# Thread: Integration. Fundamental Theorem of Calculus.

1. ## Integration. Fundamental Theorem of Calculus.

The question is asking to prove:
$\int^0_x\frac{1}{\sqrt{t^2-a^2})}=ln(x+\sqrt{x^2+a^2})-ln(a)$

So I think I understand when it is said:
$L'(x)=\frac{1}{\sqrt{a^2+x^2}}$

so that must be because the definite integral is from 0 to x so L'=f(x) or another way of saying this would be that t was replaced by x and the resulting function is L'.

In regards to R'(x) I see $ln(x+\sqrt{x^2+a^2})-ln(a)$ as $F(b)-F(a)$ per the first fundamental theorem of calculus. So then it is explained:

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Since L'(x)=R'(x), we have L(x)=R(x)+C for some constant C=L(x)-R(x). The constant C may be evaluated by assigning a value to x; the most convienient choice is x=0 which gives:
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The part above I find I am right now struggling to understand what is meant by this. L'(x)=R'(x). I thought the definite integral was = to F(b)-F(a). I went through the attached lecture notes before I tried the exercise and I think I understand a good amount of those notes although the proof of FTC1 by FTC2 is a little bit confusing to me the rest I think I was able to grasp what was being said.

Thank You!

2. ## Re: Integration. Fundamental Theorem of Calculus.

The notes you posted have a typo. It should read:

$L' = \dfrac{ d }{ dx } \int_0^x \dfrac{dt}{\sqrt{a^2+t^2}} = \dfrac{1}{\sqrt{a^2+x^2}}$

(The notes you posted show $L' = \dfrac{d}{dx} \int_0^x \dfrac{dt}{\sqrt{a^2+x^2}}$, which is wrong, since $\dfrac{1}{\sqrt{a^2+x^2}}$ does not depend on $t$.)

Also, the question you posted does not match the notes you posted. The original question should be asking to prove:

$\int_0^x \dfrac{dt}{\sqrt{a^2+t^2}} = \ln\left(x + \sqrt{a^2+x^2}\right) - \ln(a)$

Then, the proof follows by using the FTC parts 1 and 2.

Anyway, you say that you understand the derivative of the LHS. It is the derivative of the RHS that you are having trouble with.

$R'(x) = \dfrac{d}{dx}\left[\ln\left(x + \sqrt{a^2+x^2}\right) - \ln(a)\right]$

Simply take the derivative and that is $R'(x)$. You should know how to take the derivative of the natural logarithm. The derivative of $-\ln(a)$ with respect to $x$ is zero ( $\ln(a)$ is just a constant since $a$ is a constant).

Performing some algebra, you find that the derivative of the LHS and the RHS are the same. So, if you take the antiderivatives, they must differ by at most a constant.