$\int _0^1\:2xe^{x^2-1}dx$

Question

for the answer I'm getting:

$1+\frac{1}{e}$

but according to symbolab, I get:

$1-\frac{1}{e}$

steps I'm currently taking with u substitution; I get:

$\int \:e^u$

then plugging back u;

$e^{x^2-1}$ (from 0 to 1)

then I compute,

$e^{-1}+e^0$ = $1+\frac{1}{e}$

I'm not sure

You are not sure of what? The derivative of $e^{x^2-1}$ is exactly $2x e^{x^2-1}$, hence the integral is trivial to compute. — Jack D'Aurizio, Oct 21 '16 at 01:20

score 0 · Accepted Answer · answered Oct 21 '16 at 01:15

0

For the definite integral it would be $e^{1-1} - e^{0-1}$

answered Oct 21 '16 at 01:15

RJM

1 Answers1