A very simple U-sub is wrecking my brain

[jinksys]

Homework Statement

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

Homework Equations

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?

 

[Razzor7]

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

 

[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.

 

[Pjennings]

Have you thought of trying a trig substitution?

 

[Razzor7]

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

Try thinking of it in those terms.

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

EDIT: Oh.

 

[Pjennings]

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

 

[Nabeshin]

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

 

[jinksys]

Are my bounds incorrect? I thought you had to adjust your bounds when using a usub.

 

[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.

 

[jinksys]

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.

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

 

[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