Integration, U substitution help

[Dragonetti]

Homework Statement

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?

Homework Equations

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

Any pointers would be appreciated
Thanks
D

 

[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.

 

[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"

 

[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

 

[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.

 

[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.