Area between $5e^x$ and $5xe^{x^2}$ using substitution

Question

I need to find the answer to this problem:

and was told to substitute $u=x^2$. I tried that and couldn't get to the correct answer. When substituting, I get $\frac{5}{2}\; du = 5x$, and I pull out $\dfrac{5}{2}$ from the integral, and my integrand turns out to be $x^{-1}e^x + e^u$. Am I messing up somewhere? I can't get to $\dfrac{5}{2}(e-1)$ (the correct answer) no matter what I try. Any guidance or help is appreciated.

Answer 1

Split the integral into a subtraction of integrals, and substitute $u=x^2, \mathrm d u=2x\mathrm d x$ only on the second integral.$$\begin{align}\int_0^1 5(\mathrm e^x-x\mathrm e^{x^2})~\mathrm d x&=5\int_0^1 \mathrm e^x~\mathrm d x-\tfrac 52\int_0^1 \mathrm e^u\mathrm d u\\[1ex]&=\tfrac 52\int_0^1\mathrm e^x\,\mathrm d x \end{align}$$