Question

How do you integrate `int (x-1)sqrt(2-x)dx` from [1,2]?

Answer 1

-4/15

Explanation:

u substitution: u = 2-x
du = -dx
x - 1 = 1 - u,
`int_1^2(x-1)sqrt(2-x)dx = -int_1^0(1-u)sqrt(u)du = -int_1^0sqrt(u)-usqrt(u)du = -int_1^0u^(1/2)-u^(3/2)du = int_0^1u^(1/2)-u^(3/2)du`
You have to change the bounds from [1,2] to [1,0] when you u-substitute. You find the bounds for u by plugging x=1 and x=2 into the u=2-x to find u=1 and u=0 (keep the order of u=1 before u=0 because u is negatively correlated to x).

From there: `= F(1) - F(0)` where `F(x) = 2/3u^(3/2) - 2/5u^(5/2)`
`= [2/3(1)^(3/2)-2/5(1)^(5/2)]-[2/3(0)^(3/2)-2/5(0)^(5/2)]`
`= [2/3-2/5]-[0-0]`
`= 4/15`

You don't need to u-substitute back and you'll get the same answer if you do.