Question

How do you find the area bounded by `y=6x-x^2` and `y=x^2-2x`?

Answer 1

Use `A = int_a^b(y_1(x)-y_2(x))dx` where `y_1(x) >= y_2(x)`

Explanation:

Find the x coordinates of endpoints of the area.

`6x - x^2 = x^2- 2x`

`0 = 2x^2-8x`

`x = 0 and x = 4`

This means that:

`a = 0 and b = 4`

Evaluate both at 2 and observe which is greater:

`y = 6(2)-(2)^2 = 8`

`y = 2^2 - 2(2) = 0`

The first one is greater so we subtract the second from the first in the integral:

`int_0^4(6x-x^2) - (x^2 - 2x)dx =`

`int_0^4(8x-2x^2)dx =`

`{:4x^2-2/3x^3|_0^4 =`

`4(4)^2-2/3(4)^3- 4(0)^2 + 2/3(0)^2 = 64/3`