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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- February 22nd 2011, 08:28 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 - February 22nd 2011, 08:35 PMProve It
This is almost unreadable. Is this ?

- February 22nd 2011, 08:36 PMTKHunny
Just separate it. Here's the middle section.

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

Now what? - February 22nd 2011, 09:10 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 - February 22nd 2011, 09:48 PMProve It
To start with, evaluate the inner integral...

.

Here, since you are integrating w.r.t. , any function of is treated as a constant. - February 22nd 2011, 10:31 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... - February 22nd 2011, 10:36 PMmathcore
im still interested in an answer though i get 0 +cosy am i right

- February 22nd 2011, 10:37 PMProve It
Does it make more sense if it's written like this?

Can you see that there is an "inner" integral, which needs to be evaluated first, and an "outer" integral? - February 22nd 2011, 10:37 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

- February 22nd 2011, 10:38 PMmathcore
- February 22nd 2011, 10:40 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,

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. - February 22nd 2011, 10:42 PMmathcore
- February 22nd 2011, 10:46 PMAllanCuz
- February 23rd 2011, 05:09 AMTKHunny