# Part 2 of the Fundamental Theorem of Calculus

1. Sep 6, 2006

### BlackMamba

Hello,

I have a problem that I am getting stuck simplifying further.

The problem asks me to find the integral if it exists using Part 2 of the FTC.

I know that the second part of the FTC says:$\int_{a}^{b} f(x)dx = F(b) - F(a)$ Where F is the anti-derivative of f.

Here is the problem:
$\int_{9}^{4}$ x ^-1 dx

While $f(x) =$ x^ -1 is not continuous throughout it is continuous on the interval [4,9]. So therefore it does exist.

So here is my solution:
$f(x) =$ x^ -1
$F(x) = ln|x| + C$
$F(4) = ln4 + C$
$F(9) = ln9 + C$

Here is where I am having trouble simplifying. I would then use $F(b) - F(a)$

so when I do that I have $ln4 + C - ln9 + C$

Could I just write that as $ln4 - ln9 + C$ ? Do I still need to simplify further?

Last edited: Sep 6, 2006
2. Sep 6, 2006

### BlackMamba

I'm thinking that I would probably need to simplify further and write it as:

$ln 4/9 + C$

Would this be correct?

3. Sep 6, 2006

### AKG

The C's cancel out. Think about it, $\int_9 ^4x^{-1}dx$ is a single number, it doesn't depend on a constant. There is no such thing as THE antiderivative of f, but there is a family of antiderivatives. x |-> ln|x| is one of the antiderivatives. Any function of the form x |-> ln|x| + C is in the family, for any real C. Moreover, only functions of that form are in the family. The theorem should say that if F is ANY antiderivative of f, then

$$\int _a ^b f(x)dx = F(b) - F(a)$$

By choosing F(x) = ln|x| + C, you choose an arbitrary antiderivative, but even in this case, when you write F(9) = ln(9) + C and F(4) = ln(4) + C, it's the same C, so they cancel out when you do F(4) - F(9). Alternatively, you could have chosen a particular antiderivative, e.g. the one where C = 0.

Last edited: Sep 6, 2006
4. Sep 6, 2006

### ircdan

No, you forgot to flip your limits of integration. Here:

integral(x^-1 dx, x = 9 to x = 4) = -integral(x^-1 dx, x = 4 to x = 9)
= F(9) - F(4) = ln(9) + C -(ln(4) + C) = ln(9) - ln(4) = ln(9/4) = ln((3/2)^2) = 2ln(3/2)

Notice the constant of integration canceled out. This is always the case, so you can just omit it in your calculation.

5. Sep 6, 2006

### BlackMamba

Oh right, how did I miss the cancellation of the constant C? And forget to flip the limits? I kept thinking to myself it looked odd, but never thought to flip it. lol I will be sure not to forget next time. Thanks for all your help!

6. Sep 6, 2006

### BlackMamba

Actually I have a similiar problem that I could use a bit of help on as well.

I have to find the integral using the same method.

Here's the problem:
$\int_{-\pi}^{\pi} sinx dx$

So basically it boils down to:
= $-cos\pi + C - (cos\pi + C)$
= $-cos\pi - cos\pi$
= 2

Last edited: Sep 6, 2006
7. Sep 6, 2006

### ircdan

That part is wrong, you are missing a - sign, do you see where?

8. Sep 6, 2006

### BlackMamba

Uhmm not really: Let me try to break it down further.

$F(\pi) = - cos\pi + C$

$F(-\pi) = cos\pi + C$

I'm assuming for $F(-\pi)$ that since cos is already negative and that's paired with a negative $\pi$ then that combo would be positive. I'm guessing that is not correct.

9. Sep 6, 2006

### ircdan

cosx is even. This means cos(x) = cos(-x) for every x. So for you, cos(-pi) = cos(pi) = -1. I think you made the mistake and said cos(-pi) = -cos(pi), which is incorrect, since cos(-pi) = -1 and -cos(pi) = -(-1) = 1.

10. Sep 6, 2006

### ircdan

And you should get 0 as the answer btw when you do it so you can check yourself.

11. Sep 6, 2006

### BlackMamba

I think I'm getting confused. I mean I believe I understand what you are saying. I do recall learning that, but as I've proven before nothing really sticks with me. lol

So if I keep all negative signs in the equations it would look something like:

$F(\pi) = -cos\pi + C$

$F(-\pi) = -cos(-\pi) + C$

Which in this case would give me an answer of 0.

Oh you beat me to it. lol Thanks for your help. Next time I better just leave all my negative signs where they belong. :)

12. Sep 6, 2006

### d_leet

Umm... Why do you think he needs to flip the limits? The integral as written above is from 9 to 4, so the anwer would be ln(4/9)=2*ln(2/3).