evaluate the following repeated integral

int(0-pi/2) ( int(y-pi/1) cosysinxdxdy)

i have never seen a question like this please help

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- Feb 22nd 2011, 07:28 PMmathcoreRepeated integral with dx and dy
evaluate the following repeated integral

int(0-pi/2) ( int(y-pi/1) cosysinxdxdy)

i have never seen a question like this please help - Feb 22nd 2011, 07:35 PMProve It
This is almost unreadable. Is this $\displaystyle \displaystyle \int_0^{\frac{\pi}{2}}{\int_y^{\pi}{\cos{y}\sin{x} \,dx}\,dy}$?

- Feb 22nd 2011, 07:36 PMTKHunny
Just separate it. Here's the middle section.

$\displaystyle \int_{y}^{\pi}\cos(y)\sin(x)dx = \cos(y)\cdot\int_{y}^{\pi}\sin(x)dx$

cos(y) is essentially a constant under and integral in dx.

Now what? - Feb 22nd 2011, 08:10 PMmathcore
I don't know how to seperate it.

prove it, yes that is the equation

sorry my bad typo it is pi/2 in both cases - Feb 22nd 2011, 08:48 PMProve It
To start with, evaluate the inner integral...

$\displaystyle \displaystyle \int_y^{\frac{\pi}{2}}{\cos{y}\sin{x}\,dx}$.

Here, since you are integrating w.r.t. $\displaystyle \displaystyle x$, any function of $\displaystyle \displaystyle y$ is treated as a constant. - Feb 22nd 2011, 09:31 PMmathcore
ok but what do you mean by INNER integral?

i assume from what you've said you are expected to integrate wrt x then integrate the RESULT wrt y? its not the actual differentiation and integration i can't do its understanding what the question is even asking

to be honest i don't even need to solve the answer its all by the by really, i just want to know whats its asking... - Feb 22nd 2011, 09:36 PMmathcore
im still interested in an answer though i get 0 +cosy am i right

- Feb 22nd 2011, 09:37 PMProve It
Does it make more sense if it's written like this?

$\displaystyle \displaystyle \int_{0}^{\frac{\pi}{2}}{\left[\int_0^{\frac{\pi}{2}}{\cos{y}\sin{x}\,dx}\right]\,dy}$

Can you see that there is an "inner" integral, which needs to be evaluated first, and an "outer" integral? - Feb 22nd 2011, 09:37 PMmathcore
do u work out sin(pi/2) - sin(y) first then integrate or integrate then subtract? i know the top and bottom integral is add and subtract

- Feb 22nd 2011, 09:38 PMmathcore
- Feb 22nd 2011, 09:40 PMAllanCuz
You should review the chapter in your text that deals with multiple integration. Alternatively you can take a gander at the stickied thread at the top of this forum! http://www.mathhelpforum.com/math-he...on-146568.html

Essentially, if you have a series of integrals you have to evaluate them in the order of most functions in the bounds to least. In other words, if you have an integral of the form,

$\displaystyle \int_a^b \int_{f(x)}^{g(x)} H(x,y)dydx $

You need to evaluate the integral with the bounds dealing with the functions of x first, then evaluate the integral with the constant bounds. Doing it the other way around won't produce a numerical result. - Feb 22nd 2011, 09:42 PMmathcore
- Feb 22nd 2011, 09:46 PMAllanCuz
- Feb 23rd 2011, 04:09 AMTKHunny