# Math Help - Integrating an exponential function

1. ## Integrating 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

2. ## Re: Integrating an exponential function

Originally Posted by UnbeatableZoomy
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
There is no closed form solution. Though I wonder...did you mean to write $\int_1^y 2x e^{x^2} dx$ instead? This one can be done.

-Dan

3. ## Re: Integrating an exponential function

Yep, though between 1 and y

4. ## Re: 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.

5. ## Re: 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

6. ## Re: Integrating an exponential function

Well, that's a interesting development. What is the entire question asking?

7. ## Re: 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?

8. ## Re: 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.

9. ## Re: Integrating an exponential function

I would suggest reversing the order of integration. By the bounds of your integral, we have $\displaystyle 0 \leq y \leq 1$ and $\displaystyle 0 \leq x \leq y$, which implies $\displaystyle 0 \leq x \leq y \leq 1$.

From these we can determine the new bounds are $\displaystyle 0 \leq x \leq 1$ and $\displaystyle x \leq y \leq 1$. So

\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.

10. ## Re: Integrating an exponential function

Actually, no. The first integral will give $\int_0^1 (1- x)e^{x^2}dx= \int_0^1 e^{x^2}dx- \int_0^1 xe^{x^2}dx$. The second integral can be integrated as an exponential but the first cannot.

11. ## Re: Integrating an exponential function

That's why I said SHOULD :P haha