# Calculus Integrals

• Dec 6th 2006, 10:59 AM
Sucker Punch
Calculus Integrals
I have a final exam in an hour, I get understand everything but integrals. I don't have a bloody clue.

A sample question is;

Find the following integral;

$\displaystyle S (x^3 + 5x - 8)/(9x^5) d/dx$

I'm just completely lost. I must have missed the class(es) when we went over this, nothing in my notes, and I can't find anything in my textbook.

I know an hour is a little short to learn this, but if someone could go over the general idea for me, I'm hoping I can steal a mark or two on those questions.
• Dec 6th 2006, 11:55 AM
AfterShock
What in the world is the "S"? I'm assuming you're trying to do an integral sign?

So you want to find the integral of (x^3 + 5*x - 8)/(9*x^5)dx

Do you agree we can expand the above to:

1/(9*x^2) + 5/(9*x^4) - 8/(9*x^5)

Then:

Take the integral of each of the terms.

Thus,

-1/(9*x) + [-5/(27*x^3)] + [2/(9*x^4)] + C, where C is some constant.
• Dec 6th 2006, 12:38 PM
Sucker Punch
Yeah, its supposed to be that extended S integral sign thingy.

But I think I'm screwed.

I sorta get how you expanded it, but I don't know where the variable X went. And I don't even know what taking the integral means.

Oh well, hopefully I'll do well in the other sections to make up for it.
• Dec 6th 2006, 12:41 PM
ThePerfectHacker
Quote:

Originally Posted by Sucker Punch
And I don't even know what taking the integral means.

It means finding the "anti-derivative".
That is a function whose derivative is equal to the function.

For example,
$\displaystyle \int 2x dx=x^2$
Because,
$\displaystyle (x^2)'=2x$

In fact,
$\displaystyle \int 2xdx = x^2+1$
Because the constant term vanishes when you take the derivative.

Thus, all anti-derivates must be,
$\displaystyle \int 2xdx=x^2+C$
Where $\displaystyle C$ is any konstant function.