Repeated integral with dx and dy

**mathcore** · February 22nd 2011, 07:28 PM

evaluate the following repeated integral

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

i have never seen a question like this please help

**Prove It** · February 22nd 2011, 07:35 PM

This is almost unreadable. Is this $\displaystyle \int_0^{\frac{\pi}{2}}{\int_y^{\pi}{\cos{y}\sin{x} \,dx}\,dy}$ ?

**TKHunny** · February 22nd 2011, 07:36 PM

Just separate it. Here's the middle section.

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

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

Now what?

**mathcore** · February 22nd 2011, 08:10 PM

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

**Prove It** · February 22nd 2011, 08:48 PM

To start with, evaluate the inner integral...

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

Here, since you are integrating w.r.t. $\displaystyle x$ , any function of $\displaystyle y$ is treated as a constant.

**mathcore** · February 22nd 2011, 09:31 PM

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...

**mathcore** · February 22nd 2011, 09:36 PM

im still interested in an answer though i get 0 +cosy am i right

**Prove It** · February 22nd 2011, 09:37 PM

Does it make more sense if it's written like this?

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

Can you see that there is an "inner" integral, which needs to be evaluated first, and an "outer" integral?

**mathcore** · February 22nd 2011, 09:37 PM

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

**mathcore** · February 22nd 2011, 09:38 PM

Originally Posted by Prove It

Does it make more sense if it's written like this?

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

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

**AllanCuz** · February 22nd 2011, 09:40 PM

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,

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

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.

**mathcore** · February 22nd 2011, 09:42 PM

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,

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

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

**AllanCuz** · February 22nd 2011, 09:46 PM

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.....

**TKHunny** · February 23rd 2011, 04:09 AM

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.

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