Integration, U substitution help

Dragonetti
Messages
15
Reaction score
0

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



\int2x2 square root of 1-x3 dx, u = 1-x3

Any pointers would be appreciated
Thanks
D
 
Physics news on Phys.org
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.
 
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:
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
 
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.
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
12
Views
2K
Replies
28
Views
2K
Replies
2
Views
1K
Replies
12
Views
2K
Replies
3
Views
2K
Replies
44
Views
5K
Replies
22
Views
3K
Back
Top