The best way to appreciate the power of integration is to begin with what you already know how to do. Suppose you were driving in your car at a steady rate of 60 miles per hour for 5 hours. How far did you travel? The answer is easy, multiply the speed by the time to get 300 miles. But let’s think about what is going on geometrically. If you graph the speed of the car you have the graph y = 60. When you multiplied 60 by 5, you found the area of the rectangle with a height of 60 and a width of 5. You found the area under the graph in the interval of (0,5) on the x-axis.
That was an easy integral; we can make this more difficult. Suppose you are in a spaceship and you are accelerating. So your speed is not a constant of 60 mph, but it starts at zero and accelerates by 40 mph every hour. How far does the spaceship travel in 5 hours? Your speed is given by the formula y = 40x. The graph is still a straight line, but now the area under the graph makes a triangle rather than a rectangle. This is still easy to solve because we have a nice formula for the area of a triangle, which is one half the base times the height. The base is five hours, and the height is your maximum speed after five hours, which is 5 × 40 = 200. So the total distance traveled is ½ × 5 × 200 = 500 miles.
But now lets find a really difficult integral. Suppose the speed of your spaceship is given by y = 20x2. Now how far do you travel in five hours? The answer is the area under the graph. But there is no nice geometric shape that matches the area under a parabola. So how do you solve this problem? That is why we need calculus. In particular, we will use a technique called Riemann sums.
The Secret of Riemann Sums
Look at the graph of y = x3 above. Clearly the area under the graph does not make a nice geometric shape like a triangle or a rectangle. But we can approximate the area using rectangles. This is called the method of Riemann sums. The more rectangles we use, the better the approximation. And here is the secret: we can get the exact answer by using an infinite number of rectangles. And believe it or not, we do have the mathematical tools to sum up the area of an infinite number of rectangles, at least in some cases.
A Simple Beginning: y = x
Lets start really simple to make the math easy. Lets find the area under y=x from 0 to 10. We already know the answer using the formula for the area under a triangle (one half base times height); it is 50. Buts let’s find it using the method of Riemann sums. We’ll start with just a few rectangles, and work our way up.
- Let’s start with 5 rectangles. Each rectangle has a width of 10÷5 = 2. The height increases with each successive rectangle:
area = 2·2 + 2·4 + 2·6 + 2·8 + 2·10 = 60 - If we use ten rectangles of width 10÷10 = 1 we get:
area = 1·1 + 1·2 + … * 1·10 = 55. - If we use n rectangles of width 10÷n, then we get:
area = (10÷n)·1·(10÷n) + (10÷n)·2·(10÷n) + … + (10÷n)·n·(10÷n)
You may want to stare at this last example for a bit until you appreciate the pattern. The first (10÷n) is the width of each rectangle. The next two terms are the heigaht, which is given by y=x. Rectangle 2 is twice as high as rectangle 1. Rectangle 3 is three times as high. Rectangle n is n times as high as rectangle 1. Grab a pencile and work out the algebra. It works out to the following:
area = (100÷n2)(1 + 2 + 3 + 4 + … + n)
Now we are at the secret of Riemann sums. In order to find the area, we need to find a way to add up the first n integers. The solution is straightforward. Suppose we have to add up the first 100 integers. Do this by pairing them up. Add 1 and 100 to get 101. Add 2 and 99 to get 101. Add 3 and 98 to get 101. There are 50 pairs that total up to 101. So the answer is 50(101). The formula for the sum of the first n integers is ½ n (n + 1). Lets use that and rewrite the formula.
area = (100 ÷ n2) n⁄2 (n + 1)
Do the algebra (grab a pencil and work it out) and you get:
area = 50 + 50⁄n
Since we have an infinite number of rectangles, the area is 50. Exactly what we got before! We have a new technique for finding the area under the graph.
Exercises
- Use the method of Riemann sums to find the area under the graph of y = 4x from 0 to 20. Compare to the answer given by the formula for area under a triangle.
- Use the method of Riemann sums to find the area under the graph of y = 4x from 10 to 25 (hint: this does not make a triangle, but you can solve it by calculating the area for a large triangle with a base of 25 minus a small triangle with a base of 10). Compare to the answer given by the direct geometric method.
- Use the method of Riemann sums to find the area under the graph of y = 7 from 0 to 15 (hint: this is actually an easier problem since the area under the graph is a rectangle rather than a triangle. It requires a different, but easier, formula for summing up an infinite series)
