Find the interval from -2 to 0 of xf(xs^2)dx if interval from 0-4 f(u)du = 1?

Answer 1

Find Integral (-2 to 0, x f(x^2)dx ) if

Integral (0 to 4, f(u) du) = 1

First, I'm going to rearrange the integral, so that the x is next to the dx.

Integral (-2 to 0, f(x^2) x dx )

Now, we use substitution.

Let u = x^2. Then
du = 2x dx, and
(1/2)du = x dx
{Note: x dx is the tail end of our current integral, so (1/2)du will be the tail end after our substitution.}

Also, if x = -2, u = 4. If x = 0, u = 0. Therefore, our bounds of integration change subsequent to the substitution, and our integral becomes

Integral (4 to 0, f(u) (1/2) du )

Pull the (1/2) out of the integral, to get

(1/2) Integral (4 to 0, f(u) du )

It is bad form to write the bounds of integration from higher to lower number. This is equivalent to making our bounds from lower to higher, as long as we offset it by (-1).

(-1)(1/2) Integral (0 to 4, f(u) du)

BUT, we know Integral (0 to 4, f(u) du) = 1, so our answer is

(-1)(1/2)(1)

(-1/2)

Answer 2

interval from -2 to 0 of xf(x^2)dx
= interval from -2 to 0 of (1/2)f(x^2)dx^2
= interval from 4 to 0 of (1/2)f(u)du, where u = x^2
= (-1/2)interval from 0 to 4 of f(u)du
= -1/2