In the last post we learned the secret behind the technique of Riemann sums. But in order to keep the math simple, we only applied it to easy problems such as finding the area under y = x. Now we will show off the power of this new technique by finding the area under y = x2 from the interval from 0 to 10.
The first approximation uses 5 rectangles. The width of each rectangle is 10÷5. The height of rectangle one is (10÷5)2. The height of rectangle two is (2·10÷5)2, and so on. We’ve already factored the width in the formula below. Make sure that you stare at it long enough to get the idea.
area = (10÷5)[(1·10÷5)2 + (2·10÷5)2 + ... + (5·10÷5)2]
Now let’s do it for n rectangles. The width of each rectangle is 10÷n. Notice how each n replaces each 5 in the previous formula.
area = (10÷n)[1·10÷n)2 + (2·10÷n)2 + ... + (n·10÷n)2]
= (10÷n)[1·(100÷n2) + 22·(100÷n2) + ... + n2·(100÷n2)]
= (1000÷n3)[1 + 22 + 32 + ... + n2]
Now we are at crux of the problem. In the easier cases we looked at in the last post we had to sum up the first n integers. Now we have to sum up the first n integers squared. This is not quite as intuitive (although there is a fairly intuitive geometric way to visualize this using blocks) so I’ll just give the formula, which is n(n+1)(2n+1)÷6. Put that in our equation:
area = (1000÷n3)n(n+1)(2n+1)÷6
Grab an pencil and work out the algebra and you get:
area = 166.666 ( 2 + 3⁄n + 1⁄n2)
since n is infinite, the answer is 166.666 × 2 = 333.333
Our new technique has born fruit. We have now advanced our capabilities and can solve problems that we could not solve before.
Foreshadowing the Fundamental Theorem of Calculus
The answer we got is suspicious: 333.333. We were looking at a graph inside a 10×100 box and we got an answer that is exactly one third the total area. That was with the function y=x2. When we work with y = x the area under the graph is exactly half the area of the box. Hmmmmmm. If the formula were y = x3 would the area by one fourth? And would it be one-fifth for y = x4? Hold that thought, because we will revisit it with the fundamental theorem of calculus.
Let me close with a final promise. The technique of Riemann sums is a complicated and roundabout. But I hope that you worked out the arithmetic enough to understand the theory; it will deepen your understanding in the long run. After this we will never need to take a Riemann sum again. Instead we will leverage the our knowledge of differentiation by employing the fundamental theorem of calculus and get our answers much more easily.
Exercises
- Find the area under the graph y = 3x2 from 4 to 8.
- Find the area under the graph y = x3. Hint: the formula for the first n cubes is (n(n+1)/2)2.
