^n f(c_k). \end{align*} If we pass the limit, if possible, as $$n\to \infty$$ we arrive at the following definition.

Definition 4.5 If $$f$$ is integrable on $$[a,b]$$ then its $$average value$$ on $$[a,b]$$, denoted by the function $$\text{av}(f)$$, sometimes called its mean value , is $\text{av}(f) = \frac{1}{b-a}\int_a^b f(x) \, dx.$

Example 4.56 Find the average value of the function $$f(x)=3x^2-3$$ over the interval $$[0,1]$$.

Solution. The average value of the function $$f(x)=3x^2-3$$ over the interval $$[0,1]$$ is \begin{align*} \left(\frac{1}{1-0}\right) \int_0^1 (3x^2-3)\, dx & =3 \int_0^1 x^2 \, dx - \int_0^1 3 \, dx \\ & =3\left(\frac{1}{3}\right)-3(1-0) =-2. \end{align*}

Example 4.57 If $$\text{av}(f)$$ really is a typical value of the integrable function $$f(x)$$ on $$[a,b]$$, then the number $$\text{av}(f)$$ should have the same integral over $$[a,b]$$ that $$f$$ does. Show that $$\int_a^b \text{av}(f)\, dx= \int_a^b f(x) \, dx$$.

Solution. Notice that $\text{av}(f)=\frac{1}{b-a}\int_a^b f(x)\, dx:=K$ is a constant, say $$K$$. Then $\int_a^b K \, dx= K \int_a^b \, dx=K (b-a) = (b-a) \frac{1}{b-a}\int_a^b f(x)\, dx=\int_a^b f(x) \, dx$ and so yes the number $$\text{av}(f)$$ has the same integral over $$[a,b]$$ that $$f$$ does.

## Exercises

Exercise 4.28 For each of the following you are given a function $$f$$ defined on an interval $$[a,b]$$, then number $$n$$ of subintervals of equal length $$\Delta x=(b-a)/n$$, and the evaluation points $$c_k$$ in $$[x_{k-1}, x_k]$$. For each of the following, (a) sketch the graph of $$f$$ and the rectangles associated with the Riemann sum for $$f$$ on $$[a,b]$$, and (b) find the Riemann sum.   - $$f(x)=2x-3, [0,2], n=4, c_k \text{ is the midpoint}$$ - $$f(x)=-2x+1, [-1,2], n=6, c_k \text{ is the left endpoint}$$ - $$f(x)=\sqrt{x}-1, [0,3], n=6, c_k \text{ is the right endpoint}$$ - $$f(x)=2\sin x, [0,5\pi/4], n=5, c_k \text{ is the right endpoint}$$

Exercise 4.29 Use the definition of the definite integral to evaluate the following integrals.

• $$\int_{-1}^2 x^2 \, dx$$
• $$\int_{-1}^3 (x-2) \, dx$$
• $$\int_{-1}^1 (2x+1) \, dx$$
• $$\int_{-2}^1 (x^3+2x) \, dx$$

Exercise 4.30 Each of the following is given as the limit of a Riemann sum of a function $$f$$ on $$[a,b]$$. Write this expression as a definite integral on $$[a,b]$$.

• $$\lim_{n\to \infty} \sum_{k=1}^n (4 c_k-3)\Delta x, [-3,-1]$$
• $$\lim_{n\to \infty} \sum_{k=1}^n 2c_k(1-c_k)^2\Delta x, [0,3]$$
• $$\lim_{n\to \infty} \sum_{k=1}^n \frac{2c_k}{c_k^2+1} \Delta x, [1,2]$$
• $$\lim_{n\to \infty} \sum_{k=1}^n c_k(\cos c_k)\Delta x, [0,\frac{\pi}{2}]$$

Exercise 4.31 For each of the following make a sketch of $$f$$ on $$[a,b]$$ and then use the geometric interpretation of the integral to evaluate it.

• $$\int_{-2}^3 (2x+1) \, dx$$
• $$\int_{-1}^2 |x| \, dx$$
• $$\int_{-2}^2 \sqrt{4-x^2} \, dx$$
• $$\int_{0}^2 \sqrt{-x^2+2x} \, dx$$

Exercise 4.32 Given that $$\int_{-1}^3 f(x) \, dx =4$$ and $$\int_3^6 f(x) \, dx=2$$, evaluate the following integrals.

• $$\int_{-1}^3 [f(x)+g(x)] \, dx$$
• $$\int_{-1}^3 [g(x)-f(x)] \, dx$$
• $$\int_{-1}^3 [3f(x)-2g(x)] \, dx$$
• $$\int_{3}^{-1} [f(x)+5g(x)] \, dx$$

Exercise 4.33 Given that $$\int_{-2}^2 f(x) \, dx =3$$ and $$\int_0^2 f(x) \, dx=2$$, evaluate the following integrals.

• $$\int_{2}^0 f(x) \, dx$$
• $$\int_{-2}^0 [f(x)+3] \, dx$$
• $$\int_{2}^0 3f(x)\, dx-\int_0^{-2} 2f(x) \, dx$$

Exercise 4.34 For each of the following use the properties of the integral to prove that the inequality without evaluating the integral.

• $$\int_0^1 \frac{\sqrt{x^3+x}}{x^2+1}\, dx \geq 0$$
• $$\int_0^1 x^2\, dx \leq \int_0^1 \sqrt{x} \, dx$$
• $$\int_0^{\pi/4} \sin^2 x \cos x \, dx \leq \int_0^{\pi/4} \sin^2 x \, dx$$
• $$\int_0^{\pi/2} \cos x \, dx \leq \int_0^{\pi/2} (x^2+1) \, dx$$

## Fundamental Theorem of Calculus

Before we begin, let’s review the Intermediate Value Theorem and the Extreme Value Theorem.

::: {#thm- } [Intermediate Value Theorem] If $$f$$ is a continuous function on a closed interval $$[a,b]$$ and $$M$$ is a number between $$f(a)$$ and $$f(b)$$, inclusive, then there is at least one number $$c$$ in $$[a,b]$$ such that $$f(c)=M$$.

::: {#thm- } [Extreme Value Theorem] If $$f$$ is a continuous function on a closed interval $$[a,b]$$, then $$f$$ attains an absolute maximum value $$f(c)$$ for some number $$c$$ in $$[a,b]$$ and an absolute minimum value $$f(d)$$ for some number $$d$$ in $$[a,b]$$. :::

Definition 4.6 If $$f$$ is integrable on $$[a,b]$$, then the average value of $$f$$ over $$[a,b]$$ is the number $$$f_{\text{av}}=\frac{1}{b-a} \int_a^b f(x) \, dx.$$$

If we assume that $$f$$ is nonnegative, then we have the following geometric interpretation for the average value of a function over $$[a,b]$$. We see that $$f_{\text{av}}$$ is the height of the rectangle with base lying on the interval $$[a,b]$$ and having the same area as the area of the region under the graph of $$f$$ on $$[a,b]$$.

Example 4.58 Find the average value of $$f(x)=\frac{x}{\sqrt{x^2+1}}$$ over the interval $$[0,3]$$. Answer: $$\frac{\sqrt{10}-1}{3}$$

## 4.21 Mean Value Theorem for Integrals

::: {#thm- } Mean Value Theorem for Integrals If $$f$$ is continuous on $$[a,b]$$, then there exists a number $$c$$ in $$[a,b]$$ such that $f(c)=\frac{1}{b-a}\int_a^b f(x) \, dx.$ :::

Proof. Since $$f$$ is continuous on the interval $$[a,b]$$, the Extrema Value Theorem tells us that $$f$$ attains an absolute minimum value $$m$$ at some number in $$[a,b]$$ and an absolute maximum value $$M$$ at some number in $$[a,b]$$. So $$m\leq f(x) \leq M$$ for all $$x$$ in $$[a,b]$$. Hence the average value holds: $m\leq \frac{1}{b-a}\int_a^b f(x) \, dx \leq M$ Because the average value lies between $$m$$ and $$M$$, the Intermediate Value Theorem there exists at least one number $$c$$ in $$[a,b]$$ such that $f(c)=\frac{1}{b-a}\int_a^b f(x) \, dx.$ as needed.

Example 4.59 Show that the inequality $$$0\leq \int_0^1 \frac{x^5}{\sqrt[3]{1+x^4}}\, dx \leq \frac{1}{6}$$$ holds.

Solution. Let $$f(x)=\frac{x^5}{\sqrt[3]{1+x^4}}$$. Since $$f$$ is continuous on $$[0,1]$$, we know that $$f$$ must attain a maximum $$M$$ and a minimum value $$m$$. We find, $f'(x)= \frac{x^4(19 x^4+15)}{3(x^4+1)^{2/3}}$ Hence the only critical number is $$0$$. We test and find $$m=f(0)=0$$ and $$M=f(1)=1/\sqrt[3]{2}\approx 0.79$$. However, we can do better, by observing that $$1+x^4\geq 1$$ for $$x$$ in $$[0,1]$$. Then $0\leq \frac{x^5}{\sqrt[3]{1+x^4}} \leq x^5 \implies 0=\int_0^ 1 0 \, dx \leq \int_0^ 1 \frac{x^5}{\sqrt[3]{1+x^4}} \, dx \leq \int_0^ 1 x^5 \, dx=\frac{1}{6}$ as needed.

## 4.22 The First Fundamental Theorem of Calculus

Theorem 4.12 (First Fundamental Theorem of Calculus) If $$f$$ is continuous on $$[a,b]$$, then the function $$F$$ defined by $F(x)=\int_a^x f(t) \, dt, \quad a\leq x \leq b$ is differentiable on $$(a,b)$$ and $F'(x)=\frac{d}{dx} \int_a^x f(t) \, dt = f(x).$

Proof. Fix $$x$$ in $$(a,b)$$ and suppose that $$x+h$$ is in $$(a,b)$$, where $$h\neq 0$$. Then \begin{align*} F(x+h)-F(x) & =\int_a^{x+h} f(t) \, dt -\int_a^x f(t)\, dt \\ & = \int_a^x f(t)\, dt +\int_x^{x+h} f(t)\, dt -\int_a^x f(t)\, dt \\ & =\int_x^{x+h} f(t)\, dt. \end{align*} By the Mean Value Theorem for Integrals there exists a number $$c$$ between $$x$$ and $$x+h$$ such that $f(c)=\int_x^{x+h} f(t)\, dt$ Therefore, $\frac{F(x+h)-F(x)}{h}=\frac{1}{h} \int_{x}^{x+h} f(t)\, dt =f(c).$ Next, observe that as $$h$$ approaches $$0$$, the number $$c$$, which is squeezed between $$x$$ and $$x+h$$ approaches $$x$$, and by continuity, $$f(c)$$ approaches $$f(x)$$. Therefore, $F'(x)=\lim_{h\to 0} \frac{F(x+h)-F(x)}{h} =\lim_{h\to 0} \frac{1}{h} \int_x^{x+h} f(t)\, dt =\lim_{h\to 0} f(c)=f(x).$ as needed.

Example 4.60 Find the derivative of the following functions.

• $$G(x)=\int_{-1}^x t \sqrt{t^2+1} \, dt$$
• $$G(x)=\int_{0}^{x^2} t \sin t \, dt$$
• $$G(x)=\int_{2}^{\sqrt{x}} \frac{\sin t}{t} \, dt$$
• $$G(x)=\int_{x^2}^{x^3} \ln t\, dt$$, $$x>0$$
• Solution.
• $$G'(x)=x\sqrt{x^2+1}$$
• $$G'(x)=(x\sin x) (2x)=2x^2\sin x$$
• $$G'(x)=\frac{\sin x}{x} \frac{d}{dx}\left(\frac{1}{x}\right)$$
• $$G'(x)=\frac{\sin x}{x} \left(\frac{-1}{x^2}\right)$$ First we write, $G(x)=\int_{x^2}^c \ln t\, dt +\int_c^{x^3} \ln t\, dt = -\int_{c}^{x^2} \ln t\, dt +\int_c^{x^3} \ln t\, dt$ and so we find, $G'(x) = - 2x \ln x^2 +3x^2\ln x^3 = -4x \ln x+9 x^2\ln x =x(9x-4) \ln x$

as requested.

Example 4.61 Evaluate $$\lim_{h\to 0} \frac{1}{h} \int_2^{2+h} \sqrt{5+t^2} \, dt$$.

Solution. Let $$F(x)=\int_2^x \sqrt{5+t^2}\, dt$$. Then \begin{align*} F'(2) & =\lim_{h\to 0} \frac{F(2+h)-F(2)}{h} \\ & =\lim_{h\to 0} \frac{\int_0^{2+h} \sqrt{5+t^2}\, dt -\int_0^2 \sqrt{5+t^2}\, dt}{h} \\ & =\lim_{h\to 0} \frac{\int_2^{2+h} \sqrt{5+t^2}\, dt }{h}. \end{align*} Next using the Fundamental Theorem, we have $F'(x)=\frac{d}{dx} \int_2^x \sqrt{5+t^2}\, dt =\sqrt{5+x^2}$ and so $$F'(2)=3$$, hence $\lim_{h\to 0} \frac{1}{h} \int_2^{2+h} \sqrt{5+t^2} \, dt=3.$ as requested.

## 4.23 The Second Fundamental Theorem of Calculus

::: {#thm- } The Second Fundamental Theorem of Calculus If $$f$$ is continuous on $$[a,b]$$, then $\int_a^b f(x) \, dx=F(b)-F(a)$ where $$F$$ is any antiderivative of $$f$$, that is $$F'=f$$. :::

Proof. Let $$G(x)=\int_a^x f(t)\, dt$$. By $$\ref{thm2}$$, we know that $$G$$ is an antiderivative of $$f$$. If $$F$$ is any other antiderivative of $$f$$, then $$F$$ and $$G$$ must differ by a constant. In other words, $$F(x)=G(x)+C$$. To determine $$C$$, we put $$x=a$$ to obtain $F(a)=G(a)+C=\int_a^a f(t)\, dt +C=C.$ Therefore, evaluating $$F$$ at $$b$$, we have $F(b)=G(b)+C=\int_a^b f(t)\, dt +F(a)$ from which we conclude that $F(b)-F(a)=\int_a^b f(x) \, dx.$ as needed.

Example 4.62 Evaluate the following integrals.

• $$\int_{-2}^0 (2x-3)\, dx$$
• $$\int_{0}^2 (2-4u+u^2)\, du$$
• $$\int_1^2 \frac{3x^4-2x^2+1}{2x^2} \, dx$$
• $$\int_2^0 \sqrt{x}(x+1)(x-2) \, dx$$
• $$\int_0^{\pi} \sin 2x \cos x \, dx$$
• $$\int_0^{\pi} |\cos x| \, dx$$
• $$\int _0^{4\sqrt{3}} \frac{1}{x^2+16}\, dx$$
• $$\int _1^e \frac{\ln x}{x} e^{(\ln x)^2}\, dx$$

## 4.24 Exercises

Exercise 4.36 Find the derivative of the following functions.

• $$G(x)=\int_{-1}^x t \sqrt{t^2+1} \, dt$$
• $$G(x)=\int_{0}^{x^2} t \sin t \, dt$$
• $$G(x)=\int_{2}^{\sqrt{x}} \frac{\sin t}{t} \, dt$$
• $$G(x)=\int_{x^2}^{x^3} \ln t\, dt$$

Exercise 4.37 Evaluate the following integrals.

• $$\int_{-2}^0 (2x-3)\, dx$$
• $$\int_{0}^2 (2-4u+u^2)\, du$$
• $$\int_1^2 \frac{3x^4-2x^2+1}{2x^2} \, dx$$
• $$\int_2^0 \sqrt{x}(x+1)(x-2) \, dx$$
• $$\int_0^{\pi} \sin 2x \cos x \, dx$$
• $$\int0^{\pi} |\cos x| \, dx$$
• $$\int _0^{4\sqrt{3}} \frac{1}{x^2+16}\, dx$$
• $$\int _1^e \frac{\ln x}{x} e^{(\ln x)^2}\, dx$$

Exercise 4.38 Show that each of the following inequalities hold.

• $$0\leq \int_0^1 \frac{x^5}{\sqrt[3]{1+x^4}}\, dx \leq \frac{1}{6}$$
• $$0\leq \int_0^1 \frac{1}{\sqrt{4-3x+x^2}}\, dx \leq \frac{2}{3}$$

Exercise 4.39 For each of the following, find the area of the region under the graph of $$f$$ on $$[a,b]$$.

• $$f(x)=x^2-2x+2; [-1,2]$$
• $$f(x)=\frac{1}{x^2}; [1,2]$$
• $$f(x)=2+\sqrt{x+1}; [0,3]$$
• $$f(x)=\frac{1}{4+x^2}; [0,1]$$

Exercise 4.40 Find $$\frac{dx}{dy}$$ if $\int_0^x \sqrt{3+2\cos t} \, dt+\int_0^y \sin t \, dt =0.$

Exercise 4.41 Find all functions $$f$$ on $$[0,1]$$ such that $$f$$ is continuous on $$[0,1]$$ and $$\int_0^x f(t) \, dt = \int_x^1 f(t)\, dt$$ for every $$x$$ in $$(0,1)$$.

Exercise 4.42 Evaluate $$\lim_{h\to 0} \frac{1}{h} \int_2^{2+h} \sqrt{5+t^2} \, dt$$.

Exercise 4.43 Evaluate $$\int_{-1}^1 \frac{2x^5+x^4-3x^3+2x^2+8x+1}{x^2+1} \, dx$$.

Exercise 4.44 Use the identity $\frac{x \sin\left(n+\frac{1}{2}\right)}{2\sin \frac{x}{2}} = \frac{1}{2} +\cos x +\cos 2x +\cdots + \cos n x$ to show that $\int_0^\pi \frac{x \sin\left(n+\frac{1}{2}\right)}{2\sin \frac{x}{2}} \, dx =\pi.$

## 4.26 Left and Right Endpoints

The left and right rules are the most straight-forward to learn. They can be applied with either uniform width subintervals or varying width subintervals. They consist of using a Riemann sum where the subinterval representatives are chosen as the left-endpoints or the right-endpoints, respectively.

Theorem 4.13 Let $$f$$ be a continuous function on $$[a,b]$$ and let $\mathcal{P}=\left\{a=x_0,x_1,\cdots,x_{k-1,}x_k,\cdots,x_n=b\right\}$ be a partition of the interval $$[a,b].$$ Then the Riemann sum formed by using left-endpoints as the subinterval representatives $$x_k{}^*$$ is $\int_a^b f(x) \, dx \approx \sum _{k=1}^n f\left(x_k{}^*\right)\Delta x_k=\sum _{k=1}^n f\left(x_{k-1}\right)\Delta x_k$ and the Riemann sum formed by using right-endpoints as the subinterval representatives $$x_k{}^*$$ is $\int_a^b f(x) \, dx \approx \sum _{k=1}^n f\left(x_k{}^*\right)\Delta x_k=\sum _{k=1}^n f\left(x_k\right)\Delta x_k$ where in both cases $$\Delta x=x_k-x_{k-1}.$$ The first formula is called the left rule and the second formula the right rule.

Example 4.63 Consider $$f(x)=e^{-x^2}$$ on $$[0,2]$$ and let { }$$\mathcal{P}=\left\{0,\frac{1}{2},1,\frac{3}{2},2\right\}$$ be a partition of the interval $$[0,2]$$ (so $$\Delta x=\frac{1}{2}$$).

Solution. Then the Riemann sum formed by using left-endpoints as the subinterval representatives $$x_k{}^*$$ is $\int_0^2 e^{-x^2} \, dx \approx \frac{1}{2}e^{-(0)^2}+\frac{1}{2}e^{-\left(\frac{1}{2}\right)^2}+\frac{1}{2}e^{-(1)^2}+\frac{1}{2}e^{-\left(\frac{3}{2}\right)^2}\approx 1.12604$

The Riemann sum formed by using right-endpoints as the subinterval representatives $$x_k^*$$ is $\int_0^2 e^{-x^2} \, dx \approx \frac{1}{2}e^{-\left(\frac{1}{2}\right)^2}+\frac{1}{2}e^{-(1)^2}+\frac{1}{2}e^{-\left(\frac{3}{2}\right)^2}+\frac{1}{2}e^{-(2)^2}\approx 0.635198.$ Consider that as $$n$$ increases then so does the estimation of the area. Note that a somewhat accurate estimation is $$0.882081.$$ However, because a fairly large number of rectangles are needed for a good approximation there are other common techniques which do not use rectangles.

## 4.27 Midpoint Rule

The midpoint rule uses a Riemann sum where the subinterval representatives are the midpoints of the subintervals. For some functions it may be easy to choose a partition that more closely approximates the definite integral using midpoints.

::: {#thm- } Midpoint Rule Let $$f$$ be a continuous function on $$[a,b]$$ and let $\mathcal{P}=\left\{a=x_0,x_1,\cdots,x_{k-1,}x_k,\cdots,x_n=b\right\}$ be a partition of the interval $$[a,b].$$ Then the Riemann sum formed by using midpoints as the subinterval representatives $$x_k{}^*$$ is $\int_a^b f(x) \, dx \approx \sum _{k=1}^n f\left(x_k{}^*\right)\Delta x_k=\sum _{k=1}^n f\left(\frac{x_k+x_{k-1}}{2}\right)\Delta x_k$ where $$\Delta x_k=x_k-x_{k-1}.$$ :::

Example 4.64 Consider $$f(x)=e^{-x^2}$$ on $$[0,2]$$ and let $$\mathcal{P}=\left\{0,\frac{1}{2},1,\frac{3}{2},2\right\}$$ be a partition of the interval $$[0,2]$$ (so $$\Delta x=\frac{1}{2}$$). Use the midpoint rule.

Solution. Then the Riemann sum formed by using midpoints as the subinterval representatives $$x_k^*$$ is $\int_0^2 e^{-x^2} \, dx \approx \frac{1}{2}e^{-\left(\frac{1}{4}\right)^2}+e^{-\left(\frac{3}{4}\right)^2}+\frac{1}{2}e^{-\left(\frac{5}{4}\right)^2}+\frac{1}{2}e^{-\left(\frac{7}{4}\right)^2}\approx 1.16768$ Consider that as $$n$$ increases then so does the estimation of the area. Note that a somewhat accurate estimation is $$0.882081.$$ However, because a fairly large number of rectangles are needed for a good approximation there are other common techniques which do not use rectangles.

## 4.28 Trapezoidal Rule

The trapezoidal rule uses trapezoids instead of rectangles to approximate the definite interval over a closed bounded interval. By using points on the graph of the function determined by a uniform width partition of the interval the upper boundary of the trapezoid is formed. Recall that the area of a trapezoid is one half the width times the sum of the two heights; and by using this formula we can add up the areas to approximate the definite integral. Of course the more subintervals, (or said another way: the more trapezoids) the better accuracy of the estimation.

Consider $$y=-x^2+1$$ on $$[0,1]$$ and say we want to find the area. We could use rectangles as in a Riemann sum; however from inspection of the following diagram it might seem more reasonable to use trapezoids: The area is underestimated but not as much as using rectangles would (with the same partition).

::: {#thm- } Trapezoidal Rule The trapezoidal rule estimates the definite integral of $$f$$ over $$[a,b]$$ using the formula $\int_a^b f(x) \, dx \approx \frac{1}{2}\left[f\left(x_0\right)+2f\left(x_1\right)+2f\left(x_2\right)+\cdots+2f\left(x_{n-1}\right)+f\left(x_n\right)\right]\Delta x$ where $$x_k=a+k \Delta x$$ and $$\Delta x=\frac{b-a}{n}.$$ The trapezoidal rule formula can be simplified as follows \begin{align*} & \int_a^b f(x) \, dx \approx \frac{1}{2}\left[f\left(x_0\right)+2f\left(x_1\right)+2f\left(x_2\right)+\cdots.+2f\left(x_{n-1}\right)+f\left(x_n\right)\right]\Delta x \\ & =\frac{1}{2}[f(a)+2f(a+\Delta x)+2f(a+2\Delta x)+ \cdots+2f(a+(n-1) \Delta x)+f(b)]\frac{b-a}{n} \\ %& =\left(\frac{f(a)+f(b)}{2}\right)\left(\frac{b-a}{n}\right)+\left[f\left(a+\frac{b-a}{n}\right)+f\left(a+2\frac{b-a}{n}\right)+ \cdots+f\left(a+(n-1) \frac{b-a}{n}\right)\right]\frac{b-a}{n} \\ & =\left(\frac{f(a)+f(b)}{2}\right)\left(\frac{b-a}{n}\right)+\left(\frac{b-a}{n}\right)\sum _{k=1}^{n-1} f\left(a+k\frac{b-a}{n}\right) \end{align*} which is particularly nice if you want to program or pass a limit. :::

Example 4.65 Consider $$f(x)=e^{-x^2}$$ on $$[0,2]$$ and let { }$$\mathcal{P}=\left\{0,\frac{1}{2},1,\frac{3}{2},2\right\}$$ be a partition of the interval $$[0,2]$$ (so $$\Delta x=\frac{1}{2}$$). Then $\int_0^2 e^{-x^2} \, dx \approx \left(\frac{e^{-0^2}+e^{-2^2}}{4}\right)+\frac{1}{2}\sum _{k=1}^3 e^{-\left(\frac{k}{2}\right)^2}\approx 0.880619$ which is a much better estimation than using rectangles with the same partition.

Theorem 4.14 [Error in the Trapezoidal Rule] If $$f$$has a continuous second derivative on $$[a,b],$$ then the error $$E_n$$ in approximating $\int_a^b f(x) \, dx$ by the trapezoidal rule satisfies: $\left|E_n\right|\leq \frac{(b-a)^3}{12n^2}M$ where $$M$$ is the maximum value of $$|f\text{''}(x)|$$ on $$[a,b].$$

## 4.29 Simpson’s Rule

Instead of rectangles and trapezoids Simpson’s rule uses parabolic arcs. The formula obtained for Simpson’s rule requires using uniform width subintervals and an even number of them.

Consider $$y=-x^2+1$$ on $$[0,1]$$ and say we want to find the area. We could use rectangles as in a Riemann sum; however many graphs have curvature as so maybe using strips with parabolic arc will yield better estimates with less work. That is instead if rectangles or trapezoids: let’s use $$\texttt{"}$$strips$$\texttt{"}$$ with parabolic arcs. If $$\mathcal{P}=\left\{x_0, x_1,\cdots,x_n\right\}$$ is a partition of $$[a,b]$$ with $$x_0=a$$ and $$x_n=b$$ and if we pass a parabolic arc through the points, three at a time say, the points with $$x$$-coordinates $$x_0, x_1, x_2$$ then those with $$x_2, x_{3,} x_4$$ and so on. It can be shown that the area of the region under the parabolic curve $$y=f(x)$$ on the interval $$\left[x_{2k-2},x_{2k}\right]$$ has area given by $\frac{1}{3}\left[f\left(x_{2k-2}\right)+4f\left(x_{2k-1}\right)+f\left(x_{2k}\right)\right]\Delta x$ where $$\Delta x=\frac{b-a}{n}.$$ Thus, to estimate $$\int_a^b f(x) \, dx$$ we can add up the strips as follows.

::: {#thm- } Simpson’s Rule Simpson’s rule estimates the definite integral of $$f$$ over $$[a,b]$$ using the formula \begin{align*} & \int_a^b f(x) \, dx \approx \frac{1}{3}\left[f\left(x_0\right)+4f\left(x_1\right)+2f\left(x_2\right)+4f\left(x_3\right)+ \right. \\ & \qquad \qquad \qquad \qquad \qquad \left. 2f\left(x_4\right)+\cdots +4f\left(x_{n-1}\right)+f\left(x_n\right)\right]\Delta x \end{align*} where $$x_k=a+k \Delta x,$$ $$\Delta x=\frac{b-a}{n},$$ and $$n$$ is an even integer. :::

Example 4.66 Consider $$f(x)=e^{-x^2}$$ on $$[0,2]$$ and let { }$$\mathcal{P}=\left\{0,\frac{1}{2},1,\frac{3}{2},2\right\}$$ be a partition of the interval $$[0,2]$$ (so $$\Delta x=\frac{1}{2}$$). $\quad \int_0^2 e^{-x^2} \, dx \approx \frac{1}{6}\left(e^{-0^2}+4e^{-\left(\frac{1}{2}\right)^2}+2e^{-(1)^2}+4e^{-\left(\frac{3}{2}\right)^2}+e^{-(2)^2}\right)\approx 0.881812$ which is a much better estimation than using rectangles with the same partition.

::: {#thm- } [Error in the Simpson Rule] If $$f$$has a continuous fourth derivative on $$[a,b],$$ then the error $$E_n$$ (n even) in approximating $\int_a^b f(x) \, dx$ by the Simpson rule satisfies: $\left|E_n\right|\leq \frac{(b-a)^5}{180n^4}M$ where $$M$$ is the maximum value of $$\left|f^{(4)}(x)\right|$$ on $$[a,b].$$ :::

Example 4.67 Consider $$f(x)=e^{-x^2}$$ on $$[0,2]$$ and let $$\mathcal{P}=\left\{0,\frac{1}{2},1,\frac{3}{2},2\right\}$$ be a partition of the interval $$[0,2]$$. Find the error in the approximation $\int_0^2 e^{-x^2} \, dx \approx \frac{1}{6}\left(e^{-0^2}+4e^{-\left(\frac{1}{2}\right)^2}+2e^{-(1)^2}+4e^{-\left(\frac{3}{2}\right)^2}+e^{-(2)^2}\right)\approx 0.881812$

Solution. We compute

• $$f'(x)=-2 e^{-x^2} x$$,
• $$f'' (x)=-2 e^{-x^2}+4 e^{-x^2} x^2,$$
• $$f'''(x)=12 e^{-x^2} x-8 e^{-x^2} x^3$$
• $$f^{(4)}(x)=12 e^{-x^2}-48 e^{-x^2} x^2+16 e^{-x^2} x^4$$

We plot the fourth derivative obtaining: and so there is a maximum at $$x=0.$$ We obtain $$f^{(4)}(0)=12 =M.$$ Therefore, the error is less than $$\frac{32}{1804^4}(12)=0.00833333.$$