Is U Substitution the Key to Solving Tricky Integration Problems?

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Homework Help Overview

The discussion revolves around the application of u-substitution in integration problems, specifically focusing on the transformation of variables and the manipulation of integrals involving square roots and polynomial expressions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between variables during u-substitution, questioning the correctness of transformations and the resulting expressions. There are discussions about the differential dx and its relation to du, as well as the implications of different substitutions on the integral's form.

Discussion Status

The conversation is active, with participants sharing their attempts at solving the integral and clarifying their understanding of the substitution process. Some guidance has been provided regarding the calculation of differentials, but there is no explicit consensus on the final form of the integral or the correctness of all transformations discussed.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available for discussion. There is an emphasis on ensuring clarity in the steps taken during the substitution process.

robertjford80
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Homework Statement




Screenshot2012-05-23at81731PMcopy.png


Screenshot2012-05-23at81731PM.png



The Attempt at a Solution



if x2 = u - 1, and if x3 = x2 * x, then x3 should equal (u-1)x, not .5(u-1).

I'm assuming that they got u.5 because (x2+1).5 = (u-1+1).5 which is the same as u^.5
 
Last edited:
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Wait, what did you get?

I ended up getting (in terms of u):

$$\int\frac{1}{2}\sqrt{u}(u-1) du$$
 
Last edited:
if [itex]u=x^{2}+1[/itex], then what does [itex]dx[/itex] equal?
 
Ok, I see that .5du but I still believe that

if x2 = u - 1, and if x3 = x2 * x, then x3 should equal (u-1)x

so the new answer should be (u-1)x/2 u^.5
 
[itex]dx≠\frac{1}{2}du[/itex]

Show how you solve for [itex]dx[/itex].

EDIT: No, I'm sorry. Solve for [itex]du[/itex] if [itex]u=x^{2}+1[/itex].
 
du = 2xdx

du/2 = xdx
 
Yes. So then how can you write [itex]x^{3}dx=x^{2}*xdx[/itex]?
 
Ok, I get it now, sort of
 
Don't hold back if you still need help. But just make sure your questions are specific.
 
  • #10
I think I get it, we'll see if I can apply this technique to future problems, but for now I'm moving on. Thank you for your concern and helping me out.
 

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