Part 2 of the Fundamental Theorem of Calculus

[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?

Thanks in advance.

 

[BlackMamba]

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

ln 4/9 + C

Would this be correct?

 

[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.

 

[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.

 

[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!

 

[BlackMamba]

Actually I have a similar 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

 

[ircdan]

= -cos\pi + C - (cos{pi} + C)

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

 

[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.

 

[ircdan]

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.

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.

 

[ircdan]

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

 

[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. :)

 

[d_leet]

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.

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).