Integrating the Square Root: $\displaystyle \Large \int udv$

[Slimsta]

Homework Statement

http://img37.imageshack.us/img37/1237/63391287.jpg

Homework Equations

$$\displaystyle \Large \int udv$$ = uv - $$\displaystyle \Large \int vdu$$

The Attempt at a Solution

can i take this 3 out of the integral as well and make it 3/8 *$$\displaystyle \Large \int _0^8 sqrt(64 + 24x^2)dx$$?

then i do this:
let u = sqrt(64 + 24x²)
du = $$\displaystyle \Large \int _0^8 sqrt(64 + 24x^2)dx$$
let v = x
dv = dx

is this right so far?

 

[Mark44]

Homework Statement

http://img37.imageshack.us/img37/1237/63391287.jpg

Homework Equations

$$\displaystyle \Large \int udv$$ = uv - $$\displaystyle \Large \int vdu$$

The Attempt at a Solution

can i take this 3 out of the integral as well and make it 3/8 *$$\displaystyle \Large \int _0^8 sqrt(64 + 24x^2)dx$$?

Yes, you can always pull a constant out of the integrand.

then i do this:
let u = sqrt(64 + 24x²)
du = $$\displaystyle \Large \int _0^8 sqrt(64 + 24x^2)dx$$
let v = x
dv = dx

is this right so far?

No, not right. You went backwards by integrating instead of differentiating. If u = sqrt(64 + 24x²), then du = (1/2)(64 + 24x²)^-(1/2)(48x). This can be cleaned up, but I left it as I did so you can see the chain rule at work.

 

[Slimsta]

okay so i have made some changes. i think it would be easier if i do this:
$$3/8 \displaystyle \Large \int _0^8 x*sqrt(64 + 24x^2)dx$$
f=x
g'=sqrt(64 + 24x^2)
f'=1
g=$$\displaystyle \Large \int sqrt(64 + 24x^2)dx$$

to find g,
let u =sqrt(64 + 24x²)
$$\displaystyle \Large \int u dx$$
du = 1/2 (64 + 24x²)^-1/2 * 48x dx

now I am lost..
i know that i have to use the forumla
$$\displaystyle \Large \int f(x)g'(x) dx = f(x)g(x) - \int g(x)f'(x) dx$$

but how do i get g? its too complicated..

 

[Dick]

You don't need any integration by parts. Substitute u=64+24x^2. du=48*x*dx. So replace x*dx in your original integrand with du/48 and the sqrt with sqrt(u). Now there's no more x's. It's just a u-substitution.

 

[Mark44]

You want to evaluate this integral, right?
$$3/8 \int_0^8 x \sqrt{64 + 24x^2}dx$$

This is actually a pretty easy one. Use an ordinary substitution, with u = 64 + 24x². So du = ?

 

[Slimsta]

oh wow... this is pissing me off how easy that was and i did it the long and hard way. i guess the chapter just wants us to do it the hard way >:(

yeah i got the answer is 333.3333

thanks a lot for all the help! :)