## 6.1: Using Definite Integrals to Find Area and Length

Exercises 1. Find the exact area of each described region. The finite region between the curves x = y(y − 2) and x = −(y − 1)(y − 3). The region between the sine and cosine functions on the interval [ π 4 , 3π 4 ]. The finite region between x = y 2 − y − 2 and y = 2x − 1. The finite region between y = mx and y = x 2 − 1, where m is a positive constant. 2. Let f (x) = 1 − x 2 and g(x) = ax2 − a, where a is an unknown positive real number. For what value(s) of a is the area between the curves f and g equal to 2? 3. Let f (x) = 2 − x 2 . Recall that the average value of any continuous function f on an interval [a, b] is given by 1 b−a R b a f (x) dx. 3

- Find the average value of f (x) = 2 − x 2 on the interval [0, √ 2]. Call this value r.
- Sketch a graph of y = f (x) and y = r. Find their intersection point(s).
- Show that on the interval [0, √ 2], the amount of area that lies below y = f (x) and above y = r is equal to the amount of area that lies below y = r and above y = f (x).
- Will the result of be true for any continuous function and its average value on any interval? Why?

## 6.2 Using Definite Integrals to Find Volume

1. Consider the curve f (x) = 3 cos( x 3 4 ) and the portion of its graph that lies in the first quadrant between the y-axis and the first positive value of x for which f (x) = 0. Let R denote the region bounded by this portion of f , the x-axis, and the y-axis.

- Set up a definite integral whose value is the exact arc length of f that lies along the upper boundary of R. Use technology appropriately to evaluate the integral you find.
- Set up a definite integral whose value is the exact area of R. Use technology appropriately to evaluate the integral you find.
- Suppose that the region R is revolved around the x-axis. Set up a definite integral whose value is the exact volume of the solid of revolution that is generated. Use technology appropriately to evaluate the integral you find.
- Suppose instead that R is revolved around the y-axis. If possible, set up an integral expression whose value is the exact volume of the solid of revolution and evaluate the integral using appropriate technology. If not possible, explain why.

2. Consider the curves given by y = sin(x) and y = cos(x). For each of the following problems, you should include a sketch of the region/solid being considered, as well as a labeled representative slice.

Sketch the region R bounded by the y-axis and the curves y = sin(x) and y = cos(x) up to the first positive value of x at which they intersect. What is the exact intersection point of the curves? 6.2. USING DEFINITE INTEGRALS TO FIND VOLUME 353

- Set up a definite integral whose value is the exact area of R.
- Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving R about the x-axis.
- Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving R about the y-axis.
- Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving R about the line y = 2.
- Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving R about the x = −1.

3. Consider the finite region R that is bounded by the curves y = 1 + 1 2 (x − 2) 2 , y = 1 2 x 2 , and x = 0.

- Determine a definite integral whose value is the area of the region enclosed by the two curves.
- Find an expression involving one or more definite integrals whose value is the volume of the solid of revolution generated by revolving the region R about the line y = −1.
- Determine an expression involving one or more definite integrals whose value is the volume of the solid of revolution generated by revolving the region R about the y-axis.
- Find an expression involving one or more definite integrals whose value is the perimeter of the region R.

## 6.3 Density, Mass, and Center of Mass

Exercises

1. Let a thin rod of length a have density distribution function ρ(x) = 10e −0.1x , where x is measured in cm and ρ in grams per centimeter.

- If the mass of the rod is 30 g, what is the value of a? 6.3. DENSITY, MASS, AND CENTER OF MASS 363
- For the 30g rod, will the center of mass lie at its midpoint, to the left of the midpoint, or to the right of the midpoint? Why?
- For the 30g rod, find the center of mass, and compare your prediction in (b).
- At what value of x should the 30g rod be cut in order to form two pieces of equal mass?

2. Consider two thin bars of constant cross-sectional area, each of length 10 cm, with respective mass density functions ρ(x) = 1 1+x 2 and p(x) = e −0.1x .

- Find the mass of each bar.
- Find the center of mass of each bar.
- Now consider a new 10 cm bar whose mass density function is f (x) = ρ(x) + p(x).
- i. Explain how you can easily find the mass of this new bar with little to no additional work.
- ii. Similarly, compute R 10 0 x f (x) dx as simply as possible, in light of earlier computations.
- iii. True or false: the center of mass of this new bar is the average of the centers of mass of the two earlier bars. Write at least one sentence to say why your conclusion makes sense.

3. Consider the curve given by y = f (x) = 2xe−1.25x + (30 − x)e −0.25(30−x) .

- Plot this curve in the window x = 0 . 30, y = 0 . 3 (with constrained scaling so the units on the x and y axis are equal), and use it to generate a solid of revolution about the x-axis. Explain why this curve could generate a reasonable model of a baseball bat.
- Let x and y be measured in inches. Find the total volume of the baseball bat generated by revolving the given curve about the x-axis. Include units on your answer
- Suppose that the baseball bat has constant weight density, and that the weight density is 0.6 ounces per cubic inch. Find the total weight of the bat whose volume you found in (b).
- Because the baseball bat does not have constant cross-sectional area, we see that the amount of weight concentrated at a location x along the bat is determined by the volume of a slice at location x. Explain why we can think about the function ρ(x) = 0.6π f (x) 2 (where f is the function given at the start of the problem) as being the weight density function for how the weight of the baseball bat is distributed from x = 0 to x = 30. 364 6.3. DENSITY, MASS, AND CENTER OF MASS
- Compute the center of mass of the baseball bat.

## 6.4 Physics Applications: Work, Force, and Pressure

1. Let R denote the region bounded by this portion of f , the x-axis, and the y-axis. Assume that x and y are each measured in feet.

- Picture the coordinate axes rotated 90 degrees clockwise so that the positive x-axis points straight down, and the positive y-axis points to the right. Suppose that R is rotated about the x axis to form a solid of revolution, and we consider this solid as a storage tank. Suppose that the resulting tank is filled to a depth of 1.5 feet with water weighing 62.4 pounds per cubic foot. Find the amount of work required to lower the water in the tank until it is 0.5 feet deep, by pumping the water to the top of the tank.
- Again picture the coordinate axes rotated 90 degrees clockwise so that the positive x-axis points straight down, and the positive y-axis points to the right. Suppose that R, together with its reflection across the x-axis, forms one end of a storage tank that is 10 feet long. Suppose that the resulting tank is filled completely with water weighing 62.4 pounds per cubic foot. Find a formula for a function that tells the amount of work required to lower the water by h feet.
- Suppose that the tank described in is completely filled with water. Find the total force due to hydrostatic pressure exerted by the water on one end of the tank.

2. A cylindrical tank, buried on its side, has radius 3 feet and length 10 feet. It is filled completely with water whose weight density is 62.4 lbs/ft3 , and the top of the tank is two feet underground.

- Set up, but do not evaluate, an integral expression that represents the amount of work required to empty the top half of the water in the tank to a truck whose tank lies 4.5 feet above ground.
- With the tank now only half-full, set up, but do not evaluate an integral expression that represents the total force due to hydrostatic pressure against one end of the tank.

## 6.5: Improper Integrals

1. Determine, with justification, whether each of the following improper integrals converges or diverges.

- Z ∞ e ln(x) x dx
- Z ∞ e 1 x ln(x) dx
- Z ∞ e 1 x(ln(x))2 dx
- Z ∞ e 1 x(ln(x))p dx, where p is a positive real number
- Z 1 0 ln(x) x dx Z 1 0 ln(x) dx

2. Sometimes we may encounter an improper integral for which we cannot easily evaluate the limit of the corresponding proper integrals. For instance, consider R ∞ 1 1 1+x 3 dx. While it is hard (or perhaps impossible) to find an antiderivative for 1 1+x 3 , we can still determine whether or not the improper integral converges or diverges by comparison to a simpler one. Observe that for all x > 0, 1 + x 3 > x 3 , and therefore 1 1 + x 3 < 1 x 3 . It therefore follows that Z b 1 1 1 + x 3 dx < Z b 1 1 x 3 dx for every b > 1. If we let b → ∞ so as to consider the two improper integrals R ∞ 1 1 1+x 3 dx and R ∞ 1 1 x 3 dx, we know that the larger of the two improper integrals converges. And thus, since the smaller one lies below a convergent integral, it follows that the smaller one must converge, too. In particular, R ∞ 1 1 1+x 3 dx must converge, even though we never explicitly evaluated the corresponding limit of proper integrals. We use this idea and similar ones in the exercises that follow.

- Explain why x 2 + x + 1 > x 2 for all x ≥ 1, and hence show that R ∞ 1 1 x 2+x+1 dx converges by comparison to R ∞ 1 1 x 2 dx.
- Observe that for each x > 1, ln(x) < x. Explain why Z b 2 1 x dx < Z b 2 1 ln(x) dx for each b > 2. Why must it be true that R b 2 1 ln(x) dx diverges?
- Explain why q x 4+1 x 4 > 1 for all x > 1. Then, determine whether or not the improper integral Z ∞ 1 1 x · r x 4 + 1 x 4 dx converges or diverges.

## Problem Set 6

Use Integration (PDF) to do the problems below.

Use Integration (PDF) to do the problems below.

Section | Topic | Exercises |
---|---|---|

3C | Fundamental theorem of calculus | 1, 2a, 3a, 5a |

3E | Change of variables Estimating integrals | 6b, 6c |

### Solutions

*This problem set is from exercises and solutions written by David Jerison and Arthur Mattuck.*

## 6.E: Using Definite Integrals (Exercises) - Mathematics

The purpose of this lab is to introduce you to the definite integral and to Maple commands for computing definite integrals.

- Finding the mass of a body in two or three dimensions, whose density is not a constant.
- Computing net or total distance traveled by a moving object.
- Computing work involved in moving an object, compressing a gas, or pumping a liquid.
- Computing average values, e.g. centroids and centers of mass, moments of inertia, and averages of probability distributions.

Of course, when we use a definite integral to compute an area or a volume, we are adding up area or volume. So you might ask why make a distinction? The answer is that the notion of an integral as a means of computing an area or volume is much more concrete and is easier to understand.

- Many functions don't have anti-deriviatives that can be written down as formulas. Definite integrals of such function must be done using numerical techniques, which always depend on the definition of the integral as a sum.
- In many applications of the integral in engineering and science, you aren't given the function which is to be integrated but instead must derive it. The derivation always depends on the definition of the integral as a sum.

Suppose f ( x ) is a non-negative, continuous function defined on some interval [ a , b ]. Then by the area under the curve y = f ( x ) between x = a and x = b we mean the area of the region bounded above by the graph of f ( x ), below by the x axis, on the left by the vertical line x = a , and on the right by the vertical line x = b .

The rectangular area approximations we will describe in this section depend on subdividing the interval [ a , b ] into subintervals of equal length. For example, dividing the interval [0,4] into four uniform pieces produces the subintervals [0,1], [1,2], [2,3], and [3,4]. The next step is to approximate the area above each subinterval with a rectangle, with the height of the rectangle being chosen according to some rule. In particular, we will consider the following rules: left endpoint rule The height of the rectangle is the value of the function f ( x ) at the left-hand endpoint of the subinterval.

right endpoint rule The height of the rectangle is the value of the function f ( x ) at the right-hand endpoint of the subinterval.

midpoint rule The height of the rectangle is the value of the function f ( x ) at the midpoint of the subinterval.

The Maple student package has commands for visualizing these three rectangular area approximations. To use them, you first must load the package via the with command. Then try the three commands given below. Make sure you understand the differences between the three different rectangular approximations. Take a moment to see that the different rules choose rectangles which in each case will either underestimate or overestimate the area. Even the midpoint rule, whose rectangles usually have heights more appropriate than those for the left endpoint and right endpoint rules, will produce an approximation which is not exact.

There are also Maple commands leftsum , rightsum , and middlesum to sum the areas of the rectangles, see the examples below. Note the use of evalf to obtain numerical answers.

The Maple commands above use the short-hand summation notation. Since there are only four terms in the sums above, it isn't hard to write them out explicitly, as follows.

However, if the number of terms in the sum is large, summation notation saves a lot of time and space.

It should be clear from the graphs that adding up the areas of the rectangles only approximates the area under the curve. However, by increasing the number of subintervals the accuracy of the approximation can be increased. All of the Maple commands described so far in this lab permit a third argument to specify the number of subintervals. The default is 4 subintervals. See the example below for the area under y = x from x =0 to x =2 using the rightsum command with 4, 10, 20 and 100 subintervals. (As this region describes a right triangle with height 2 and base 2, this area can be easily calculated to be exactly 2.) Try it yourself with the leftsum and middlesum commands.

Since, in this trivial example, we knew that the area is exactly 2, it appears that, as the number of subintervals increases, the rectangular approximation becomes more accurate.

In the example in the previous section, we saw that increasing the number of subintervals gave a better approximation to the area. In such a case, it seems reasonable that taking a limit as the number of subintervals goes to infinity should give the exact answer. This is exactly the idea in defining the definite integral.

When this limit as the number of subintervals goes to infinity exists, we have special notation for this limit and write it as

We will learn later on that if the function f ( x ) is continuous on the interval [ a , b ], then this limit always exists and we can write

where A is the area under the curve y = f ( x ) between x = a and x = b .

The Fundamental Theorem of Calculus (FTOC) provides a connection between the definite integral and the indefinite integral (a.k.a. the antiderivitave ). That is, the FTOC provides a way to evaluate definite integrals if an anti-derivative can be found for f ( x ). We'll spend a lot of time later in the course developing ways to do this, but for now we'll let Maple do the work.

The basic maple command for performing definite and indefinite integrals is the `int` command. The syntax is very similar to that of the `leftsum` and `rightsum` commands, except you don't need to specify the number of subintervals. This should make sense, if you recall that the definite integral is defined as a limit of a rectangular sum as the number of subintervals goes to infinity. In the section on rectangular approximations, we used two examples. The first was the function on the interval [0,4] and the second was the function y = x on the interval [0,2]. We would express the areas under these two curves with our integral notation as

Using Maple, we would compute these two definite integrals as shown below.

Notice that Maple gives an exact answer, as a fraction. If you want a decimal approximation to an integral, you just put an `evalf` command around the `int` command, as shown below.

In the exercises, we'll ask you to compare rectangular approximations to integrals. In doing so, you'll need to apply several commands to the same function. To save typing and prevent errors, you can define the function as a function or an expression in Maple first and then use it in subsequent `int`, `leftsum`, etc. commands. For example, suppose you were given the function on the interval . Then you can define this function in Maple with the command

and then use this definition to save typing as shown below.

You can also simply give the expression corresponding to f ( x ) a label in Maple, and then use that label in subsequent commands as shown below. However, notice the difference between the two methods. You are urged you to choose one or the other, so you don't mix the syntax up.

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## How to estimate a definite integral using Taylor series

I have a problem that asks Evaluate to five decimal places using the taylor series for the definite integral. $int_<1>^ <2>frac

the taylor series representation of: $frac

so I know now to integrate that statement which gives me the following series to solve for

as you can see that is only the first 6 terms for the series being integrated and then we would solve those for 2 minus those solved at 1 to have an answer. My question is: is there an easy way to know how many terms I need to go to find this answer. I know with alternating series you go until the place value you are looking to do doesnt change its answer anymore with more terms added, but with this kind of series I do not know how far to take this.

## 3 The Definite Integral

In the preceding section we defined the area under a curve in terms of Riemann sums:

However, this definition came with restrictions. We required to be continuous and nonnegative. Unfortunately, real-world problems don’t always meet these restrictions. In this section, we look at how to apply the concept of the area under the curve to a broader set of functions through the use of the definite integral.

### Definition and Notation

The definite integral generalizes the concept of the area under a curve. We lift the requirements that be continuous and nonnegative, and define the definite integral as follows.

If is a function defined on an interval the definite integral of *f* from *a* to *b* is given by

provided the limit exists. If this limit exists, the function is said to be integrable on or is an integrable function .

The integral symbol in the previous definition should look familiar. We have seen similar notation in the chapter on Applications of Derivatives, where we used the indefinite integral symbol (without the *a* and *b* above and below) to represent an antiderivative. Although the notation for indefinite integrals may look similar to the notation for a definite integral, they are not the same. A definite integral is a number. An indefinite integral is a family of functions. Later in this chapter we examine how these concepts are related. However, close attention should always be paid to notation so we know whether we’re working with a definite integral or an indefinite integral.

Integral notation goes back to the late seventeenth century and is one of the contributions of Gottfried Wilhelm Leibniz , who is often considered to be the codiscoverer of calculus, along with Isaac Newton. The integration symbol ∫ is an elongated S, suggesting sigma or summation. On a definite integral, above and below the summation symbol are the boundaries of the interval, The numbers *a* and *b* are *x*-values and are called the limits of integration specifically, *a* is the lower limit and *b* is the upper limit. To clarify, we are using the word *limit* in two different ways in the context of the definite integral. First, we talk about the limit of a sum as Second, the boundaries of the region are called the *limits of integration*.

We call the function />the integrand , and the *dx* indicates that />is a function with respect to *x*, called the variable of integration . Note that, like the index in a sum, the variable of integration is a dummy variable , and has no impact on the computation of the integral. We could use any variable we like as the variable of integration:

Previously, we discussed the fact that if is continuous on then the limit exists and is unique. This leads to the following theorem, which we state without proof.

If is continuous on then *f* is integrable on

Functions that are not continuous on may still be integrable, depending on the nature of the discontinuities. For example, functions with a finite number of jump discontinuities on a closed interval are integrable.

It is also worth noting here that we have retained the use of a regular partition in the Riemann sums. This restriction is not strictly necessary. Any partition can be used to form a Riemann sum. However, if a nonregular partition is used to define the definite integral, it is not sufficient to take the limit as the number of subintervals goes to infinity. Instead, we must take the limit as the width of the largest subinterval goes to zero. This introduces a little more complex notation in our limits and makes the calculations more difficult without really gaining much additional insight, so we stick with regular partitions for the Riemann sums.

Use the definition of the definite integral to evaluate Use a right-endpoint approximation to generate the Riemann sum.

We first want to set up a Riemann sum. Based on the limits of integration, we have and For let be a regular partition of Then

Since we are using a right-endpoint approximation to generate Riemann sums, for each *i*, we need to calculate the function value at the right endpoint of the interval The right endpoint of the interval is and since *P* is a regular partition,

Thus, the function value at the right endpoint of the interval is

Then the Riemann sum takes the form

Using the summation formula for we have

Now, to calculate the definite integral, we need to take the limit as We get

Use the definition of the definite integral to evaluate Use a right-endpoint approximation to generate the Riemann sum.

## Welcome!

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## 3 Answers 3

The two cases $a, bleq -1$ and $2leq a, b$ can be excluded by noting that those integrals can't get positive, since the function is negative on $[a, b]$. All other cases can be covered in one fell swoop by saying that $int_<-1>^af(x) dx$ and $int_b^2f(x)dx$ are both non-negative, so $ int_<-1>^2f(x)dx = int_a^bf(x)dx + int_<-1>^af(x)dx + int_b^2f(x)dx $ shows that $int_<-1>^2f(x)dx geq int_a^bf(x)dx$ (keeping in mind that $int_p^qf(x)dx = -int_q^pf(x)dx$).

Let $a$ be a fixed number. Consider the function

$F(t)= int_a^t (2+x-x^2) mathrm dx$

Observe that $F'$ is positive on $(-1,2)$ and negative on $(2,infty)$.

This implies that $t=2$ is a local maximum for $F$.

Now similarly, you can consider $G(t)= int_t^2(2+x-x^2) mathrm dx = -int_2^t(2+x-x^2) mathrm dx$

and observe that $t=-1$ is a local max for $G$.

Your reasoning on the negativity of the integrand is correct. Solving for intersections between $f$ and the x-axis, as well as the fact that $frac

Another relatively straight forward solution can be given by defining

$g(a,b) = int_a^bf(x)dx = int_a^0 f(x)dx + int_0^b f(x)dx = int_0^b f(x)dx - int_0^a f(x)dx $

and by the fundamental theorem of calculus

$frac

## Methods of Integration

A great deal of integration tricks exist for evaluating definite integrals exactly, but there still exist many integrals for each of which there does not exist a closed-form expression in terms of elementary mathematical functions. For instance, the integral

may not be evaluated without numerical methods.

Many numerical methods exist to approximate integrals to any degree of accuracy, but their efficiency is wholly dependent upon the function and interval in question. When possible, an exact calculation is always preferred.

Partial Riemann sums, the trapezoid rule, and Simpson's rule all seek to provide a geometric approximation of the region in question. Other approaches, like those using Chebyshev's formula, seek to model the function in question (at least, in the relevant interval) with functions that are easier to integrate.

Determining good approximations for definite integrals is one of the principal aims of numerical analysis.