Integral more complicated than it should be, fear I got homework question wrong

[midwestudent]

Homework Statement

Given the definite integral
$$\int_{0}^{2}{x^2 \over \sqrt{x^2 + 16}}$$

find the average using the mean value theorum for integrals

Homework Equations

Here's the tricky part: Although I employed it as a last resort (see later), we haven't been taught integration by parts yet. We also aren't at antiderivatives for inverse trig functions, or any transcendtal functions.

The hammer which you get to swing at this problem that's supposed to be a nail is u substitution,
$$\int{f(g(x))g'(x)} == \int{u du}$$

We also know about symmetry, reducing even functions to half the domain and odd functions to zero when their intervals cancel out, and periodicity.

The Attempt at a Solution

I first looked at the far easier problem,
$$\int{x \over \sqrt{x^2 + 16}}$$
which can be solved easily with
$${1 \over 2} \int{2x \over \sqrt(x^2 + 16}}$$
since that reduces to
$${1 \over 2} \int{u^{-1 \over 2} du}$$

So I thought it would be a good idea to split up that x^2 into x * x, to get
$${1 \over 2} \int{x u^{-1 \over 2} du}$$
But I don't know how to break it up past that point without using integration by parts.
Integration by parts also involves introducing so many complicated steps that I'm positive I made a mistake somewhere, so I'll post everything I have and please correct at will.

$$u = x^2 + 16, du = 2x dx$$
$$dv = x du == x(2x) dx== 2x^2 dx, v = \int_{0}^{2}{2x^2 dx} == ({2 \over 3}) (2^3) == {16 \over 3}$$
$$w = u^{-1 \over 2}, dw = {2x \over 2 \sqrt{x^2 + 16}} dx == {x \over \sqrt{x^2 + 16}} dx$$
$$({1 \over 2}) ({1 \over 2}) \int_{0}^{2}{x u^{-1 \over 2} du}$$
$${1 \over 4} \int_{0}^{2}{w dv}$$
$$({16 \over 3}) ({1 \over \sqrt{20}}) - {1 \over 4} \int_{0}^{2}{v dw}$$
$${16 \over 3 \sqrt{20}} - {1 \over 4}\int_{0}^{2}{{16 \over 3} {x \over \sqrt{x^2 + 16}}dx}$$
At this point, I remembered that 2x is the derivative of u, and there's 2 twos in 16/3 with 4/3 left over, and if you multiply that 4/3 by 3/2 you get another 2x, soooo (I told you this all seemed too complicated for a homework problem)
$${1 \over 4}\int_{0}^{2}{2 dw} + {1 \over 4}\int_{0}^{2}{2 dw} + ({1 \over 4})({2 \over 3})\int_{0}^{2}{2 dw}$$
$$== {2 \over 3}\int_{0}^{2}{2 dw} == {2 \over 3}\int_{0}^{2}{u^{-1 \over 2} du}$$
$${16 \over \sqrt{20}} - {2 \over 3}\int_{0}^{2}{u^{-1 \over 2} du}$$
$${16 \over \sqrt{20}} - ({2 \over 3})({1 \over 2}) \sqrt{2^2 + 16}$$

which is a math problem involving subtracting one number with an irrational numerator from one number with an irrational denominator. Which, to me, is a red flag saying I did something wrong.

So I'm looking for
A.)Corrections to any mistakes I may have made which will help me anyway, and
B.)The "right" way to do this problem, without integration by parts

One thing (I'm not great with integration by parts yet), I'm not really sure I should have pulled v through and subbed in 2 and 0 when I did, or if I should have waited and just left it as a variable.

 

[hwmaltby]

Try using $$x=4\tan{\theta}$$ :).

 

[foxjwill]

Try using $$x=4/tan{/theta}$$ :).

Commands in TeX start with a backslash ("\"), not a forward slash ("/").

 

[Mark44]

And need to be inside a [ tex] or [ itex] tags, like this (but without the space after the [ character).
[ itex]x = 4 tan(\theta) [ /itex]

 

[hwmaltby]

Oh, yes, I wrote it in tex tags, but it disappeared for some reason =(. I was using mobile physicsforums, so I messed up. It's edited now...

 

[midwestudent]

Thanks for the help :)

While we're on the subject of latex, I noticed neither

\\

nor

\linebreak

works. Is the only way to make a line break closing the tex tag and then hitting enter?

 

[hwmaltby]

You're welcome =).

I generally make new tex tags, but I believe there is a way using \vspace[]. However, you should probably ask someone more experienced than I =).