# Definite Integration by Substitution

1. Sep 28, 2011

### theRukus

1. The problem statement, all variables and given/known data
$\int^2_1 6x\sqrt{x-1}dx$

2. Relevant equations

3. 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: Sep 28, 2011
2. Sep 28, 2011

### dynamicsolo

Umm, could it be that you forgot to find the antiderivative function...?

3. Sep 28, 2011

### ArcanaNoir

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?

4. Sep 28, 2011

### dynamicsolo

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!

5. Sep 28, 2011

### ArcanaNoir

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?)

6. Sep 28, 2011

### dynamicsolo

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.