Integration Help: Leibniz's Rule Problem

[elle]

Integration help please!

Hi, I was wondering if anyone can help me with the following integration problem.

I've provided part of my solution so far but its not the same as the answer given.

Question has been stated at the top of my solution and I am supposed to use Leibniz's rule to show that the I'(y) = 4y ln (2y).

I can't seem to get the answer so if anyone would be kind enough to point out my mistake, I'd be very grateful! thanks very much for your time

*Also in my solution a part of my first line of working has been cut off. There is supposed to be a limit of x = y^2 next to the ln (x + y^2) | [ if you know wot i mean ]

http://i26.photobucket.com/albums/c109/mathsnerd/911d4376.jpg"

 

[BerkMath]

Remeber that ln(a)-ln(b)=ln(a/b), as well as the other properties of the natural log function.

 

[elle]

Thanks i got it nows!

Oh I also got another question to ask. I've not done integration since last year so my mind is totally blank Anyways, I've again provided part of my solution but I'm stuck at the next integration...I don't know what method to use! It's probably really easy but as I said my mind's a blank..sigh. I tried it and my final answer was nothing near the solution given which is 48. Help very much appreciated!

* Question requires the use of double integrals

http://i26.photobucket.com/albums/c109/mathsnerd/72cd4951.jpg"

 

[BerkMath]

Your first integration is wrong. int(sqrt(xy-y^2),x)=2/3*(x-y)*sqrt(xy-y^2). Try it from there and see what you get now.

 

[elle]

Your first integration is wrong. int(sqrt(xy-y^2),x)=2/3*(x-y)*sqrt(xy-y^2). Try it from there and see what you get now.

It's wrong? Ack no wonder How did you get 2/3*(x-y)*sqrt(xy-y^2)? Can you show me please? I don't understand how you got that answer...

 

[BerkMath]

You are letting the y bother you. First do this to make it more clear:
sqrt(xy-y^2)=sqrt(y(x-y))=sqrt(y)sqrt(x-y), so int(sqrt(xy-y^2),x)=int(sqrt(y)sqrt(x-y),x), now take out sqrt(y) sinc we may view it as a constant. We have sqrt(y)*int(sqrt(x-y),x). This is just a standard substitution problem. let u=x-y, then du=dx, then y^(-1/2)*int((x-y)^(-1/2),x)=y^(-1/2)*int(u^(-1/2),u)=y^(-1/2)*2/3*u^(3/2)=y^(-1/2)*2/3*(x-y)^(3/2)=2/3*(x-y)sqrt(xy-y^2), where in the last equality we made use of the fact u^(3/2)=u*u^(1/2).