How do you go about integrating 2e^(x^2) between y and 1? According to wolfram alpha it gives an error though I would assume there can be an answer expressed to this in one form or another.
Thanks
If you mean integrating 2e^(x^2) from 1 to y, then there is no closed form like topsquark mentioned.
However, if you are integrating 2xe^(x^2) from 1 to y, then use a direct substution. Let u(x) = x^2. Then du = 2x dx. Then u(1) = 1 and u(y) = y^2.
Integrating 2xe^(x^2) from 1 to y is the same as integrating e^u from 1 to y^2.
Thus, the integral is e^u | from 1 to y^2.
Thus, e^(y^2) - e^1.
Thus, e^(y^2) - e.
I would suggest reversing the order of integration. By the bounds of your integral, we have $\displaystyle \displaystyle 0 \leq y \leq 1$ and $\displaystyle \displaystyle 0 \leq x \leq y$, which implies $\displaystyle \displaystyle 0 \leq x \leq y \leq 1$.
From these we can determine the new bounds are $\displaystyle \displaystyle 0 \leq x \leq 1$ and $\displaystyle \displaystyle x \leq y \leq 1$. So
$\displaystyle \displaystyle \begin{align*} \int_0^1{ \int_0^y{ 2e^{x^2} \,dx } \,dy } &= \int_0^1{ \int_x^1{ 2e^{x^2} \,dy } \,dx } \end{align*}$
This should be able to be evaluated.