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Math Help - Integrals

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    Integrals

    Suppose f(x) is differentiable on (−+),
    F(x)=xf(t)dt

    With the boundaries being from 1/x to 0.

    for x/=0. then, what is F(x)?

    Sorry for the sloppy notation, I'm not sure the math code for integrals and such. I'm not sure how to go about solving this question. I need to ultimately find the derivative, but first I believe I need to integrate. However, I do not know f(t), so how can I integrate it? As well, there are two different variables... So confused. Any help would be greatly appreciated...
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    Quote Originally Posted by Hellreaver View Post
    Suppose f(x) is differentiable on (−+),
    F(x)=xf(t)dt

    With the boundaries being from 1/x to 0.

    for x/=0. then, what is F(x)?

    Sorry for the sloppy notation, I'm not sure the math code for integrals and such. I'm not sure how to go about solving this question. I need to ultimately find the derivative, but first I believe I need to integrate. However, I do not know f(t), so how can I integrate it? As well, there are two different variables... So confused. Any help would be greatly appreciated...
    First off, is this the integral

    F(x) = \int_{\frac{1}{x}}^0 x f(t)\, dt
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    That is it, except that the upper and lower bounds are switched, so 1/x is the upper bound and 0 is the lower. Thanks, I have no idea how to write the questions out like that.
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    Quote Originally Posted by Hellreaver View Post
    Suppose f(x) is differentiable on (−+),
    F(x)=xf(t)dt

    With the boundaries being from 1/x to 0.

    for x/=0. then, what is F(x)?

    Sorry for the sloppy notation, I'm not sure the math code for integrals and such. I'm not sure how to go about solving this question. I need to ultimately find the derivative, but first I believe I need to integrate. However, I do not know f(t), so how can I integrate it? As well, there are two different variables... So confused. Any help would be greatly appreciated...
    F(x) = \int^{\frac{1}{x}}_0 x f(t)\, dt = x \int^{\frac{1}{x}}_0 f(t)\, dt = x \cdot G(x).

    Let w = \frac{1}{x}: ~ G(w) = \int^{w}_0 f(t)\, dt.

    To differentiate G(x) apply the chain rule and Fundamental Theorem of Calculus:

    \frac{dG}{dx} = \frac{dG}{dw} \cdot \frac{dw}{dx} = f(w) \cdot \frac{dw}{dx}.

    It is left for you to substitute back w = \frac{1}{x} to get the final expression for \frac{dG}{dx}.

    Knowing \frac{dG}{dx} you can now easily use the product rule to differentiate F(x).

    Then differentiate the result again.
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    Quote Originally Posted by mr fantastic View Post
    F(x) = \int^{\frac{1}{x}}_0 x f(t)\, dt = x \int^{\frac{1}{x}}_0 f(t)\, dt = x \cdot G(x).

    Let w = \frac{1}{x}: ~ G(w) = \int^{w}_0 f(t)\, dt.

    To differentiate G(x) apply the chain rule and Fundamental Theorem of Calculus:

    \frac{dG}{dx} = \frac{dG}{dw} \cdot \frac{dw}{dx} = f(w) \cdot \frac{dw}{dx}.

    It is left for you to substitute back w = \frac{1}{x} to get the final expression for \frac{dG}{dx}.

    Knowing \frac{dG}{dx} you can now easily use the product rule to differentiate F(x).

    Then differentiate the result again.
    I am lost at the chain rule and Fundamental Theorem part of this. Can you elaborate on what you did here?
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  6. #6
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    Quote Originally Posted by Hellreaver View Post
    I am lost at the chain rule and Fundamental Theorem part of this. Can you elaborate on what you did here?
    Read this: The Second Fundamental Theorem of Calculus
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    Here is a little tid bit to remember.

    \frac{d}{dx}\int_{a}^{g(x)}f(t)dt=f(g(x))g'(x)

    And:

    \frac{d}{dx}\int_{h(x)}^{g(x)}f(t)dt=f(g(x))g'(x)-f(h(x))h'(x)

    Now, prove each of these.

    The former is what Mr F is getting at.

    Hopefully, you know what the chain rule is by now?. That is one of the first things to learn in calc. It is invaluable.
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    Sorry if I'm being slow, but what is the f(w) value supposed to be? I know that w=1/x, but how do I derive the function? Is it just f'(1/x)?
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  9. #9
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    Quote Originally Posted by Hellreaver View Post
    Sorry if I'm being slow, but what is the f(w) value supposed to be? I know that w=1/x, but how do I derive the function? Is it just f'(1/x)?
    All you do is substitute w = \frac{1}{x} into the result I gave you: \frac{dG}{dx} = f\left( \frac{1}{x}\right) \cdot \left(- \frac{1}{x^2}\right).

    This gives you the derivative that you need when applying the product rule to find \frac{dF}{dx}.
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