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Calculus 2 Homework Problem Bugging Me

C

CallTheFBI

Banned
Ok, I have been working on this homework problem for awhile now. It is right here. Take a look at it first before you read on. The first two lines are the actual problem and underneath are my feeble attempts to solve it. Now in the back of the book where the answers are it just says: let u = x ^ m. Which is what I did as you can see and things aren't looking too pretty. The (e^x)/(x) integral cannot be solved by me or my TI-89 calculator and that is why I left it in that form. If you can tell me what I am doing wrong that would be great.
 
just after glancing at it...try doing intergration by parts again on the (e^x)/x...i don't know if that will help but it might. im tired
 
Originally posted by: hdeck
just after glancing at it...try doing intergration by parts again on the (e^x)/x...i don't know if that will help but it might. im tired
Click to expand...

Tried that, just got another nasty integral of (e^x)(ln x) that neither my TI-89 nor I can solve.

 
you are doing the intergration by parts incorrectly.

integrate(udv) = uv - integrate(vdu)

u = x^m
dv = e^x

using that, you get:

du = mx^(m-1)dx
v = e^x

uv = (x^m)(e^x)
integrate(vdu) = (e^x)(mx^(m-1))dx

i love integration by parts! one of my professors told me that the phrase "udv = uv - vdu" is universally known. haha
 
Shaka totally got it right. You seem to have screwed up integration by parts somewhere. Don't worry, its easy to do. I'm happy to have finally seen a math problem on AT that I could do though.

R
 
OMG, this problem was 10x easier than I thought. Was looking way too deeply into the problem. All they want is for me to use the integration by parts formula to show why it equals that. I solved that problem easy but there is one that says use this formula to solve this integral. LONG problem involving a ton of algebra, finally finished though. I'll post the solution once I get it entered into my equation editor.
 
Originally posted by: RossGr
I am not sure why you have

dv = (e^x)/x

it should simply be dv = e^x dx so v = e^x
Click to expand...

Notice how u = x^m so there is an x^(-1) left over for the dv part. It is: (x^m)(x^(-1))(e^x). See why now? Don't worry about it though, I was approaching the problem incorrectly, making it 10x harder on myself.
 
u = x^m


dv = e^x dx

du = m x^ (m-1) dx

v= e^x

Int (udv) = uv - Int v du

= x^m e^x - Int (e^x m x ^(m-1) dx
= x^m e^x - m Int( e^x x^(m-1) dx

There is no x^-1 floating around that I can see. You must take the u and dv from the original problem statement.
 
Originally posted by: RossGr
u = x^m


dv = e^x dx

du = m x^ (m-1) dx

v= e^x

Int (udv) = uv - Int v du

= x^m e^x - Int (e^x m x ^(m-1) dx
= x^m e^x - m Int( e^x x^(m-1) dx

There is no x^-1 floating around that I can see. You must take the u and dv from the original problem statement.
Click to expand...

I know but silly me was trying to solve the integral on the right hand side of the equal sign. That's why.

 
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