# Thread: Can you integrate this?

1. ## Can you integrate this?

$\displaystyle {e}^{ {x}^{2 } }$?

with respect to x

The full question is $\displaystyle \iint {e}^{{ x}^{ 2} }$ dx dy by choosing the best order of integration

Limits for outside integral is $\displaystyle 2\sqrt{ln2}$ and 0
Inside integral is $\displaystyle \sqrt{ln 2}$ and $\displaystyle \frac{y}{ 2}$

$\displaystyle {e}^{ {x}^{2 } }$?

with respect to x

The full question is $\displaystyle \iint {e}^{{ x}^{ 2} }$ dx dy by choosing the best order of integration

Limits for outside integral is $\displaystyle 2\sqrt{ln2}$ and 0
Inside integral is $\displaystyle \sqrt{ln 2}$ and $\displaystyle \frac{y}{ 2}$
This integral has no closed form solution. You'll have to do it numerically.

-Dan

Yup. I was wrong. I didn't get back in time to edit my post. Sorry!

3. Originally Posted by topsquark
This integral has no closed form solution. You'll have to do it numerically.

-Dan
How? do you mean re-write the exponential as a ln?

4. I can't make out your LaTeX there. Is this your integral?

$\displaystyle \int_{0}^{2\sqrt{\ln(2)}}\int_{y/2}^{\sqrt{\ln(2)}}e^{x^{2}}\,dx\,dy?$

In this case, reverse the order of integration, to obtain

$\displaystyle \int_{0}^{\sqrt{\ln(2)}}\int_{0}^{2x}e^{x^{2}}\,dy \,dx,$

and go from there. To get the limits, draw the region in the xy plane, and figure it out.

5. Originally Posted by Ackbeet

$\displaystyle \int_{0}^{2\sqrt{\ln(2)}}\int_{y/2}^{\sqrt{\ln(2)}}e^{x^{2}}\,dx\,dy?$

In this case, reverse the order of integration, to obtain

$\displaystyle \int_{0}^{\sqrt{\ln(2)}}\int_{0}^{2x}e^{x^{2}}\,dy \,dx,$

and go from there. To get the limits, draw the region in the xy plane, and figure it out.
Thanks yes it is.

The order is not my problem its integrating that with repsect to x? You cant do it.

Thanks yes it is.
The order is not my problem its integrating that with repsect to x? You cant do it.
Once you have $\displaystyle \int_0^{2x} {e^{x^2 } dy} = \left. {ye^{x^2 } } \right|_{y = 0}^{2x} = 2xe^{x^2 }$ the rest is trivial.

7. Originally Posted by Plato
Once you have $\displaystyle \int_0^{2x} {e^{x^2 } dy} = \left. {ye^{x^2 } } \right|_{y = 0}^{2x} = 2xe^{x^2 }$ the rest is trivial.
???

We've still got to integrate $\displaystyle 2xe^{x^2 }$ with respect to x

???

We've still got to integrate $\displaystyle 2xe^{x^2 }$ with respect to x
My bad i've been a numpty

u = x^2
du = 2xdx

integrate e^u du

= e^u = e ^ x^ 2

put the limits in =

e (ln2)^0.5 - e^0 = 1