Chapter 4. Integration

4.1 Definite and indefinite integrals

Because any formula can be differentiated symbolically to find another formula, the main motivation for doing derivatives numerically would be if the function to be differentiated wasn't known in symbolic form. A typical example might be a two-person network computer game, in which player A's computer needs to figure out player B's velocity based on knowledge of how her position changes over time. But in most cases, it's numerical integration that's interesting, not numerical differentiation.

As a warm-up, let's see how to do a running sum of a discrete function using Yacas. The following program computers the sum 1+2+…+100 discussed to on page 7. Now that we're writing real computer programs with Yacas, it would be a good idea to enter each program into a file before trying to run it. In fact, some of these examples won't run properly if you just start up Yacas and type them in one line at a time. If you're using Adobe Reader to read this book, you can do Tools>Basic>Select, select the program, copy it into a file, and then edit out the line numbers.

Example 1

  \nn n := 1;
  \nn sum := 0;
  \nn While (n<=100) [
  \nn   sum := sum+n;
  \nn   n := n+1;
  \nn ];
  \nn Echo(sum);

The semicolons are to separate one instruction from the next, and they become necessary now that we're doing real programming. Line 1 of this program defines the variable n, which will take on all the values from 1 to 100. Line 2 says that we haven't added anything up yet, so our running sum is zero do far. Line 3 says to keep on repeating the instructions inside the square brackets until n goes past 100. Line 4 updates the running sum, and line 5 updates the value of n. If you've never done any programming before, a statement like n:=n+1 might seem like nonsense --- how can a number equal itself plus one? But that's why we use the := symbol; it says that we're redefining n, not stating an equation. If n was previously 37, then after this statement is executed, n will be redefined as 38. To run the program on a Linux computer, do this (assuming you saved the program in a file named sum.yacas):

   % yacas -pc sum.yacas 
  5050

Here the \verb@One way of stating this result is

$sum_{n=1}^{100} n = 5050 qquad .$

The capital Greek letter Σ, sigma, is used because it makes the “s” sound, and that's the first sound in the word “sum.” The n=1 below the sigma says the sum starts at 1, and the 100 on top says it ends at 100. The n is what's known as a dummy variable: it has no meaning outside the context of the sum. Figure a shows the graphical interpretation of the sum: we're adding up the areas of a series of rectangular strips. (For clarity, the figure only shows the sum going up to 7, rather than 100.)

discrete-sum

a / Graphical interpretation of the sum 1+2+…+7.

Now how about an integral? Figure b shows the graphical interpretation of what we're trying to do: find the area of the shaded triangle. This is an example we know how to do symbolically, so we can do it numerically as well, and check the answers against each other. Symbolically, the area is given by the integral. To integrate the function $xdot(t)=t$ , we know we need some function with a t² in it, since we want something whose derivative is t, and differentiation reduces the power by one. The derivative of t² would be 2t rather than t, so what we want is x(t)=t²/2. Let's compute the area of the triangle that stretches along the t axis from 0 to 100: x(100)=100^/2=5000.

integral-of-t

b / Graphical interpretation of the integral of the function $xdot(t)=t$ .

Figure c shows how to accomplish the same thing numerically. We break up the area into a whole bunch of very skinny rectangles. Ideally, we'd like to make the width of each rectangle be an infinitesimal number dx, so that we'd be adding up an infinite number of infinitesimal areas. In reality, a computer can't do that, so we divide up the interval from t=0 to t=100 into H rectangles, each with finite width dt=100/H. Instead of making H infinite, we make it the largest number we can without making the computer take too long to add up the areas of the rectangles.

riemann

c / Approximating the integral numerically.

Example 2

  \nn tmax := 100;
  \nn H := 1000;
  \nn dt := tmax/H;
  \nn sum := 0;
  \nn t := 0;
  \nn While (t<=tmax) [
  \nn   sum := N(sum+t*dt);
  \nn   t := N(t+dt);
  \nn ];
  \nn Echo(sum);

In example 50, we split the interval from t=0 to 100 into H=1000 small intervals, each with width dt=0.1. The result is 5,005, which agrees with the symbolic result to three digits of precision. Changing H to 10,000 gives 5,000.5, which is one more digit. Clearly as we make the number of rectangles greater and greater, we're converging to the correct result of 5,000.

In the Leibniz notation, the thing we've just calculated, by two different techniques, is written like this:

$int_{0}^{100} t :dt = 5,000$

It looks a lot like the Σ notation, with the Σ replaces by a flattened-out letter “S.” The t is a dummy variable. What I've been casually referring to as an integral is really two different but closely related things, known as the definite integral and the indefinite integral.

Definition of the indefinite integral

If $xdot$ is a function, then a function x is an indefinite integral of $xdot$ if, as implied by the notation, $dx/dt=xdot$ .

Interpretation: Doing an indefinite integral means doing the opposite of differentiation. All the possible indefinite integrals are the same function except for an additive constant.

Example 3

◊ Find the indefinite integral of the function $xdot(t)=t$ .

◊ Any function of the form

x(t)=t²/2+c ,

where c is a constant, is an indefinite integral of this function, since its derivative is t.

Definition of the definite integral

If $xdot$ is a function, then the definite integral of $xdot$ from a to b is defined as

$int_{a}^{b} xdot(t) dt$

$= lim_{Hrightarrowinfty} sum_{i=0}^{H} xdotleft(a+iDelta tright)Delta t qquad ,$

where Δ t=(b-a)/H.

Interpretation: What we're calculating is the area under the graph of $xdot$ , from a to b. (If the graph dips below the t axis, we interpret the area between it and the axis as a negative area.) The thing inside the limit is a calculation like the one done in example 50, but generalized to a≠ 0. If H was infinite, then Δ t would be an infinitesimal number dt.

4.2 The fundamental theorem of calculus

The fundamental theorem of calculus

Let x be an indefinite integral of $xdot$ , and let $xdot$ be a continuous function (one whose graph is a single connected curve). Then

$int_{a}^{b} xdot(t) dt = x(b)-x(a) qquad .$

Interpretation: In the simple examples we've been doing so far, we were able to choose an indefinite integral such that x(0)=0. In that case, x(t) is interpreted as the area from 0 to t, so in the expression x(b)-x(a), we're taking the area from 0 to a, but subtracting out the area from 0 to b, which gives the area from a to b. If we choose an indefinite integral with a different c, the c's will just cancel out anyway in the difference x(b)-x(a).

The fundamental theorem is proved on page 152.

Example 4

◊ Interpret the indefinite integral

$int_{1}^{2} frac{1}{t}:dt qquad .$

graphically; then evaluate it it both symbolically and numerically, and check that the two results are consistent.

hyperbola-area

d / The indefinite integral $int_1^2 (1/t)dt$ .

◊ Figure d shows the graphical interpretation. The numerical calculation requires a trivial variation on the program from example 50:

  a := 1;
  b := 2;
  H := 1000;
  dt := (b-a)/H;
  sum := 0;
  t := a;
  While (t<=b) [
    sum := N(sum+(1/t)*dt);
    t := N(t+dt);
  ];
  Echo(sum);

The result is 0.693897243, and increasing H to 10,000 gives 0.6932221811, so we can be fairly confident that the result equals 0.693, to 3 decimal places.

Symbolically, the indefinite integral is x=ln t. Using the fundamental theorem of calculus, the area is ln 2-ln 1≈ 0.693147180559945.

Judging from the graph, it looks plausible that the shaded area is about 0.7.

This is an interesting example, because the natural log blows up to negative infinity as t approaches 0, so it's not possible to add a constant onto the indefinite integral and force it to be equal to 0 at t=0. Nevertheless, the fundamental theorem of calculus still works.

4.3 Properties of the integral

Let f and g be two functions of x, and let c be a constant. We already know that for derivatives,

$frac{der}{dx}(f+g) = frac{df}{dx} + frac{dg}{dx}$

$text{and} frac{der}{dx}(cf) = cfrac{df}{dx} qquad .$

But since the indefinite integral is just the operation of undoing a derivative, the same kind of rules must hold true for indefinite integrals as well:

$int(f+g)dx = int f dx + int g dx$

$text{and} int(cf)dx = c int f dx qquad .$

And since a definite integral can be found by plugging in the upper and lower limits of integration into the indefinite integral, the same properties must be true of definite integrals as well.

Example 5

◊ Evaluate the indefinite integral

$int (x+2sin x) dx qquad .$

◊ Using the additive property, the integral becomes

$int x dx + int 2sin x dx qquad .$

Then the property of scaling by a constant lets us change this to

$int x dx + 2 int sin x dx qquad .$

We need a function whose derivative is x, which would be x²/2, and one whose derivative is sin x, which must be -cos x, so the result is

$frac{1}{2}x^2 - 2 cos x + c qquad .$

4.4 Applications

Averages

In the story of Gauss's problem of adding up the numbers from 1 to 100, one interpretation of the result, 5,050, is that the average of all the numbers from 1 to 100 is 50.5. This is the ordinary definition of an average: add up all the things you have, and divide by the number of things. (The result in this example makes sense, because half the numbers are from 1 to 50, and half are from 51 to 100, so the average is half-way between 50 and 51.)

Similarly, a definite integral can also be thought of as a kind of average. In general, if y is a function of x, then the average, or mean, value of y on the interval from x=a to b can be defined as

$bar{y} = frac{1}{b-a} int_a^b y:dx qquad .$

In the continuous case, dividing by b-a accomplishes the same thing as dividing by the number of things in the discrete case.

Example 6

◊ Show that the definition of the average makes sense in the case where the function is a constant.

◊ If y is a constant, then we can take it outside of the integral, so

$bar{y} = frac{1}{b-a} y int_a^b 1:dx$

$= frac{1}{b-a} y left.xright|_a^b$

$= frac{1}{b-a} y (b-a)$

$= y$

Example 7

◊ Find the average value of the function y=x² for values of x ranging from 0 to 1.

$bar{y} = frac{1}{1-0} int_0^1 x^2:dx$

$= left.frac{1}{3}x^3right|_0^1$

$= frac{1}{3}$

The mean value theorem

If the continuous function y(x) has the average value $bar{y}$ on the interval from x=a to b, then y attains its average value at least once in that interval, i.e., there exists ξ with a<ξ<b such that $y(xi)=bar{y}$ .

The mean value theorem is proved on page 159. The special case in which $bar{y}=0$ is known as Rolle's theorem.

Example 8

◊ Verify the mean value theorem for y=x² on the interval from 0 to 1.

◊ The mean value is 1/3, as shown in example 55. This value is achieved at $x=sqrt{1/3}=1/sqrt{3}$ , which lies between 0 and 1.

Work

In physics, work is a measure of the amount of energy transferred by a force; for example, if a horse sets a wagon in motion, the horse's force on the wagon is putting some energy of motion into the wagon. When a force F acts on an object that moves in the direction of the force by an infinitesimal distance dx, the infinitesimal work done is dW=Fdx. Integrating both sides, we have $W=int_a^b Fdx$ , where the force may depend on x, and a and b represent the initial and final positions of the object.

Example 9

◊ A spring compressed by an amount x relative to its relaxed length provides a force F=kx. Find the amount of work that must be done in order to compress the spring from x=0 to x=a. (This is the amount of energy stored in the spring, and that energy will later be released into the toy bullet.)

◊

$W = int_0^a F dx$

$= int_0^a kx dx$

$= left.frac{1}{2}kx^2right|_0^a$

$= frac{1}{2}ka^2$

The reason W grows like a², not just like a, is that as the spring is compressed more, more and more effort is required in order to compress it.

Probability

Mathematically, the probability that something will happen can be specified with a number ranging from 0 to 1, with 0 representing impossibility and 1 representing certainty. If you flip a coin, heads and tails both have probabilities of 1/2. The sum of the probabilities of all the possible outcomes has to have probability 1. This is called normalization.

globe

e / Normalization: the probability of picking land plus the probability of picking water adds up to 1.

So far we've discussed random processes having only two possible outcomes: yes or no, win or lose, on or off. More generally, a random process could have a result that is a number. Some processes yield integers, as when you roll a die and get a result from one to six, but some are not restricted to whole numbers, e.g., the height of a human being, or the amount of time that a uranium-238 atom will exist before undergoing radioactive decay. The key to handling these continuous random variables is the concept of the area under a curve, i.e., an integral.

single-die

f / Probability distribution for the result of rolling a single die.

Consider a throw of a die. If the die is “honest,” then we expect all six values to be equally likely. Since all six probabilities must add up to 1, then probability of any particular value coming up must be 1/6. We can summarize this in a graph, f. Areas under the curve can be interpreted as total probabilities. For instance, the area under the curve from 1 to 3 is 1/6+1/6+1/6=1/2, so the probability of getting a result from 1 to 3 is 1/2. The function shown on the graph is called the probability distribution.

two-dice

g / Rolling two dice and adding them up.

Figure g shows the probabilities of various results obtained by rolling two dice and adding them together, as in the game of craps. The probabilities are not all the same. There is a small probability of getting a two, for example, because there is only one way to do it, by rolling a one and then another one. The probability of rolling a seven is high because there are six different ways to do it: 1+6, 2+5, etc.

If the number of possible outcomes is large but finite, for example the number of hairs on a dog, the graph would start to look like a smooth curve rather than a ziggurat.

What about probability distributions for random numbers that are not integers? We can no longer make a graph with probability on the y axis, because the probability of getting a given exact number is typically zero. For instance, there is zero probability that a person will be exactly 200 cm tall, since there are infinitely many possible results that are close to 200 but not exactly two, for example 199.99999999687687658766. It doesn't usually make sense, therefore, to talk about the probability of a single numerical result, but it does make sense to talk about the probability of a certain range of results. For instance, the probability that a randomly chosen person will be more than 170 cm and less than 200 cm in height is a perfectly reasonable thing to discuss. We can still summarize the probability information on a graph, and we can still interpret areas under the curve as probabilities.

human-height

h / A probability distribution for human height.

But the y axis can no longer be a unitless probability scale. In the example of human height, we want the x axis to have units of meters, and we want areas under the curve to be unitless probabilities. The area of a single square on the graph paper is then

$(text{unitless area of a square})$

$= (text{width of square}$

$text{with distance units})$

$times (text{height of square}) qquad .$

If the units are to cancel out, then the height of the square must evidently be a quantity with units of inverse centimeters. In other words, the y axis of the graph is to be interpreted as probability per unit height, not probability.

Another way of looking at it is that the y axis on the graph gives a derivative, dP/dx: the infinitesimally small probability that x will lie in the infinitesimally small range covered by dx.

Example 10

◊ A computer language will typically have a built-in subroutine that produces a fairly random number that is equally likely to take on any value in the range from 0 to 1. If you take the absolute value of the difference between two such numbers, the probability distribution is of the form dP/dx=k(1-x). Find the value of the constant k that is required by normalization.

◊

$1 = int_0^1 k(1-x):dx$

$= left.kx-frac{1}{2}kx^2right|_0^1$

$= k-k/2$

$k = 2$

self-check:

Compare the number of people with heights in the range of 130-135 cm to the number in the range 135-140.

(answer in the back of the PDF version of the book)

average

i / The average can be interpreted as the balance point of the probability distribution.

When one random variable is related to another in some mathematical way, the chain rule can be used to relate their probability distributions.

laser

j / Example 59.

Example 11

◊ A laser is placed one meter away from a wall, and spun on the ground to give it a random direction, but if the angle u shown in figure j doesn't come out in the range from 0 to π/2, the laser is spun again until an angle in the desired range is obtained. Find the probability distribution of the distance x shown in the figure. The derivative dtan^-1z/dz=1/(1+z²) will be required (see example 65, page 86).

◊ Since any angle between 0 and π/2 is equally likely, the probability distribution dP/du must be a constant, and normalization tells us that the constant must be dP/du=2/π.

The laser is one meter from the wall, so the distance x, measured in meters, is given by x=tan u. For the probability distribution of x, we have

$frac{dP}{dx} = frac{dP}{du}cdotfrac{du}{dx}$

$= frac{2}{pi}cdotfrac{der tan^{-1}x}{dx}$

$= frac{2}{pi(1+x^2)}$

Note that the range of possible values of x theoretically extends from 0 to infinity. Problem 7 on page 102 deals with this.

If the next Martian you meet asks you, “How tall is an adult human?,” you will probably reply with a statement about the average human height, such as “Oh, about 5 feet 6 inches.” If you wanted to explain a little more, you could say, “But that's only an average. Most people are somewhere between 5 feet and 6 feet tall.” Without bothering to draw the relevant bell curve for your new extraterrestrial acquaintance, you've summarized the relevant information by giving an average and a typical range of variation. The average of a probability distribution can be defined geometrically as the horizontal position at which it could be balanced if it was constructed out of cardboard, i. This is a different way of working with averages than the one we did earlier. Before, had a graph of y versus x, we implicitly assumed that all values of x were equally likely, and we found an average value of y. In this new method using probability distributions, the variable we're averaging is on the x axis, and the y axis tells us the relative probabilities of the various x values.

For a discrete-valued variable with n possible values, the average would be

$bar{x} = sum_{i=0}^n x:P(x) qquad ,$

and in the case of a continuous variable, this becomes an integral,

$bar{x} = int_a^b x:frac{dP}{dx} : dx qquad .$

Example 12

◊ For the situation described in example 58, find the average value of x.

◊

$bar{x} = int_0^1 x:frac{dP}{dx} : dx$

$= int_0^1 x cdot 2(1-x):dx$

$= 2int_0^1 (x-x^2):dx$

$= 2left. left(frac{1}{2}x^2-frac{1}{3}x^3right) right|_0^1$

$= frac{1}{3}$

Sometimes we don't just want to know the average value of a certain variable, we also want to have some idea of the amount of variation above and below the average. The most common way of measuring this is the standard deviation, defined by

$sigma = sqrt{int_a^b (x-bar{x})^2:frac{dP}{dx} : dx} qquad .$

The idea here is that if there was no variation at all above or below the average, then the quantity $(x-bar{x})$ would be zero whenever dP/dx was nonzero, and the standard deviation would be zero. The reason for taking the square root of the whole thing is so that the result will have the same units as x.

Example 13

◊ For the situation described in example 58, find the standard deviation of x.

◊ The square of the standard deviation is

$sigma^2 = int_0^1 (x-bar{x})^2:frac{dP}{dx} : dx$

$= int_0^1 (x-1/3)^2cdot2(1-x) : dx$

$= 2 int_0^1 left(-x^3+frac{5}{3}x^2-frac{7}{9}x+frac{1}{9}right) : dx$

$= frac{1}{18} qquad , text{so the standard deviation is} sigma = frac{1}{sqrt{18}}$

$approx 0.236$

Homework Problems

1. Write a computer program similar to the one in example 52 on page 72 to evaluate the definite integral

$int_0^1 e^{x^2} qquad .$

(solution in the pdf version of the book)

2. Evaluate the integral

$int_0^{2pi} sin x :dx qquad ,$

and draw a sketch to explain why your result comes out the way it does. (solution in the pdf version of the book)

3. Sketch the graph that represents the definite integral

$int_0^2 -x^2+2x qquad ,$

and estimate the result roughly from the graph. Then evaluate the integral exactly, and check against your estimate. (solution in the pdf version of the book)

4. Make a rough guess as to the average value of sin x for 0<x<π, and then find the exact result and check it against your guess. (solution in the pdf version of the book)

5. Show that the mean value theorem's assumption of continuity is necessary, by exhibiting a discontinuous function for which the theorem fails. (solution in the pdf version of the book)

6. Show that the fundamental theorem of calculus's assumption of continuity for $xdot$ is necessary, by exhibiting a discontinuous function for which the theorem fails. (solution in the pdf version of the book)

7. Sketch the graphs of y=x² and $y=sqrt{x}$ for 0≤ x≤ 1. Graphically, what relationship should exist between the integrals $int_0^1 x^2:dx$ and $int_0^1 sqrt{x}:dx$ ? Compute both integrals, and verify that the results are related in the expected way.

8. Evaluate $intsqrt{bxsqrt{x}}dx$ , where b is a constant. (solution in the pdf version of the book)

9. In a gasoline-burning car engine, the exploding air-gas mixture makes a force on the piston, and the force tapers off as the piston expands, allowing the gas to expand. (a) In the approximation F=k/x, where x is the position of the piston, find the work done on the piston as it travels from x=a to x=b, and show that the result only depends on the ratio b/a. This ratio is known as the compression ratio of the engine. (b) A better approximation, which takes into account the cooling of the air-gas mixture as it expands, is F=kx^-1.4. Compute the work done in this case.

hw-piston

k / Problem 9.

10. A certain variable x varies randomly from -1 to 1, with probability distribution dP/dx=k(1-x²).
(a) Determine k from the requirement of normalization.
(b) Find the average value of x.
(c) Find its standard deviation.

11. Suppose that we've already established that the derivative of an odd function is even, and vice versa. (See problem 29, p. 50.) Something similar can be proved for integration. However, the following is not quite right.

Let f be even, and let $g=int f(x)dx$ be its indefinite integral. Then by the fundamental theorem of calculus, f is the derivative of g. Since we've already established that the derivative of an odd function is even, we conclude that g is odd.

Find all errors in the proof. (solution in the pdf version of the book)

12. A perfectly elastic ball bounces up and down forever, always coming back up to the same height h. Find its average height.

hw-holditch

l / Problem 13.

13. The figure shows a curve with a tangent line segment of length 1 that sweeps around it, forming a new curve that is usually outside the old one. Prove Holditch's theorem, which states that the new curve's area differs from the old one's by π. (This is an example of a result that is much more difficult to prove without making use of infinitesimals.)