Definite Integration by Substitution

Click For Summary

Homework Help Overview

The problem involves evaluating the definite integral \(\int^2_1 6x\sqrt{x-1}dx\) using substitution. The original poster attempts to apply the substitution \(u = x - 1\) and transform the integral accordingly.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the necessity of finding the antiderivative after substitution and the importance of reverting back to the original variable before evaluating the definite integral. There is also a focus on the legitimacy of changing the bounds during substitution.

Discussion Status

Several participants have provided guidance on the steps that may have been overlooked, such as the need to compute the antiderivative and the correct handling of bounds. There is an ongoing exploration of different methods for managing bounds during substitution.

Contextual Notes

Some participants express differing educational backgrounds regarding the handling of bounds in substitution, indicating a variety of approaches to the problem. There is a recognition of common pitfalls in integration, particularly the transformation of limits.

theRukus
Messages
49
Reaction score
0

Homework Statement


[itex]\int^2_1 6x\sqrt{x-1}dx[/itex]

Homework Equations


The Attempt at a Solution


Let [itex]u=x-1[/itex].
Then, [itex]u+1=x[/itex],
and [itex]du=dx[/itex].

Continued from problem statement,
[itex]=6 \int^1_0 (u+1)u^{\frac{1}{2}}du[/itex]
[itex]=6 \int^1_0 u^{\frac{3}{2}} + u^{\frac{1}{2}}du[/itex]
[itex]=6(1^{\frac{3}{2}} + 1^{\frac{1}{2}})[/itex]
[itex]=6(2)[/itex]
[itex]=12[/itex]

My Web submission program (for University) has told me this is wrong.

What am I doing wrong?

Thank you for any help!
 
Last edited:
Physics news on Phys.org
Umm, could it be that you forgot to find the antiderivative function...?
 
you have to put x-1 back in for u after you integrate and before you solve it for the definite numbers. Also, why did you change the bounds from 1 to 2, to 0 to 1?
 
ArcanaNoir said:
you have to put x-1 back in for u after you integrate and before you solve it for the definite numbers. Also, why did you change the bounds from 1 to 2, to 0 to 1?

theRukus is doing the definite integral under a u-substitution, so the limits must be changed from x = 1 to x = 2 into u = 0 to u = 1 . That is perfectly legitimate: the problem is that the integration hasn't actually been carried out; tR just put 1 ( and 0 ) into the integrand!
 
dynamicsolo said:
theRukus is doing the definite integral under a u-substitution, so the limits must be changed from x = 1 to x = 2 into u = 0 to u = 1 . That is perfectly legitimate:

interesting! I was never taught that, so I always just used a and b temporarily until I put the original variable back in, and with it the original bounds.

Yes, integrating an integral is a good idea. (why is it so hard to see the obvious sometimes?)
 
ArcanaNoir said:
interesting! I was never taught that, so I always just used a and b temporarily until I put the original variable back in, and with it the original bounds.

Working with the transformed definite integral under transformed limits of integration* can save a bit of writing. You are basically still doing the calculations you would have done to evaluate the definite integral in x anyway, but you don't have to take the time writing out the antiderivative in x to set up the evaluation...

*Common integration error: forgetting to transform those limits... so don't forget.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
3
Views
2K