In this chapter you will learn about:

• Antiderivatives and the indefinite integral;
• Integrating a variety of functions;
• Some applications of integral calculus;
• Finding the area under a curve;
• Definite integrals with applications;
• Multiple integrals.

The previous chapter was devoted to the first major branch of calculus—differential calculus. This chapter is devoted to the second major branch—integral calculus. The two branches have different natures: differential calculus has procedures that make it possible to differentiate any continuous function; integral calculus has no such general procedures—every problem presents a fresh puzzle. Nevertheless, integral calculus is essential to all of the sciences, engineering, economics, and in fact to every discipline that deals with quantitative information.

There are two routes to introducing the concepts of integration. Although they start in different directions, they finally meld and create a single entity. If they were marked by road signs, the first would be “Antiderivatives and the indefinite integral” while the second would be “Area under a curve and the definite integral.”

3.1 Antiderivative, Integration, and the Indefinite Integral

306

The Antiderivative:

The goal of this section is to learn some techniques for integration, sometimes called antidifferentiation.

In this section we generally designate a function by . The concept of the antiderivative is fundamental to the process of integration and is easily explained. When a function is differentiated to give , then is an antiderivative of , that is,

This notation describes the defining property of the antiderivative, although only in terms of the derivative , not itself.

The antiderivative is usually written in the form:

The expression is also called the integral of . The symbol is known as the integration symbol; it represents the inverse of differentiation.

To summarize the notation, if F′(x) = f(x), then F(x) is the _____________ or _________________________.

307

Indefinite Integral:

Often one can find the antiderivative simply by guesswork. For instance, if , then . To prove this, note that

However, is not the only antiderivative of ; , where is a constant, is also an antiderivative because

In fact, a constant can always be added to a function without changing its derivative. If is an antiderivative of , then all the antiderivatives of can be denoted by writing

where is an arbitrary constant.

If it is useful, we can write the defining equation for the antiderivative in terms of a differential: . (If you need to review differentials, see frame 264.) Then we can describe all the antiderivatives by the notation

Because of the arbitrary constant, the definition is imprecise. For this reason this integral is called the indefinite integral.

In summary, the integral of the differential of a function is equal to the function plus a constant.

308

Because this first meaning of integration is the inverse of differentiation, for every differentiation formula in Chapter 2, there is a corresponding integration formula here. Thus from Chapter 2, frame 211,

so by the definition of the indefinite integral,

Now you try one. What is Choose the answer: [cos x + c | −cos x + c | sin x cos x + c | none of these]

Make sure you understand the correct answer (you can check the result by differentiation).

309

Now try to find these integrals:

1. , .
2. 

310

If you had difficulty with these problems, recall the definition of the indefinite integral. If then

In order to find , we must find an expression that when differentiated yields the given function . For instance, the derivative of is

by the formula for differentiating in frame 180. Hence, is an integral of . But the integral is indefinite because we could add any constant to it without changing its defining property: . Thus, including the integration constant , we find . (Note that this formula does not work for . That case will be discussed later.)

Likewise, by frame 235,

But the integral is indefinite because we could add any constant. Thus

311

A table with some common integrals is given in the next frame. You can check the truth of any of the equations by confirming that We will shortly use this method to verify some of the equations.

312

Table of Integrals:

The following integral table is a list of antiderivatives of some common functions. The integrals occur in many applications and are worth getting to know. For simplicity, in the integration table the arbitrary integration constants are omitted; and are constants.

Table of Integrals

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.

For convenience this table is repeated as Table 2: Integrals on page 288.

313

Let's see if you can check some of the formulas in the table. Show that integral formulas 10 and 16 are correct.

314

To prove that , we must show that .

1. 10. , .

1. 16. , .

3.2 Some Techniques of Integration

315

Change of Variable:

Often an unfamiliar function can be converted into a familiar function having a known integral by using a technique called change of variable. The method applies to integrating a “function of a function.” (Differentiation of such a function was discussed in frame 198. It is differentiated using the chain rule.) For example, can be written , where . With the following rule, the integral with respect to the variable can be converted into another integral, often simpler, depending on the variable .

Let's see how this works by applying it to a few problems.

316

Consider the problem of evaluating the integral

Let , or , and . Hence . Using the rule for change of variable, , the integral becomes

To prove that this result is correct, note that

as required.

Try the following somewhat tricky problem. If you need a hint, see frame 317. Evaluate

317

Let . Then , and by the rule for change of variable,

The integral becomes

318

Here is an example of a simple change of variable. The problem is to calculate . If we let , then the integral is , which is easy to integrate. Using , we have

To see whether you have caught on, evaluate (You may find the integral table in frame 312 helpful.)

319

Here is the answer:

If you obtained this result, go right on to 320. Otherwise, continue. If we let , then and

From formula 16 of frame 312 we have

Hence

Let's check this result:

as required. (We have used the chain rule here.)

320

Try to evaluate , where a and b are constants. The integral table in frame 312 may be helpful.

321

If we let , then and

322

We have seen how to evaluate an integral by changing the variable from to , where is some constant. Often it is possible to simplify an integral by substituting still other quantities for the variable.

Here is an example. Evaluate:

Suppose we let . Then , and

Try to use this method for evaluating the integral:

323

Taking , then and

324

Integration by Parts:

A technique known as integration by parts is sometimes helpful. Let and be any two functions of . Then the rule for integration by parts is

Here is the proof: using the product rule for differentiation,

Now integrate both sides of the equation with respect to .

But , and after transposing, we have

Here is an example: Find .

Let , . Then it is easy to see that , and . Note we have dropped the constant for simplicity. Thus

325

Try to use integration by parts to find .

326

To find use the formula for integration by parts. Because we now know that (frame 312, formula 7), we let , , so that , . Then,

327

Use the method of integration by parts to find the integral:

328

Here is the answer:

If you want to see how this is derived, continue here. Otherwise, go on to 329.

Let us make the following substitution and , and integrate by parts. Thus , , and the integral is

329

In integration problems it is often necessary to use a number of different integration “tricks of the trade” in a single problem.

Try the following (b is a constant):

330

The answers are

If you did both of these correctly, you are doing fine—jump ahead to frame 332. If you missed either problem, go to frame 331.

331

If you missed (a), you may have been confused by the change in notation from to . Remember is merely a symbol for some variable. All the integration formulas could be written replacing the with or or whatever you wish. Now for (a) in detail:

For problem (b), let , :

(The last step uses formula 10, frame 312.) Therefore,

You could also have solved this problem using integration by parts.

332

Method of Partial Fractions:

A useful manipulation from elementary algebra is to combine two simple fractions into one function. The method of partial fractions involves reversing this process in which you split a function into a sum of fractions with simpler denominators.

To illustrate this method, here is a simple example. Consider the function

Because , we can write

where and , which are yet to be defined, are called undetermined coefficients. Combining terms yields

Equate coefficient of like powers (this is comparable to solving a system of linear equations). Therefore and . Thus .

Now integrate in terms of simpler integrals that you already know how to evaluate see using formula 6, frame 312.

333

Now try this problem. Use the method of partial fractions to evaluate the integral .

334

First note that . Then write

Next, compare coefficients: and . Solve these two equations with the result that . Therefore

Hence the integral is

(The last step uses formula 6, frame 312.)

3.3 Area Under a Curve and the Definite Integral

The first section of this chapter focused on the techniques of integration, i.e. finding antiderivatives, all of which are embodied in the term “indefinite integral.” Just as differentiation is useful for many applications besides finding slopes of curves—for instance, calculating rates of growth or finding maxima and minima—so integral calculus has many applications such as finding the volumes of solids or finding the distance traveled by a body moving with a velocity . These applications all rely on the second type of integration, called the definite integral. This originated in the geometric problem of finding the area under a curve. We start by explaining what “area under a curve” means.

335

Area under a Curve:

To illustrate what is meant by “the area under a curve,” here is a graph of the simplest of all curves—a straight line given by .

What is the area between the line , and the ‐axis between the interval and ?

336

The area in the rectangle is the product of the base, , and the height, 3. Thus the area is .

There is an important convention in the sign of the area. In the drawing at the left, because is positive, the area is positive. However, the area under the graph at the right is negative because the height is , which is negative. Thus, areas can be positive or negative.

337

A key operation in integral calculus is to find the area under the graph of an arbitrary function bordered by the ‐axis between and , and and . Our procedure will be to divide the interval into slices of width , and then watch what happens as increases and the width decreases.

Let's approximate the area under a curve shown in the figure below left using the above procedure.

First we divide the area into a number of strips of equal widths by drawing lines parallel to the vertical axis. The figure shows four such strips. Each strip has an irregular top: but we can divide each strip's area into two sections: a rectangular shape and an approximately triangular shape. Suppose we label the strips 1, 2, 3, 4. The width of each strip is

The height of the first rectangular shape is , where is the value of at the beginning of the first strip. Similarly, the height of the second rectangular shape is where . The third and fourth rectangular shapes have heights and , respectively, where and .

The height of the first triangular shape is , where is the slope of evaluated at . The heights of the other triangles are similarly described.

You should be able to write an approximate expression for the area of any of the strips. Below write the approximate expression for the area of strip number 3, ΔA₃,

The symbol ≈ means “approximately equal to.”

338

The area of rectangular shape in strip number 3 is . The triangular shape has area . The approximate area of strip 3 is then

Now try to write an approximate expression for A, the total area of all four strips.

339

An approximate expression for the total area is simply the sum of the areas of all the strips. In symbols, because A = ΔA₁ + ΔA₂ + ΔA₃ + ΔA₄, we have

We could also write this as

∑ is the Greek letter Sigma, which corresponds to the English letter S and stands here for the sum. The symbol means

where is an index and the counting goes from to .

340

Suppose we divide the area into more strips each of which is narrower, as shown in the drawings. Evidently our approximation gets better and better.

If we divide the area into N strips, then

where . Now, if we take the limit where Δx → 0, the approximation becomes an equality. Thus,

Such a limit is so important that it is given a special name and symbol. It is called the definite integral and is written . This expression looks similar to the antiderivative , from frame 327, and as we shall see in the next frame, it is related. However, it is important to remember that the definite integral is defined by the limit described above:

(Incidentally, the integral symbol also evolved from the letter S and, like sigma, it was chosen to stand for sum.)

341

With this definition for the definite integral, the discussion in the last frame shows that the area under the curve between and is equal to the definite integral.

The function is called the integrand. The points and are called the limits of the integral. This usage has nothing to do with ; here “limit” simply means the boundary.

The process of evaluating is often spoken of as “integrating from to ,” and the expression is called the “integral of from to .” Caution: the indefinite and definite integral both employ the integral symbol , and so they can easily be confused. They are entirely different: the definite integral is a number and is equal to the area under the curve between limits a and b. In contrast the indefinite integral is a function — the antiderivative of the integrand.

342

The Area Function:

For a given function , we can introduce a new area function by introducing a variable for the upper limit of the definite integral in frame 341, and to avoid confusion we are renaming the integration variable

The area function and the function are closely related; the derivative of the area is simply . Hence

Recall from frame 306 that this means that is an antiderivative of the function as we will now show.

343

To illustrate that let's look at some simple areas that one can calculate directly.

The area under the curve , where is a constant, is Differentiating,

Find the area under between and .

Prove to yourself that

344

One way to calculate the area above is to think of it as the difference of the area of two right triangles. Using area = ½ base × height, we have

345

To see why , consider how the area changes as increases by an amount . , where is the narrow strip shown.

Can you find an approximate expression for ΔA?

This represents a major step in the development of integral calculus, so don't be disappointed if the result eludes you.

346

The answer we want is

If you wrote this, go on to 347. If you would like a more detailed discussion, read on.

The following argument may appear similar to our definition of the definite integral (frame 340), but here we are proving that the area function is an antiderivative.

Let's take a close look at the area . As you can see, the area is a long narrow strip, which we divided into a rectangle and a triangle with a small piece remaining. In the figure above we just show the rectangle and triangle. As before (frame 338) the area of the strip is then

where the small term in the above expression represents the part of the area not included in the triangle or rectangle. For a sufficiently small value of , the two last terms can be ignored compared to the first term. Therefore in the limit as

This supports the conjecture in frame 342 that the area function is an antiderivative,

347

To summarize this section, we have found an expression for the area under a curve defined by that satisfies the equation Thus, if we can find a function whose derivative is , we can find the area up to a constant that remains to be determined to find the exact area.

348

The second meaning of integration, the definite integral, was introduced by the problem of finding the area under a curve, , from to ; this involved finding the limit of an infinite sum. (Recall that this area is a number and not a function.)

Finding the limit of an infinite sum requires a tedious calculation. Fortunately, we can calculate a definite integral by a far simpler method using the techniques of integration for indefinite integrals. The result is that the area is

where is any antiderivative of , i.e. . What makes this possible is explicitly calculating the constant . Instead of calculating a limit, we are now evaluating an antiderivative at the endpoints of the interval.

Let's begin with the area function associated with a function from to ,

Note that our area function is a function of the variable , which is the upper limit of the integral. That's why, to avoid confusion with the variable , we chose a different variable, , for integration. In frame 346 we showed that .

When , the area is zero

Let be any antiderivative of . Then the area function is equal to

where is an arbitrary constant. Set . Then

Therefore . Hence the area under the curve is equal to

Now set . Then the definite integral corresponds to the area under the curve of the function for . This can be determined by calculating the value of any antiderivative of at and subtracting the value of at .

The definite integral is commonly written as

349

Fundamental Theorem of Calculus:

The significance of the result in the last frame is that determining the area does not require calculating the limit of the summation in the definition of the definite integral (which can be a difficult calculation) but merely evaluating any antiderivative at the endpoints of the interval.

In frame 336, we discussed the convention for determining the sign for the area. Note that a definite interval can be positive or negative for some interval. If a function is negative in some interval, then the definite integral evaluated in that interval will be negative.

Summary: We have shown that for any continuous function ,

where .

This relation is so extraordinary that it is called the fundamental theorem of calculus.

350

To see how all this works, we will find the area under the curve between and .

In this example an antiderivative is because . Therefore

Note that the definite integral yields a number. Furthermore, we no longer need to introduce an undetermined constant whenever we evaluate an expression such as .

351

Can you find the area under the curve , between the points and ?

352

The solution is straightforward. An antiderivative for the function is . Therefore

353

Find the area under the curve between and .

354

An antiderivative for the function is . Therefore

Let's introduce a little labor‐saving notation. Frequently we have to find the difference of an expression evaluated at two points, as . This is often denoted by

For example, . Using this notation, the solution to this problem is written

Note that this area is negative.

355

Here is another practice problem:

The graph shows a plot of . Find the area between the curve and the ‐axis from and .

• Answer: [5 | ¼ | 4 | | | none of these]

356

Here is how to do the problem: one antiderivative is . You can check that . Therefore

357

To help remember the definition of definite integral, try writing it yourself. Write an expression defining the definite integral of between the limits and .

358

The answer is

Congratulations if you wrote this or an equivalent expression. If you wrote

your statement is true, but it is not the definition of a definite integral. The result is true because both sides represent the same thing—the area under the curve of between and . It is an important result because without it the definite integral would be much more difficult to calculate; however, it is not true by definition.

The definite integral appears merely to provide a second way to find the area under a curve. To compute the area, we were led back to the indefinite integral, but we could have found the area directly from the indefinite integral. The importance of the definite integral arises from its definition as the limit of a sum. The process of dividing a system into small elements and then adding them together is a powerful technique that is applicable to more problems than finding the area under a curve. These invariably lead to definite integrals, which we can evaluate in terms of indefinite integrals by using the fundamental theorem of calculus (frame 349).

359

Can you prove that

360

The proof that is as follows:

Reversing the limits of the integral yield,

361

Which of the following expressions correctly gives ?

362

The answer is

It is easy to see why this result is true by inspecting the figure.

The integral yields the total area under the curve, from 0 to , which is the sum of the area between 0 to , and between and . But is negative, because is negative in that region. By symmetry, the two areas just add to 0. However, you should be able to find or separately. Try this problem:

363

The answer is

If you forgot the integral, you can find it in frame 312. In evaluating at the limits, , and .

364

Here is a graph of the function .

Can you find the shaded area under the curve between the origin and ?

• Answer: [e^−x | 1 − e^−x | x + e^−x | x + e^−x − 1]

365

Here is the solution to 364.

The area is bounded by a vertical line through . Our result gives as a variable that depends on . If we choose a specific value for , we can substitute it into the above formula for and obtain a specific value for . We have obtained a definite integral in which one of the boundary points is left as a variable.

366

Let's evaluate one more definite integral before going on. Find:

(If you need to, use the integral tables, frame 312.)

• Answer: [0 | 1 | ∞ | π | π/2 | none of these]

367

From the integral table, frame 312, we see that

Therefore,

Because , we have that . Similarly, sin⁻¹ 0 = 0. Thus, the integral has the value .

A graph of is shown above. Although the function is discontinuous at x = 1, the area under the curve is perfectly well defined.

3.4 Some Applications of Integration

368

In this section we are going to apply integral calculus to a few problems.

In Chapter 2 we learned how to find the velocity of a particle if we know its position in terms of time. Now we can reverse the procedure and find the position from the velocity. For instance, we are in an automobile driving along a straight road through thick fog. To make matters worse, our mileage indicator is broken. Instead of watching the road all the time, let's keep an eye on the speedometer. We have a good watch along, and we make a continuous record of the speed starting from the time when we were at rest. The problem is to find how far we have gone. More specifically, given , how do we find the change in position of the automobile , called the displacement, when we were at rest? Because the automobile is traveling in the same direction, the change of position is equal to the distance traveled.

Try to work out a method.

369

Because , we must have (as was shown in frame 275). Now let us integrate both sides from the initial point to the final point in order to find the change in position of the automobile

where we have replaced the integration variable by .

370

Another way to understand this problem is to look at it graphically. Here is a plot of as a function of .

In time the distance traveled is . The total distance traveled is thus equal to the area under the curve between the initial time and the time of interest, and this is . There may be some confusion because the same symbol , which is the dependent variable of the function, appears both as an endpoint of the integral and as the integration variable in the integrand . The integration variable is what we call a dummy variable and we can denote it by any symbol, for example, . Thus the integral is written as

Often this distinction between the dependent variable of the function and the integration variable is not explicit but implicit, and the symbol is used in both places, as we will do in a few frames.

371

Suppose an object moves with a velocity that continually decreases in the following way, ( and are positive constants).

At the object is at the origin; . Which of the following is the distance, , the object will have moved after an infinite time (or, if you prefer, after a very long time)?

• Answer:

372

Here is the solution to the problem of frame 371.

We are interested in the . Because e^−bt → 0 as t → ∞, we have

Although the object never comes completely to rest, its velocity gets so small that the total distance traveled is finite.

373

Not all integrals give finite results. For example, try this problem.

A particle starts from the origin at with a velocity , where and are constants. How far does it travel as ?

373

374

It is easy to see that problem 373 leads to an infinite integral.

(The last step uses formula 6, frame 312.)

Because as , we see that as . In this case, the particle is always moving fast enough so that its motion is unlimited. Or, alternatively, the area under the curve increases without limit as .

375

The Method of Slices:

Integration can be used for many tasks besides calculating the area under a curve. For example, it can be used to find the volumes of solids of known geometry. A general method for this is explained in frame 395. However, one can calculate the volume of symmetric solids by a simple extension of methods we have learned already. In the next few frames we are going to find the volume of a right circular cone.

The height of the cone is , and the radius of the base is . We will let represent distance vertically from the base.

Our method of attack, called the method of slices, is similar to that used in frame 378 to find the area under a curve. We will slice the body into a number of discs whose volume is approximately that of the cone in the figure (the cone has been approximated by ten circular discs). Then we have

where is the volume of one of the discs. In the limit where the height of each disc (and hence the volume) goes to 0, we have

In order to evaluate this, we need an expression for .

376

Because we are going to take the limit where , we will represent the volume element by from the start.

A section of the cone is shown in the figure, which for our purposes is represented by a disc. The radius of the disc is and its height is . Try to find an expression for in terms of . (You will have to find in terms of .)

377

The answer is:

If you got this answer, go on to 378. If you want to see how to derive it, read on.

The volume of this disc is the product of the area and height. Thus, . Our remaining task is to express in terms of . The diagram shows a cross-section of the cone.

Because and are corresponding edges of similar triangles, it should be clear that

Thus,

378

We now have an integral for V.

Try to evaluate this.

379

The result is Congratulations, if you obtained the correct answer. Go on to 380.

Otherwise, read below:

380

Let's find the volume of a sphere. Can you write an integral that will give the volume of the hemisphere?

(The slice of the hemisphere shown in the drawing may help you in this.)

381

The answer is

If you wrote this, go ahead to frame 382. Otherwise, read on.

In order to calculate the volume of the sphere, consider a disk of thickness and radius , located at a distance from the center of the sphere as shown in the drawing in frame 280. The differential volume of the disk between and is . By the Pythagorean theorem , therefore .

As you add up the disks, the limits of the variable are to . The volume integral is

382

Now go ahead and evaluate the integral

383

The answer is

384

Here's a problem that involves finding the area under a curve and the volume of a solid of revolution generated by the curve.

What is the area under the curve , for the range to ?

385

Here is how to solve the problem. The area under the curve is the integral

386

Now rotate that area about the ‐axis to generate the volume shown in the diagram.

Can you use the method of slices to write an integral that will give the volume? (The slice in the drawing may help you visualize this.)

387

The answer is . If you wrote this, go ahead to frame 388. Otherwise, read on.

Consider a disk of thickness and radius , located at a distance from the origin as shown in the diagram. The differential volume for the disk is

The limits of the variable are to . Now you can set up the volume integral

388

Now go ahead and evaluate the integral.

389

The answer is

3.5 Multiple Integrals

390

The subject of this section—multiple integrals—introduces some new concepts and enables us to apply calculus to a world of problems that involve multiple variable calculus, in contrast to single variable calculus.

The integrals we have discussed so far, of the form , have had a single independent variable, usually called . Double integrals are similarly defined for two independent variables, and . In general, multiple integrals are defined for an arbitrary number of independent variables, but we will only consider two. Note that up to now has often been the dependent variable: . In this section, however, along with will always be independent variables, and will be the dependent variable. Thus, is a function of two variables.

In frame 349 the definite integral of between and was defined by

The double integral is similarly defined, but with two independent variables. There are, however, some important differences. For a single definite integral the integration takes place over a closed interval between and on the ‐axis. In contrast, the integration of takes place over a closed region in the plane.

To define the double integral, divide the region into smaller regions each of area .

Let be an arbitrary point inside the region . Then in analogy to the integral of a single variable, the double integral is defined as

391

The double integral is often evaluated by taking to be a small rectangle with sides parallel to the and axes. The procedure is first evaluate the sum and limit along one direction and then along the other. Consider the upper portion of the region in the plane to be bounded by the curve , while the lower portion is bounded by , as in the diagram.

If we let , then

This is a complicated expression, but it can be simplified by carrying it out in two separate steps. Let us insert some brackets to clarify the separate steps.

The first step is to carry out the operation within the brackets. Note that is not altered as we sum over in the brackets. This corresponds to summing over the crosshatched strip in the diagram with treated as approximately a constant. The quantity in square brackets is then merely a definite integral of the variable , with treated as a constant. Note that although the limits of integration, and , are constants for a particular value of , they are in general non‐constant functions of . The quantity in square brackets can then be written as

This quantity will no longer depend on , but it will depend on both through the integrand and the limits , . Consequently,

In calculations it is essential that one first evaluate the integral in the square brackets while treating as a constant. The result is some function, which depends only on . The next step is to calculate the integral of this function with respect to , treating now as a variable.

392

Multiple integrals are most easily evaluated if the region is a rectangle whose sides are parallel to the and coordinate axes, as shown in the drawing.

The double integral is

As an exercise to test your understanding, how would the above expression be written if the integration over were to be carried over before the integration over ?

393

When is chosen to be integrated first, the double integral becomes

This can be found merely by interchanging the and operations in evaluating the double integral. (Note that the integration limits must also be interchanged.)

To see how this works, let us evaluate the double integral of over the rectangle in the plane bounded by the lines , , , and . The double integral is equal to the iterated integral.

Alternatively we could have written

Evaluate each of the above expressions. The answers should be the same.

394

Integrating first over and then over yields

Integrating first over and then over yields

395

Just as the equation defines a curve in the two‐dimensional plane, the equation defines a surface in the three‐dimensional space because that equation determines the value of for any values assigned independently to and .

We can easily see from the above definition of the double integral that is equal to the volume of space under the surface and above the region . In this case is the height of column above . Therefore, is approximately equal to the volume of that column. The sum of all these columns is then approximately equal to the volume under the surface. In the limit as , the sum defining the double integral becomes equal to the volume under the surface and above , so

Calculate the volume under the surface defined by and above the rectangle whose sides are determined by the lines , , , and .

396

The answer is 36. If you obtained this result, go to frame 397. If not, study the following.

The iterated integral could just as well have been evaluated in the opposite order.

397

The bottom of this plow‐shaped solid is in the form of an isosceles triangle with base and height . When oriented along the axes as shown, its thickness is given by , where is a constant. The problem is to find an expression for the volume.

398

The volume is . Read on if you want an explanation; otherwise go to 399.

The base of the object forms a triangle, as shown. The integral can be carried out with respect to and in either order. We shall integrate first over .

From the drawing, , so that , and

The integral can also be evaluated in reverse order. The calculation is simplified by making use of symmetry; the volume is twice the volume over the upper triangle. Thus

where . The answer is the same, .

Conclusion to Chapter 3

399

Well, here you are at the last frame of Chapter 3. At this point you should understand the principles of integration and be able to do some integrals. With practice your repertoire will increase. Don't be afraid to use integral tables—everyone does. You can find them online, for instance, https://en.wikipedia.org/wiki/Lists_of_integrals.

Summary of Chapter 3

3.1 Antiderivative, Integration, and the Indefinite Integral (frames 306–314)

When a function is differentiated to give , then is an antiderivative of , that is, If is an antiderivative of , then all the antiderivatives of can be denoted by writing

where is arbitrary constant. The expression is also called the integral of . The symbol is known as the integration symbol, and represents the inverse of differentiation. It is important not to omit this constant. Otherwise the answer is incomplete.

Indefinite integrals are often found by hunting for an expression that when differentiated gives the integrand . Thus from the earlier result that

we have that

By starting with known derivatives as in Table 1, a useful list of integrals can be found. Such a list is given in frame 312 and for convenience is repeated in Table 2. You can reconstruct the most important of these formulas from the differentiation expressions in Table 1. More complicated integrals can often be found in large integral tables.

3.2 Some Techniques of Integration (frames 315–334)

Often an unfamiliar function can be converted into a familiar function having a known integral by using a technique called change of variable which is related to the chain rule of differentiation and uses the relation

Another valuable technique is integration by parts, as described by the relation proved in frame 324.

Frequently a number of different integration procedures are used in a single problem as illustrated in frames 329–331.

The method of partial fractions (frame 332) involves splitting a function into a sum of fractions with simpler denominators that can be integrated by other methods.

3.3 Area under a Curve and Definite Integrals (frames 335–367)

The area under the curve of a function between and can be found by dividing the area into N narrow strips parallel to the y‐axis, each of area , and summing the strips. In the limit as the width of each strip approaches zero, the limit of the sum approaches the area under the curve.

This infinite sum is called a definite integral and is denoted by

The area function has the variable for the upper limit of the definite integral in frame 341, and the integration variable is renamed .

The area function is an antiderivative of the function (frame 346), . Therefore, it is an indefinite integral.

where is any particular antiderivative of , and is an arbitrary constant. If we want to know the area bounded by and some value , the constant can be evaluated by noting that if , then the area is zero, so and . Therefore, The area under the curve between and is then

This result is called the fundamental theorem of integral calculus (frame 349). The significance of the result in the last frame is that determining the area does not require calculating the limit of the summation in the definition of the definite integral but merely evaluating any antiderivative at the endpoints of the interval.

3.4 Some Applications of Integration (frames 368–389)

If we know v(t), the velocity of a particle as a function of t, we can obtain the position of the particle as a function of time by integration. We saw earlier that , so , and if we integrate both sides of the equation from the initial point to the final point , we have the change of position of the particle, called the displacement,

where we have replaced the integration variable by .

Applications of integration in finding volumes of symmetric solids are given in frames 375–389.

3.5 Multiple Integrals (frames 390–399)

Multiple integrals may be defined for an arbitrary number of independent variables. We discuss two variables because the procedures for an arbitrary number are merely generalizations of those that apply to two independent variables. The double integral over a region in the plane of the function is defined as

as discussed in frames 390–391. The double integral can be evaluated by integrating over one variable while holding the other variable constant, in either order.

Continue to Chapter 4.