Page 36 - Calc for the Clueless

5. Our task is now to add all these up and then multiply everything by ∆x.

A. If we multiply out f(x 1), we see that the number we get from this term is 3 + 4(3) + 7 = 28. We see that
2
every term, if we were to multiply them out, would have a 28. Since there are n terms, the sum would be 28n.

B. Let's look at the ∆x terms. f(x 1) gives us 6(1 ∆x) + 4(1 ∆x) = 10(1 ∆x). f(x 2) gives us 6(2 ∆x) + 4(2 ∆x) =
10(2 ∆x). Similarly, f(x 3) = 10(3 ∆x). And f(x n) = 10(n ∆x). Adding and factoring, we get 10 ∆x(1 + 2 + 3 + ...
+ n).

2
C. Looking at the (∆x) terms, we would get

Factoring, we would get

6. Now, adding everything up, multiplying by ∆x, hoping everything fits on one line, we get

7. Substituting the formulas in the beginning and remembering ∆x = 3/n, we get

8. Using the distributive law, we get three terms:

A.

B.

C.

9. Adding A + B + C, we get a value for the area of 84 + 45 + 9 = 138.

Note

By letting n go to infinity, we are doing two things: chopping up the interval 3 < x < 6 into more and more
rectangles and, since ∆x = 3/n, making each rectangle narrower and narrower.

Wow!!!!!!!!! Are we grateful for the fundamental theorem!