# Repeated integral with dx and dy

• Feb 22nd 2011, 07:28 PM
mathcore
Repeated integral with dx and dy
evaluate the following repeated integral

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

• Feb 22nd 2011, 07:35 PM
Prove 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 PM
TKHunny
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 PM
mathcore
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 PM
Prove It

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

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 PM
mathcore
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 PM
mathcore
im still interested in an answer though i get 0 +cosy am i right
• Feb 22nd 2011, 09:37 PM
Prove 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 PM
mathcore
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 PM
mathcore
Quote:

Originally Posted by Prove 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?

yes if i saw that question i would know exactly what to do but i didnt realise you could take it in turn like that solve one by one i didnt know that was what u had to do
• Feb 22nd 2011, 09:40 PM
AllanCuz
Quote:

Originally Posted by mathcore
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...

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 PM
mathcore
Quote:

Originally Posted by AllanCuz
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.

i dont have a 'text' i am not in a college or school

yourr trying to kill me with that link
i use what ever resources i can find to learn i am handed these questions and told the websites to look for info
• Feb 22nd 2011, 09:46 PM
AllanCuz
Quote:

Originally Posted by mathcore
i dont have a 'text' i am not in a college or school
i use what ever resources i can find to learn i am handed these questions and told the websites to look for info

That's not why I posted a link to a turorial on multiple integration or anything.....
• Feb 23rd 2011, 04:09 AM
TKHunny
Quote:

Originally Posted by mathcore
i didnt realise you could take it in turn like that solve one by one i didnt know that was what u had to do

It is one of the great results of the calculus, that you CAN do it a piece at a time. Intuition may suggest otherwise for so complex an operation.