1. ## integral

compute the double integral

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

how can I compute sin(y^2)

2. Originally Posted by yzc717
compute the double integral

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

how can I compute sin(y^2)
Can you explain the limits more clearly.

CB

3. Hello,

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

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

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

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

So the integral is now :

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

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

And that's easier

4. Originally Posted by Moo

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

And that's easier

that's what I got, I suddenly forget how to compute the sin(y^2)...

5. The inner integral will give y.

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

For this one : substitution !