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

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- Apr 15th 2013, 06:33 AMUnbeatableZoomyIntegrating an exponential function
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 - Apr 15th 2013, 07:05 AMtopsquarkRe: Integrating an exponential function
- Apr 15th 2013, 07:09 AMUnbeatableZoomyRe: Integrating an exponential function
Yep, though between 1 and y

- Apr 15th 2013, 07:17 AMmathguy25Re: Integrating an exponential function
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. - Apr 15th 2013, 07:28 AMUnbeatableZoomyRe: Integrating an exponential function
I didn't mention this but the question was actually a double integral which finished with dxdy so I think it involved a change of order

- Apr 15th 2013, 08:51 AMmathguy25Re: Integrating an exponential function
Well, that's a interesting development. What is the entire question asking?

- Apr 15th 2013, 09:02 AMMINOANMANRe: Integrating an exponential function
UNBEATABLE....

you are posting an exponential function and then you are talking for a double integral? What really do you want? - Apr 16th 2013, 03:29 AMUnbeatableZoomyRe: Integrating an exponential function
The question asks you to evaluate the integral between 1 and 0 of the integral between 1 and y of 2e^(x^2) dxdy.

Word isn't working and I haven't yet learned Latex thus why I've just described it in a sentence. - Apr 16th 2013, 03:36 AMProve ItRe: Integrating an exponential function
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. - Apr 16th 2013, 09:41 AMHallsofIvyRe: Integrating an exponential function
Actually, no. The first integral will give $\displaystyle \int_0^1 (1- x)e^{x^2}dx= \int_0^1 e^{x^2}dx- \int_0^1 xe^{x^2}dx$. The second integral can be integrated as an exponential but the first cannot.

- Apr 16th 2013, 11:20 PMProve ItRe: Integrating an exponential function
That's why I said SHOULD :P haha