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Thread: Integration. Fundamental Theorem of Calculus.

  1. #1
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    Integration. Fundamental Theorem of Calculus.

    Integration. Fundamental Theorem of Calculus.-question.png
    Integration. Fundamental Theorem of Calculus.-answer.png

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

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

    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 $\displaystyle ln(x+\sqrt{x^2+a^2})-ln(a)$ as $\displaystyle F(b)-F(a)$ per the first fundamental theorem of calculus. So then it is explained:

    --------
    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:
    ----------

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

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

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

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

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

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

    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.

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

    Simply take the derivative and that is $\displaystyle R'(x)$. You should know how to take the derivative of the natural logarithm. The derivative of $\displaystyle -\ln(a)$ with respect to $\displaystyle x$ is zero ($\displaystyle \ln(a)$ is just a constant since $\displaystyle a$ 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.
    Last edited by SlipEternal; Dec 20th 2013 at 07:22 PM.
    Thanks from sepoto and HallsofIvy
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