While revising, I came across this question:
Let $h(x)=\int_0^{x^2}e^{x+t}dt$.
Find $h'(1)$.
I tried using the substitution $t=u^2$, then $dt=2u du$. The integral becomes $h(x)=\int_0^x{e^{x+u^2}}\cdot 2udu$.
Then by the Fundamental Theorem of Calculus, $h '(x)=e^{x+x^2}\cdot 2x$, and hence $h'(1)=2e^2$.
However, the correct answer is $3e^2-e$.
May I know what is wrong with the above approach?
Thanks a lot.