Question

Question #e2528

Answer 1

I don't know what the mistake was, but you should have `A_2 = 3/4`. (It is a triangle with base `1/2` and height `1`.)

Explanation:

If I understand what you're trying to do with "horizontal rectangles" You need:

`y` varies from `-1` to `0`

The greater `x` (the one on the right) is `x=1/2(3-y)`

The lesser `x` is `x = -2y`

The area is #int_-1^0 (x_"right" - x_"left") dy, or

`int_-1^0 (1/2(3-y)-(-2y) dy = int_-1^0 (3/2+3/2y) dy`

`= {: 3/2y+3/4y^2]_-1^0`

`= (0)-(-3/2+3/4) = 3/4`