Can you integrate this?

**adam_leeds** · May 2nd 2011, 08:08 AM

$<br /> {e}^{ {x}^{2 } } <br />$ ?

with respect to x

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

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

**topsquark** · May 2nd 2011, 08:21 AM

Originally Posted by adam_leeds

$<br /> {e}^{ {x}^{2 } } <br />$ ?

with respect to x

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

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

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!

**adam_leeds** · May 2nd 2011, 08:24 AM

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?

**Ackbeet** · May 2nd 2011, 08:33 AM

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

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

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

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

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

**adam_leeds** · May 2nd 2011, 08:47 AM

Originally Posted by Ackbeet

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

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

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

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

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.

**Plato** · May 2nd 2011, 09:02 AM

Originally Posted by adam_leeds

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

Once you have $\int_0^{2x} {e^{x^2 } dy} = \left. {ye^{x^2 } } \right|_{y = 0}^{2x} = 2xe^{x^2 }$ the rest is trivial.

**adam_leeds** · May 2nd 2011, 09:05 AM

Originally Posted by Plato

Once you have $\int_0^{2x} {e^{x^2 } dy} = \left. {ye^{x^2 } } \right|_{y = 0}^{2x} = 2xe^{x^2 }$ the rest is trivial.

???

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

**adam_leeds** · May 2nd 2011, 09:16 AM

Originally Posted by adam_leeds

???

We've still got to integrate $2xe^{x^2 }$ 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

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