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Math Help - Fundamental Theorem Of Calculus

  1. #1
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    Fundamental Theorem Of Calculus

    It makes sense when I read the statements but I can't seem to apply it without getting totally lost.
    If I have a problem that looks something like:

    d/dx of the integral of some function f(x) evaluated from -1 to sin(x) then how do I apply FToC to solve it?
    Also what if the bounds are 2x to x^2 or something along those lines?


    I also have a related question that I am dead stuck on but an answer to the above would be appreciated more than a solution to the below if you don't have time or something.

    2 plus the integral of (f(t)/(t^3))dt evaluated from a to x equals 6x^-5. Find a value for a and a function f(t) so that this is true.
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    Quote Originally Posted by Vashti View Post
    It makes sense when I read the statements but I can't seem to apply it without getting totally lost.
    If I have a problem that looks something like:

    d/dx of the integral of some function f(x) evaluated from -1 to sin(x) then how do I apply FToC to solve it?

    [snip]
    From the chain rule and the FTC it can be shown that:

    \frac{d}{dx} \left [ \int_{a}^{g(x)} f(t) \, dt \right] = f(g(x)) \, g'(x).

    In your case a = -1 and g(x) = \sin x.
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  3. #3
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    Quote Originally Posted by Vashti View Post
    [snip]d/dx of the integral of some function f(x) evaluated from -1 to sin(x) then how do I apply FToC to solve it?
    Also what if the bounds are 2x to x^2 or something along those lines?

    [snip]
    From the chain rule and the FTC it can be shown that

    \frac{d}{dx} \left [ \int_{h(x)}^{g(x)} f(t) \, dt \right] = f(g(x)) \, g'(x) - f(h(x)) \, h'(x) .

    In your case h(x) = 2x and g(x) = x^2.
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    if it's not too much trouble could someone prove that for me? You say to use the chain rule but I'm not sure how...
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    p.s. thanks so much you've helped me a lot already.
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  6. #6
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    Quote Originally Posted by Vashti View Post
    [snip]

    2 plus the integral of (f(t)/(t^3))dt evaluated from a to x equals 6x^-5. Find a value for a and a function f(t) so that this is true.
    2 + \int_{a}^{x} \frac{f(t)}{t^3} \, dt = \frac{6}{x^5} .... (1)

    Differentiate both sides of (1) with respect to x using the rule in post #2:

    \frac{f(x)}{x^3} = -\frac{30}{x^6} \Rightarrow f(x) = .....

    Now substitute your answer for f(x) into (1), do the required integration and solve for a.
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    Quote Originally Posted by mr fantastic View Post
    From the chain rule and the FTC it can be shown that

    \frac{d}{dx} \left [ \int_{h(x)}^{g(x)} f(t) \, dt \right] = f(g(x)) \, g'(x) - f(h(x)) \, h'(x) .

    [snip]
    \frac{d}{dx} \left [ \int_{h(x)}^{g(x)} f(t) \, dt \right] = \frac{d}{dx} \left [ \int_{h(x)}^{a} f(t) \, dt + \int_{a}^{g(x)} f(t) \, dt \right]


    = \frac{d}{dx} \left [ -\int_{a}^{h(x)} f(t) \, dt + \int_{a}^{g(x)} f(t) \, dt \right]


    = - \frac{d}{dx} \left [\int_{a}^{h(x)} f(t) \, dt \right] + \frac{d}{dx} \left[ \int_{a}^{g(x)} f(t) \, dt \right]


    Now apply the result given in post #2 to each term. To prove the result in post #2:

    Let w = g(x). Then you have:


    \frac{d}{dx} \left [ \int_{a}^{g(x)} f(t) \, dt \right] = \frac{d}{dw} \left [ \int_{a}^{w} f(t) \, dt \right] \, \times \, \frac{dw}{dx} = f(w) \, g'(x) = f(g(x)) \, g'(x).
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    you are my hero! thank you so so much!
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    Quote Originally Posted by Vashti View Post
    you are my hero!
    Of course... Did you ignore his job is to save the world ?
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