Definite Integration by Substitution

Click For Summary
The discussion focuses on the process of definite integration by substitution, specifically addressing an integral problem. The user initially attempted the integration but received feedback indicating their solution was incorrect. Key points include the necessity of changing the limits of integration when performing u-substitution, which the user did correctly, but they failed to compute the antiderivative properly. Other participants emphasize the importance of completing the integration step before substituting back to the original variable. The conversation highlights common mistakes in definite integrals and the value of transforming limits accurately.
theRukus
Messages
49
Reaction score
0

Homework Statement


\int^2_1 6x\sqrt{x-1}dx

Homework Equations


The Attempt at a Solution


Let u=x-1.
Then, u+1=x,
and du=dx.

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

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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
6K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
3
Views
2K