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Math Help - Fundamental Theorem of Calculus

  1. #1
    mei
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    Unhappy Fundamental Theorem of Calculus

    Let g \in C^1 such that g(0)=0 and g'(0)=1

     \phi(x)=\int g(t).dt

    upper limit:  log x
    lower limit: x^2-1

    I have to calculate \phi' and \phi'' and I have been reading about the fundamental theorem of calculus but I'm not too sure of how I can apply it to calculate \phi' and \phi''. Every example I look at has some expression instead of just g(t) like I have.

    (I apologize for not knowing how to put the limits in the integral.)

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by mei View Post
    Let g \in C^1 such that g(0)=0 and g'(0)=1

     \phi(x)=\int g(t).dt

    upper limit:  log x
    lower limit: x^2-1

    I have to calculate \phi' and \phi'' and I have been reading about the fundamental theorem of calculus but I'm not too sure of how I can apply it to calculate \phi' and \phi''. Every example I look at has some expression instead of just g(t) like I have.

    (I apologize for not knowing how to put the limits in the integral.)

    Thanks in advance.
    \phi(x) = \int_{x^2-1}^{\log{x}} g(t) \, dt

    \phi'(x) = g(\log{x}) \cdot \frac{1}{x} - g(x^2-1) \cdot 2x
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  3. #3
    Super Member Random Variable's Avatar
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    Quote Originally Posted by skeeter View Post
    \phi(x) = \int_{x^2-1}^{\log{x}} g(t) \, dt

    \phi'(x) = g(\log{x}) \cdot \frac{1}{x} - g(x^2-1) \cdot 2x
    But how do you incorporate the given conditions?
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  4. #4
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    Quote Originally Posted by Random Variable View Post
    But how do you incorporate the given conditions?
    good question ... you think the OP has posted all the info regarding the problem?
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  5. #5
    Super Member Random Variable's Avatar
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    Quote Originally Posted by skeeter View Post
    good question ... you think the OP has posted all the info regarding the problem?
    I'm leaning towards "no."
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  6. #6
    mei
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    Quote Originally Posted by Random Variable View Post
    I'm leaning towards "no."
    I have posted all the info that was given me.
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  7. #7
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    Quote Originally Posted by mei View Post
    Let g \in C^1 such that g(0)=0 and g'(0)=1

     \phi(x)=\int g(t).dt

    upper limit:  log x
    lower limit: x^2-1

    I have to calculate \phi' and \phi'' and I have been reading about the fundamental theorem of calculus but I'm not too sure of how I can apply it to calculate \phi' and \phi''. Every example I look at has some expression instead of just g(t) like I have.

    (I apologize for not knowing how to put the limits in the integral.)

    Thanks in advance.
    The "fundamental theorem of calculus" says that if F(x)= \int_a^x f(t)dt then F'(x)= f(x).

    Here you have an integral but the lower limit is a function of x, not a constant, and the upper limit is a function of x, not x. We can handle the first problem by separating this into two integrals:
    \phi(x)= \int_{x^2-1}^a g(t)dt+ \int_a^{log x} g(t)dt
    where "a" is any specific number. Then
    \phi(x)= \int_a^{log x} (t)dt- \int_a^{x^2-1}g(t)dt
    since swapping the upper and lower limits of integration multiplies the integral by -1.

    We take care of the problem with the upper limit by introducting two new variables: let u(x)= x^2- 1 and v= log(x) so
    \int_a^{x^2-1} g(t)dt= \int_a^u g(t)dt= f(u), \int_a^{log(x)}= \int_a^v g(t)dt= h(v), and \phi(x)= f(u)- h(v).

    Now, use the chain rule to find \frac{d\phi}{dx} and \frac{d^2\phi}{d\phi^2}.
    \frac{d\phi}{dx}= \frac{df}{du}\frac{du}{dx}- \frac{dh}{dv}\frac{dv}{dx}. Of course, \frac{df}{du}= g(u) and \frac{dh}{dv}= g(v).

    You "put the limits in the integral" by using _{ } for the lower integral and ^{ } for the upper limit. In fact, you can see the code for any LaTex on this forum by clicking on it.

    Click on \int_{x^2-1}^{log(x)} g(t)dt and see what you get.
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  8. #8
    mei
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    Thank you so much!
    I think I was having a bit of a problem on how I was supposed to incorporate the chain rule onto that, but now I get it!

    And also thank you for explaining the "nice" way of writing integrals in this forum.
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