# Double integrals problem

• May 5th 2009, 12:02 PM
chella182
Double integrals problem
Well, my problem isn't with the double integral bit, it's just with the integration once I've switched the limits et cetera. The original problem is:
$\displaystyle \int_{0}^{1}\int_{\sqrt{y}}^{1}2x^5y\cos{(xy^2)}dx dy$
...and I'm told to solve it by changing the order of integration. I get:
$\displaystyle \int_{0}^{1}\int_{0}^{x^2}2x^5y\cos{(xy^2)}dydx$
I know this forum's not here to check my work, but whether my new limits are correct or not, I still don't know how to integrate $\displaystyle 2x^5y\cos{(xy^2)}$ with respect to y, taking x as a constant.

Can anyone help?
• May 5th 2009, 12:11 PM
TheEmptySet
Quote:

Originally Posted by chella182
Well, my problem isn't with the double integral bit, it's just with the integration once I've switched the limits et cetera. The original problem is:
$\displaystyle \int_{0}^{1}\int_{\sqrt{y}}^{1}2x^5y\cos{(xy^2)}dx dy$
...and I'm told to solve it by changing the order of integration. I get:
$\displaystyle \int_{0}^{1}\int_{0}^{x^2}2x^5y\cos{(xy^2)}dydx$
I know this forum's not here to check my work, but whether my new limits are correct or not, I still don't know how to integrate $\displaystyle 2x^5y\cos{(xy^2)}$ with respect to y, taking x as a constant.

Can anyone help?

I didn't check your change of order of inegration but use this sub to evaluate the integral

$\displaystyle u=xy^2 \implies du=2xydy$

and we get

$\displaystyle \int_{0}^{1}\int_{0}^{x^2}2x^5y\cos{(xy^2)}dydx$

$\displaystyle \int_{0}^{1}\int x^4\cos(u)dudx$
• May 5th 2009, 12:15 PM
redsoxfan325
Quote:

Originally Posted by chella182
Well, my problem isn't with the double integral bit, it's just with the integration once I've switched the limits et cetera. The original problem is:
$\displaystyle \int_{0}^{1}\int_{\sqrt{y}}^{1}2x^5y\cos{(xy^2)}dx dy$
...and I'm told to solve it by changing the order of integration. I get:
$\displaystyle \int_{0}^{1}\int_{0}^{x^2}2x^5y\cos{(xy^2)}dydx$
I know this forum's not here to check my work, but whether my new limits are correct or not, I still don't know how to integrate $\displaystyle 2x^5y\cos{(xy^2)}$ with respect to y, taking x as a constant.

Can anyone help?

Your reordering of the integration is correct. Proceeding from there, let $\displaystyle u=y^2$. Thus $\displaystyle du = 2y\,dy$

You now have $\displaystyle \int_0^1\int_0^{x^2}x^5\cos(xu)\,du = \frac{1}{x}\cdot x^5\sin(xu) = \left[x^4\sin(xy^2)\right]_0^{x^2} = x^4\sin(x^5)$

So we have $\displaystyle \int_0^1 x^4\sin(x^5)\,dx$

Let $\displaystyle u=x^5$ and $\displaystyle du=5x^4$. Now you have:

$\displaystyle \frac{1}{5}\int_0^1\sin(u)\,du = -\frac{1}{5}\cos(u) = \left[-\frac{1}{5}\cos(x^5)\right]_0^1 = \boxed{\frac{1}{5}-\frac{1}{5}\cos(1)}$
• May 5th 2009, 12:34 PM
chella182
I totally don't understand the second line of your working, sorry :( where does the $\displaystyle 2y$ disappear to? And the $\displaystyle x^4$ in the second integral? And I don't get where the $\displaystyle \frac{1}{x}$ has come from (Worried)
• May 5th 2009, 12:36 PM
Pinkk
$\displaystyle 2y$ is the derivative of $\displaystyle y^{2}$. All he did was a u-substitution.
• May 5th 2009, 12:37 PM
chella182
No, I know that, where does it disappear to though?
• May 5th 2009, 12:41 PM
Pinkk
Into the $\displaystyle du$. It never "disappears", it's just part of his new variables $\displaystyle u$ and $\displaystyle du$. And the $\displaystyle \frac{1}{x}$ is part of the integration of $\displaystyle cos(xy)$.

The antiderivative of $\displaystyle cos(xy)$ where $\displaystyle x$ is treated as a variable is $\displaystyle \frac{sin(xy)}{x}$.
• May 5th 2009, 12:42 PM
chella182
So because the $\displaystyle 2y$ appears in the $\displaystyle du$ it's not in the problem anymore? :S I'm well confused.

I get where the $\displaystyle \frac{1}{x}$ comes from now though, cheers.
• May 5th 2009, 12:44 PM
Pinkk
It's just u-substitution.

Just replace the $\displaystyle y^{2}$ with a $\displaystyle u$ and $\displaystyle 2ydy$ with a $\displaystyle du$. Don't let the fact that it's a double integral throw you off. It's just a straightforward u-substitution.
• May 5th 2009, 12:48 PM
chella182
I know I just... don't get it :S urgh. I'll go see my lecturer tomorrow if I get chance.
• May 5th 2009, 12:58 PM
Pinkk
Let's look at a single integral:

$\displaystyle \int 2ycos(y^{2})\,\,dy$

Let $\displaystyle u=y^{2}$ and $\displaystyle du=2ydy$.

Now just substitute:

$\displaystyle \int cos(u)\,\,du$

Now just integrate:

$\displaystyle \int cos(u)\,\,du = sin(u)$

Now just resubstitute for $\displaystyle u=y^{2}$:

$\displaystyle sin(y^{2})$.

Take the derivative to clarify:

$\displaystyle \frac{d}{dy}sin(y^{2})=cos(y^{2})\cdot \frac{d}{dy}y^{2} = 2ycos(y^{2})$

All you're doing is substituting one type of variable for another. The actual function isn't changed, just the way it's being represented.
• May 5th 2009, 01:03 PM
chella182
Ohhh right, I get it. All you needed to do to explain was say $\displaystyle dy=\frac{du}{2y}$ so the $\displaystyle 2y$'s cancel :p
• May 5th 2009, 01:05 PM
Pinkk
It hasn't disappeared. $\displaystyle du=2ydy$

I'm just representing $\displaystyle 2ydy$ as $\displaystyle du$ so I can have an easier integral to evaluate.

And yes, you can think of it that way if you want. Whatever clicks with you. :)
• May 5th 2009, 01:08 PM
chella182
It has cancelled out therefore it's disappeared in my mind. I do understand the whole thing, all I didn't get was why that disappeared 'cause no one was telling me where it'd gone. Although this does seem rather more complicated than examples we've done in lectures :S