integrating a fraction of u

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March 3rd 2009, 09:21 PM
ckylek

integrating a fraction of u

Im having an issue with:

(u^(3/2))/2

i know that once i evaluate it i am supposed to add +1 to r (which is 3/2 in this case) and that would give me u^(5/2) but I thought i was also supposed to multiply 5/2 times the denominator which would give me 10/2 or 5.......

However I am confused because the answer is supposed to be:

(1/10)(u)^5/2 and when i do it my way i get 5(u)^5/2

What am I doing wrong? Any help would be great, thanks!
March 3rd 2009, 09:32 PM
Jameson

Is the original expression $\frac{u^{\frac{3}{2}}}{2}$ ?

You are right you add 1, or 2/2, to the exponent and divide by the new exponent. If you do that you get $\frac{u^{\frac{5}{2}}}{2*\frac{5}{2}}$

So I find that both you and your other answer are wrong. (Thinking)
March 3rd 2009, 09:41 PM
ckylek

Quote:

Originally Posted by Jameson

Is the original expression $\frac{u^{\frac{3}{2}}}{2}$ ?

You are right you add 1, or 2/2, to the exponent and divide by the new exponent. If you do that you get $\frac{u^{\frac{5}{2}}}{2*\frac{5}{2}}$

So I find that both you and your other answer are wrong. (Thinking)

Haha your right i meant from what I have up there i get (1/5)(u)^5/2 but the solutions is supposed to be (1/10)(u)^5/2, so im guessing i messed up somewhere else in the problem because this is only part of it the original problem which was:

Find the following indefinite integral:

(x)(square root of 2x-1)dx

sorry i dont know how to make it look prettier
answer is supposed to be:
(1/10)(2x-1)^(5/2) + (1/6)(2x-1)^(3/2) +c
March 3rd 2009, 09:50 PM
Jameson

If you mean $\int x\sqrt{2x-1}dx$ , then I again find your solution wrong. At least Mathmatica says so.
March 3rd 2009, 09:53 PM
ckylek

Quote:

Originally Posted by Jameson

If you mean $\int x\sqrt{2x-1}dx$ , then I again find your solution wrong. At least Mathmatica says so.

Yes that was the equation, was I correct in putting 1/5 instead of 1/10 and i got 1/3 instead of 1/6?
March 3rd 2009, 09:59 PM
Jameson

Quote:

Originally Posted by ckylek

Yes that was the equation, was I correct in putting 1/5 instead of 1/10 and i got 1/3 instead of 1/6?

I think I know why you're off from the solution by 1/2. When you let u=2x-1, you must notice that du=2dx, thus dx = (1/2)du. So if you forgot to factor in this constant from your change in variables, you would be off by 1/2.
March 3rd 2009, 10:06 PM
ckylek

Quote:

Originally Posted by Jameson

I think I know why you're off from the solution by 1/2. When you let u=2x-1, you must notice that du=2dx, thus dx = (1/2)du. So if you forgot to factor in this constant from your change in variables, you would be off by 1/2.

yes sir that was it, i was suspecting a hidden 2 or something, thanks again for all your help. I really appreciate it.