How to prove that $\int_{0}^{4}xe^{x^2}$ is $\frac{1}{2}\int_{0}^{16}e^u$

Question

How to prove that $\int_{0}^{4}xe^{x^2}$ is equal to the definite integral of $\frac{1}{2}\int_{0}^{16}e^u$?

My confusion/what I know so far

My professor asked our class this and I was very confused because I didn't understand how the integral value goes from 4 to 16?

I also believe, if i'm not mistaken, that $U=x^2$? and $dU=2x$ $dx$

Answer 1

Let $u=x^2$. Then $du=2xdx$.

When $x=0$, $u=0$.

when $x=4$, $u=16$.

$$\int_0^4xe^{x^2}dx=\frac{1}{2}\int_0^{16}e^udu$$

On L.H.S, $0$ and $4$ are values of $x$.

On R.H.S, $0$ and $16$ are values of $u$ corresponding to $x=0$ and $x=4$ respectively.

Answer 2

Yes, you are right. Now substitute your limits into the $u=x^2$, and you'll get your new limits in terms of $u$.