# Homework Help: Need some help with some various integrals (studying for finals)

1. Nov 13, 2014

### leo255

< Mentor Note -- thread moved from General Math to the Homework Help forums >Hi all,

Calc II finals is 4-5 weeks away...We're on Taylor Series right now, but I wanted to get started early on studying for the final. I have a few questions that are confusing me that I took from a final exam I saw online (not my college):

http://www.dawsoncollege.qc.ca/publ...ciplines/math/exams/201-203-dw-winter2012.pdf

Problem 1:

Find f(x) given f prime (x) = ( 2x^(2/3) - 3x ) / x , and f(8) = 4.

I got 3x^(2/3) - 3x

I think the integral is correct, but am not sure what to do with the f(8) = 4. What does it mean in relation to this problem?

Problem 2:

Integral of (2x+3) * (sin(x/2))

-Here, I used Int. by parts --> u = 2x+3, du = 2dx, dV = sin(x/2), V = -2cos(x/2)

I ended up with -4xcos(x/2) - 6cos(x/2) + 8sin(x/2) + C <--This is incorrect/partially correct, as per the answers (on the bottom of that page [I can also post them here]).

Problem 2B:

The Integral of (15+4x-x^2) / (x-1)(x^2 + 5)
<---This seems/seemed like a partial fractions problem. I have:

A(x^2 + 5) + B(x-1) = 15+4x - x^2

Let x = 1, 6A = 18 --> A = 3.
I am stuck here. I can't make x into anything to get rid of the 5. What am I missing here? System of Equations? Completing the square? Would appreciate the guidance.

These are what I'm working on. This class has been hard for me, so I'm just trying to work hard and chug through problems.

Thanks.

Last edited by a moderator: Nov 13, 2014
2. Nov 13, 2014

### Mentallic

Your integral is almost correct, you're just missing the constant +c at the end. So now that you have

$$f(x)=3x^{2/3}-3x+c$$

You can now apply f(8)=4 to find the value of the constant.

Why is it incorrect? It looks good to me. The only thing I would change would be to factorize the two cos(x/2) terms.

When you use partial fractions, the reason you choose constants such as A and B is because it's implied that the denominator has linear factors. What you're actually doing is choosing a polynomial of degree one less than the factor in your denominator. So the way you should decompose the above integral is into

$$\frac{15+4x-x^2}{(x-1)(x^2+5)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+5}$$

We chose Bx+C because this is a general linear equation (one degree less than the quadratic in the denominator).

Now, for your other problem in not knowing what to plug in:

If you've studied complex numbers at all, you are allowed to choose x such that $x^2+5=0$ which would be $x=\pm\sqrt{5}i$. But if you haven't or aren't comfortable with complex numbers, you're allowed to choose any value of x which will turn it into an equation with A,B,C unknown, but since you already found the value of A easily, you'll be left with two equations in the other 2 unknowns B and C, which you can of course solve simultaneously.
So choose something simple like x=0, or whatever else you feel will make the numbers easy to work with. You wouldn't choose x=100 for example, but any x will give the same result for the unknowns if you've done everything correctly up to that point.

Good luck!

3. Nov 13, 2014

### leo255

Thanks for the help! Yeah, I see that for the integration by parts question that I did, they just left it as (2x+3)(-2cos(x/x)), so that's probably where I was confused.

As far as the partial fractions problem:

putting together a system of equations, I'm getting:

A + B = -1
C - B = 4
5A - C = 15

I did this after plugging x = 0 into the equation to get rid of Bx, and got that C = 0, and later got that B = -4. As you can see, the system of equations above will still yield the same results.

The answer for the problem is 3ln(x-1) - 2ln(x^2 + 5) + C --> Obviously A is correct at 3, but I'm pretty sure I got B and C wrong.

4. Nov 13, 2014

### Mentallic

You can always check for yourself to confirm if you've gotten it right or not. Do your values satisfy the result? i.e. does

$$15+4x-x^2\equiv A(x^2+5)+(Bx+C)(x-1)$$

given your values of A,B,C? I used the equivalent symbol $\equiv$ in this case to denote that it is true for all x.