Homework Help: A very simple U-sub is wrecking my brain

1. Jul 19, 2009

jinksys

1. The problem statement, all variables and given/known data

Integral Sqrt[16-x^2]*x dx from 0 to 2.

2. Relevant equations

3. The attempt at a solution

First thing I do is a u-sub, so

u=16-x^2

du=-2xdx

I also my bounds change to upper=12 lower=16.

After I integrate and swap out my U, I get

-1/3*(16-x^2)^(3/2) from 16 to 12.

Am I using the wrong method?

2. Jul 19, 2009

Razzor7

This is probably a stupid question, but why did you change the limits of integration?

3. Jul 19, 2009

Nabeshin

Umm, this is actually a trig-substitution problem (which I guess is a class of u-substitution).

Try thinking of it in those terms.

4. Jul 19, 2009

Pjennings

Have you thought of trying a trig substitution?

5. Jul 19, 2009

Razzor7

You sure it's necessary? I get an answer that matches my calculator without incorporating any trig.

EDIT: Oh.

6. Jul 19, 2009

Pjennings

You could do the problem using a u substitution, it's just easier to use the trig substitution.

7. Jul 19, 2009

Nabeshin

I actually missed the *x in the problem the first glance through, but I still would go with trig substitution. Necessary, maybe not.

8. Jul 19, 2009

jinksys

9. Jul 19, 2009

Nabeshin

You adjust if you keep the equation in terms of u, but if you substitute back in for x after integration, you retain the original x boundaries. Seems you substituted back into x and changed the boundaries, which doesn't work.

10. Jul 19, 2009

jinksys

This clears up a lot. Thanks. I got the correct answer now!

11. Jul 19, 2009

HallsofIvy

That is one reason why it is a good idea to write the variable in the limits of integration:
$$\int_{x= 0}^2 \sqrt{16- x^2} xdx$$
after the substitution $u= 16- x^2$ becomes
$$-\frac{1}{2}\int_{u=16}^{12}u^{1/2}du$$