# integral

• May 16th 2009, 10:45 PM
yzc717
integral
compute the double integral

$\int\int sin(y^2) dydx$ from [0,sqrt(pi)]x[x,sqrt(pi)]

how can I compute sin(y^2)
• May 17th 2009, 12:35 AM
CaptainBlack
Quote:

Originally Posted by yzc717
compute the double integral

$\int\int sin(y^2) dydx$ from [0,sqrt(pi)]x[x,sqrt(pi)]

how can I compute sin(y^2)

Can you explain the limits more clearly.

CB
• May 17th 2009, 12:41 AM
Moo
Hello,

One can guess that $0\leq x \leq \sqrt{\pi}$ and $x\leq y\leq\sqrt{\pi}$

that is $\int_0^{\sqrt{\pi}}\int_x^{\sqrt{\pi}} \sin(y^2) ~dy ~dx$

Reverse the order of the integrals (which is possible since the integrand is positive) :
$0\leq x\leq y\leq \sqrt{\pi} \Rightarrow$ y ranges from 0 to $\sqrt{\pi}$

$0\leq x\leq y\leq \sqrt{\pi} \Rightarrow$ x varies from 0 to y.

So the integral is now :

$\int_0^{\sqrt{\pi}}\left(\int_0^y \sin(y^2) ~dx\right)~dy$

$=\int_0^{\sqrt{\pi}}\left(\sin(y^2)\int_0^y dx\right)~dy$

And that's easier :p
• May 17th 2009, 12:47 AM
yzc717
Quote:

Originally Posted by Moo

$=\int_0^{\sqrt{\pi}}\left(\sin(y^2)\int_0^y dx\right)~dy$

And that's easier :p

that's what I got, I suddenly forget how to compute the sin(y^2)...(Doh)(Doh)
• May 17th 2009, 01:11 AM
Moo
The inner integral will give y.

Which finally gives $\int_0^{\sqrt{\pi}} y\sin(y^2)~dy$

For this one : substitution !