# Integration, U substitution help

1. Jul 22, 2011

### Dragonetti

1. The problem statement, all variables and given/known data

Hi I am having a few problems with the below u substitution can anyone help,
In particular what to do with the integral of the u substitution?

2. Relevant equations

$\int$2x2 square root of 1-x3 dx, u = 1-x3

Any pointers would be appreciated
Thanks
D

2. Jul 22, 2011

### Stengah

Once you have u just take the derivative of it and substitute it back in for dx. The u is right and I can see that it will work out just fine. Try reviewing an example from your notes or the textbook.

3. Jul 23, 2011

### nickalh

Can you show us the work you've done so far?
The next step is to find du/dx.
Create something in the integral which we can replace (...)dx and put in du.
Can you show what you get for du/dx?

Please read the section on https://www.physicsforums.com/showthread.php?t=414380"

Last edited by a moderator: Apr 26, 2017
4. Jul 23, 2011

### Dragonetti

Hi Thanks for the help much appriciated,

I think I have worked it out, answer below;

-4/9(1-x3)3/2 du?

I have differentiated it and I get back to original answer.

Just one thing, if my original limits were 2,1 would my new limits be -7,0?

Thanks again
Dominic

5. Jul 23, 2011

### Stengah

You got it except you don't need the du at the end. And actually you need your +C as well. But I don't know what you mean with your question regarding limits.

6. Jul 23, 2011

### nickalh

Yes, that's right.
Check your answer here
A word of caution about that site- use it only to check your work, or help you through a particularly difficult integral. It's too easy to put it in there first & think one is learning.

When to drop the du?
When one actually does the integral operation. Both the integral sign & du disappear on the same step, when one does the integral itself, after the prep. work(substituting u, constant multipliers, etc.) but before evaluating a definite integral.

About +C, stengah is half right. Without limits of integration it needs a +C. Of course with limits of integration drop the +C.

When you do a definite integral with limits of integration, 2 & 1, the new limits of integration would be -7 & 0. The convention is to mention the bottom limit first, because if you see it graphically, the bottom number is on the left.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook