| 19 | Introduction to Integrals |
Now that you’ve gained some familiarity with differentiation, it’s time to look at the reverse process, called integration. A math professor once said, “Differentiation is like navigating down the Mississippi. Integration is like navigating up the Mississippi and its tributaries. It’s the same river system, but a different process!” For most people, integration is more difficult than differentiation, and it’s easier to “get lost.” That professor knew what he was talking about! But with study, integration gets easier.
ADDING UP THE PIECES
Slope, found by differentiation, uses a vanishingly small change in x to produce a vanishingly small change in y. If you add an arbitrarily large number of these tiny changes, you have x and y, respectively.
If f’(x) is equal to y’ (two names for the same derivative), then the reverse process gives you the original f(x) or y. This process is expressed with the integral sign, which is an old-fashioned long letter S like the ones you see on violins and cellos. They use the long letter S because integration is an exalted form of summation, a sort of super sum. To abbreviate that, you need a super S!
THE ANTIDERIVATIVE
Imagine an object moving in a straight line at variable speed. The function that defines its instantaneous speed is the derivative of the function that defines the distance it has traveled from the starting point (its cumulative displacement). Reversing this, the function that defines the cumulative displacement is the antiderivative of the function that defines the instantaneous speed.
For functions denoted by lowercase italic letters such as f, g, or h, the antiderivative is customarily denoted by the uppercase italic counterpart such as F, G, or H. The concept of the antiderivative is the basis for integration. When you take the antiderivative of a particular function f, you find the function F that, when differentiated, will give you the function f again. Integration takes this concept and goes a little bit further, spawning an infinite number of different functions, any one of which will give you the original function back when you differentiate. You’ll see why in a minute.
From the differentiating you have done, you can start on integration just by taking the reverse process, or antiderivative. Here is a simple example:
so therefore
Here’s another example of a derivative and its corresponding antiderivative:
so therefore
PATTERNS IN CALCULATIONS
When you review the study of mathematics, always begin with a positive process, then reverse it to produce a negative process. After counting, you got into addition, which was reversed to make subtraction. After shortening multiple addition to make multiplication, it was reversed to parallel multiple subtraction, making division. Then indices brought powers, and you reversed that process to find roots.
In each situation, what began as a negative process, searching for a question to produce an answer, later developed into a positive approach to eliminate the search. Integration has a similar relationship to differentiation. They’re opposites. One process “undoes” the other (Fig. 19-1).
Figure 19-1
In mathematics, every “positive” process has a reverse or “negative” process that undoes it.
THE CONSTANT OF INTEGRATION
When you differentiate a function, you find its slope at a point or at a sequence of points. However, saying that a road has a certain slope (1 in 16, for example) doesn’t state how high the road is. It could be at sea level or on top of a mountain, or at any one of an infinite number of elevations in between.
Looked at mathematically, the three equations graphed in Fig. 19-2 each begin with y = x3 − 12x. Then there is a constant that is shown as 8, 0 (nothing shown), or −8. All three equations give the same derivative because the derivative of a constant is always equal to 0 (that is, nothing):
and
and
Figure 19-2
The constant of integration determines how high or low a curve is, but does not affect the shape of the curve.
By reversing the process, you have no direct means to find the constant. Again, think of the Mississippi and its branches! Going upstream, you don’t know which tributary to take. You didn’t have that sort of trouble with differentiation. You must leave room for an unknown constant, called the constant of integration. If the road began at sea level (constant of integration equal to 0), then integrating over any distance would find the height of the latest point above sea level. But if the road began at 860 ft above sea level, integrating the same way would give you an answer that would be off by 860 ft unless you took that constant into account. In general, you can say this:
where c can be any real number. If you say f(x) = 3x2 − 12, then you can write the antiderivative like this, without the constant:
But you must write the indefinite integral with the constant:
The integral is called “indefinite” because you don’t know the value of the constant c unless you have more specific, or definite, information about the problem.
Note the infinitesimal dx after the statement of the integral. This little value, called the differential, must always be written after integrals. If the variable were something else, such as z, you would write dz.
DEFINITE INTEGRALS
The indefinite integral, shown above and in Fig. 19-2, has mainly a theoretical value. A definite integral “adds it up” between specific values. It specifies that the curve starts at a certain point and follows it to a new point.
The definite integral is written the same way as the indefinite integral, but numbers or variables are put against the long S, one to its lower right and the other to its upper right. The figure at the lower right represents where the defined interval begins, and the figure at the upper right represents where the interval ends. Next, the expression for the antiderivative is put in square brackets with the limits outside, one at the lower right and the other at the upper right. (Alternatively, a vertical bar can be placed to the right of the antiderivative, and the limits can be written to the right of the bar.) Then you subtract the value at the lower limit from the value at the upper limit. An example is shown in Fig. 19-3.
Figure 19-3
The definite integral shows the area between a curve and the independent variable axis (in this case the x axis) over a defined interval.
In the graph of Fig. 19-3, the indefinite integral is plotted with the constant of integration set at 0. That’s the antiderivative function. When substituting in the lower value, you could make the starting point by inserting a constant of integration that would make the point equal to 0. However, it is not necessary, because you subtract this value from the upper value. Whatever you make the constant, it disappears when you subtract one value from the other.
By substituting x = −2 and x = 1, the second produces −11 and the first produces 16. By subtracting the first from the second, the change is −27. Substituting values x = 2 and x = 3, the same process produces the change of y in this range as 7.
FINDING AREA BY INTEGRATION
A most useful application for integration is for finding areas. If y is a succession of vanishingly small elements in an area, the sum of these elements over a certain range of the curve that is represented by this function will be equal to the area under the curve, which consists of a huge number of tiny strips (Fig. 19-4A). You go from the arbitrarily small to the arbitrarily large. In this context, the word “arbitrary” means as extreme as your imagination can make things—and then even a little more extreme than that! It’s a way of “sneaking up on infinity.”
To demonstrate how this method works, take the shaded area in Fig. 19-4B. The equation of the upper side is
Figure 19-4
At A, the area under a curve can be found by adding up the areas of a huge number of vanishingly small vertical strips. At B, the area of a simple geometric object, in this case a trapezoid, is found by integration.
The antiderivative (working the derivative formula backward) is
The indefinite integral would be
By making substitutions for x = 2 and x = 10, the limits for the definite integral, and subtracting, the area is 80. This will be true no matter what the value of c happens to be (it subtracts from itself), so you can just get rid of it when making the subtraction. Checking it by the geometric formula for the area of a trapezoid proves that you have the right answer.
AREA OF A CIRCLE
The area you find by means of integration does not have to be “under a curve.” In Fig. 19-5, two methods are illustrated for finding the area of a circle by means of integration.
In the method shown by Fig. 19-5A, the element of area is a wedge from the center of the circle to the circumference, taken at angle (in circular measure) x. The element has an area equal to 1/2 the base length times the height, because as the wedge gets arbitrarily narrow, its shape approaches a triangle. The height of this arbitrarily narrow slice is so close to r that you can think of its as actually being equal to r. The base length, according to similar reasoning, is equal to r dx. Using the familiar formula for the area of a triangle, you find that the area of the element is (1/2)r2 dx. Integrating produces (1/2)r2x because r can be treated as a constant. (You’re not integrating with respect to r, but with respect to the variable x!) For one complete revolution around the circle, the lower limit of x is 0 rad, and the upper limit is 2π rad. Substituting and subtracting give you the well-known formula for the area of a circle, πr2.
The other method, shown in Fig. 19-5B, uses an arbitrarily thin ring at radius x from the center. Here, the area of the element is the length of the ring 2πx times its thickness dx. Integrating that from 0 to r gives the same well-known result. If you like, you can work through the derivation for yourself and think of it as “extra credit.”
Figure 19-5
At A, the area of a circle can be found by breaking it up into a huge number of tiny wedges and integrating once around. At B, the area can be found by breaking the circle into a huge number of tiny concentric rings and integrating from the center to the circumference.
CURVED AREAS OF CYLINDERS AND CONES
With cylinders and cones, you can find the area of the curved part of the surface in two ways, as you did with the circle. For the cylinder (Fig. 19-6A), the length of the element is 2πr and its width is dx. Integrating from 0 to h (the height) produces what, in this case, is fairly obvious: 2πrh. For the cone (Fig. 19-6B), instead of being 2πrh, it is πrl, where l is the slant height of the cone, measured from the apex down to any point on the circumference of the base.
Figure 19-6
At A, integration is used to find the area of the curved surface of a cylinder. At B, integration is used to find the area of the curved surface of a cone.
SURFACE AREA OF A SPHERE
To find the surface area of a sphere using integration, imagine a ring that makes its way around the sphere from one pole to the other, as if it were a “moving zone of latitude” going from the earth’s north pole to its south pole. Measure the position of the ring according to the angle x with respect to a reference axis from pole to pole as shown in Fig. 19-7. By taking angle x from 0 to π rad, the ring will cover the entire area of the sphere’s surface. The circumferential length of the element is 2πr sin x. Its width is r dx, so its area, described as a function a(x), is equal to
Figure 19-7
Integration can be used to determine the surface area of a sphere.
Now consider the indefinite integral ∫a(x) dx as follows:
You can consider the quantity 2πr2 as a constant because you’re integrating not with respect to r, but with respect to x. Now evaluate the integral from 0 to π. Ignore the constant of integration; it cancels out in the subtraction anyway. When x = π, cos x = −1, so minus times minus makes a plus. When x = 0, cos x = 1, so you end up with −cos x = −1. Again, minus times minus makes a plus. If you follow the calculations in Fig. 19-7, you will see that the final answer is equal to 4πr2. You should recognize this as the formula for the surface area of a sphere, based on its radius r.
VOLUMES OF WEDGES AND PYRAMIDS
Integration can be used to find the volumes of three-dimensional objects. An example, in which the general formula for the volume of a wedge is determined, is shown in Fig. 19-8. Just as the vanishingly small element of an area is a line or curve of width dx, so the vanishingly small element of a volume is a surface having thickness dx. If we take the volume of this wedge, the area of the element or “slice,” based on the distance x from the apex of the wedge, can be defined as a function a(x) such that
Figure 19-8
Integration can be used to find the volume of a wedge.
where l is the distance from the apex to the “base” (shown as the flat, vertically oriented face at the right in Fig. 19-8), w is the width of the base, and t is the depth of the base. The thickness of the element is dx, so its volume is a(x) dx, or (wt/l)x dx. Integrating the function a(x) with respect to x produces this result:
Evaluating the expression [wt/(2l)]x2 from 0 to 1, you can easily calculate that the volume of the whole wedge is equal to wtl/2.
Figure 19-9 illustrates a similar method applied to a pyramid. To make the formula more general, A is used to represent the area of the base, and an element or “slice” is taken at a distance x from the apex. You can follow along with the calculations next to the drawing.
Figure 19-9
Integration can be used to find the volume of a pyramid.
VOLUMES OF CONES AND SPHERES
The method for finding the volume of a pyramid can also be applied to find the volume of a cone. Its base is a circle whose area A is πr2, as shown in Fig. 19-10A. Notice that the height h is the vertical height, measured perpendicular to the base, rather than the slant height, which is measured along the curved surface.
The volume of the disk-shaped element dx is πr2dx. The radius of the disk-shaped element is equal to the radius of the base of the whole cone times x/h, or (r/h)x. The surface area of one side of this thin disk (either the top or the bottom) can therefore be expressed as a function a(x) like this:
The volume of the disk-shaped element is (A/h2)x2 dx. To find the volume of the whole cone, first note the indefinite integral:
You know by now that you can get rid of the constant of integration when you evaluate this from x = 0 to x = h. Therefore the volume V of the entire cone is
It checks out! You should recognize this as the familiar formula for the volume of a cone having base radius r and vertical height h.
Another method to find the volume of a cone uses an element that is a cylindrical “shell” of radius x, as shown in Fig. 19-10B. Just follow along with the calculations in the figure.
Figure 19-10
At A, integration can be used to find the volume of a cone by slicing the cone into thin horizontal disks. At B, the cone is sliced into thin-walled, hollow cylinders.
The simplest method of using integration to find the volume of a sphere (Fig. 19-11) takes thin, disk-shaped “slices” of the sphere, with each slice perpendicular to a reference axis running from pole to pole. The value of x, the variable with respect to which you perform the integration, goes from − r to r. You can follow along with the calculations in the figure to see how it works.
Figure 19-11
Integration can be used to find the volume of a sphere by slicing the sphere into thin disks perpendicular to the reference axis.
QUESTIONS AND PROBLEMS
This is an open-book quiz. You may refer to the text in this chapter (and earlier ones, too, if you want) when figuring out the answers. Take your time! Consider all values given as exact, so you don’t have to worry about significant figures. The correct answers are in the back of the book. This quiz contains questions relevant to both this chapter and the previous one for review.
1. Find the derivatives of the products of functions u(x) and v(x), abbreviated here as u and v, under the following conditions.
(a) u = 2x + 3 and v = x3
(b) u = x2 − 6x + 4 and v = 3 sin x
(c) u = sin x and v = cos x
(d) u = x5− 4 and v = −2x3 + 2
2. Find the derivatives of the quotients of functions u(x) and v(x), abbreviated here as u and v, under the following conditions.
(a) u = −3x −6 and v = x2 + 2
(b) u = x2 + 2x and v = sin x
(c) u = sin x and v = 3 + cos x
(d) u = 3x4 − 4 and v = −2x2
3. In rectangular coordinates, the slope m of a line whose equation is y = mx + b, where b is a constant, is equal to dy/dx. Find the slope of a line tangent to a circle centered at the origin and having a radius of 3 units, when the point on the circle through which the tangent line passes corresponds to the following angles counterclockwise from the x axis.
(a) 10°
(b) 55°
(c) 105°
(d) 190°
(e) 300º
(f) 1 rad
(g) 2 rad
(h) 4 rad
4. Suppose an oscilloscope shows a waveform like the one in Fig. 19-12. Assume each vertical division represents exactly 1 volt (1 V), and each horizontal division represents exactly 1 millisecond (1 ms). The peak signal amplitudes are exactly plus and minus 5 V. The ramps (slanted lines) in the waveform are all straight. Assign t = 0 for the time at the origin (the center of the display). Now imagine that this signal is passed through a circuit called a differentiator, which produces an output waveform that is the derivative of the input waveform. Draw a graph of the output waveform. Include the maximum and minimum amplitudes.
Figure 19-12
Illustration for problems 4 through 7.
5. Suppose the waveform in Fig. 19-12 is passed through two successive differentiator circuits. Draw a graph of the output waveform from the second differentiator. Include the maximum and minimum amplitudes.
6. Suppose the waveform in Fig. 19-12 on page 277 is passed through a circuit called an integrator, whose output is the mathematical antiderivative of the input. Draw a graph of the output waveform. Include the maximum and minimum amplitudes.
7. Suppose the waveform in Fig. 19-12 on page 277 is passed through two successive integrators. Draw a graph of the output waveform, including the maximum and minimum amplitudes.
8. Evaluate the following indefinite integrals and check your results by differentiation.
(a) ∫4x3 dx
(b) ∫6x5 dx
(c) ∫9x8 dx
(d) ∫x4 dx
(e) ∫2 cos x d
(f) ∫ −4 sin x dx
9. Evaluate the integrals from Prob. 8 as definite integrals, with limits of −1 and 1. Assume angle measures in parts (e) and (f) to be in radians.
10. Figure 19-13 shows a quadratic function—that is, a function of the following form:
Figure 19-13
Illustration for problems 10 and 11.
where a, b, and c are constants. You can determine the values of these constants based on the points shown in the graph. Find the area under this curve between the following pairs of vertical lines on the x axis.
(a) x = −2 and x = 0
(b) x = 0 and x= 1
(c) x = − 1 and x = 2
(d) x = − 1 and x = 4
11. Find the derivative of the function whose graph is shown in Fig. 19-13. Then find the area under the resulting curve (or straight line, as the case might be) between the same pairs of vertical lines as in Prob. 10.
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