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Thread: Finding the derivative of an integral with respect to X

  1. #1

    Question Finding the derivative of an integral with respect to X

    Hi guys I'm doing my homework and I've run into a couple problems that I don't quite understand, and the textbook doesn't give me much information to form an intuitive understanding of what's going on here so I don't know where to begin.

    The question is of the form:

    Using the fundamental theorem of calculus part I, find the derivative:

    f(x) = (from cos x to x^5) ∫ sin(u) du

    Initially I thought perhaps I'm supposed to split the integral into two parts with a constant going to the respective x function, and then solve each part accordingly, but since cos x and x^5 are variable I don't see how to determine a point to separate them at which makes me think my approach is wrong. Any insight would be greatly appreciated, thanks

  2. #2
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    You don't have to do the integral at all- the problem asks you to find the derivative of the integral and those two operations are inverse to each other: the "Fundamental Theorem of Calculus" says that [tex]\frac{d}{dx}\int_a^x f(t)dt= f(x)[/tex]. Letting, say, u= cos(x) and [tex]v= x^5[/tex], we can break that into two integrals, as you suggest, and then apply the "chain rule" to each derivative. The upshot of all that is "Liebniz's rule":
    [tex]\frac{d}{dx}\int_{u(x)}^{v(x)} f(t, x) dx= \frac{dv}{dx} f(v(x), x)- \frac{du}{dx}f(u(x), x)+ \int_{u(x)}^{v(x)} \frac{\partial f(x, t)}{\partial x}dt[/tex]

    In this problem, since the integrand does not depend on x, that reduces to
    [tex]\frac{d}{dx}\int_{cos(x)}^{x^5} sin(u)du= \frac{d x^5}{dx} sin(x^5)- \frac{d cos(x)}{dx} sin(cos(x))[/tex]

    (As far as "separating" the upper and lower limits is concerned, it NOT necessary that the point be between the two limits:
    [tex]\int_a^b f(u)du= \int_a^c f(u)du+ \int_c^b f(u)du[/tex] for any c, whether between a and b, less than a, or larger than b.)
    Last edited by HallsofIvy; 09-12-2014 at 05:23 PM.

  3. #3
    I'm currently doing a course that is calculus of a single variable. The calculus tools currently at my disposal are a chapter on Riemann's sums, and an introduction to the first and second parts of the fundamental theorem of calculus. I noticed that the part that looks like gibberish to me doesn't apply since the integrand doesn't depend on x. But given the bare bones toolbox at my disposal how would I have come to the answer you have?

  4. #4
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    Quote Originally Posted by ZibZabZabbityDoo View Post
    I'm currently doing a course that is calculus of a single variable. The calculus tools currently at my disposal are a chapter on Riemann's sums, and an introduction to the first and second parts of the fundamental theorem of calculus. I noticed that the part that looks like gibberish to me doesn't apply since the integrand doesn't depend on x. But given the bare bones toolbox at my disposal how would I have come to the answer you have?
    You are going to need something like Leibniz's integral rule, so if you don't have it in your tool box, you will need to prove it. It can be proved using the first part of the fundamental theorem of calculus.

    Basically what it says for this situation is
    [tex]\frac{d}{dx}\int_{a}^{g(x)} f(u)du = \frac{d g(x)}{dx} f(g(x))[/tex]
    which is a simplification of what HallsofIvy said.

    To prove this, let
    h(x) = [tex]\int_{a}^{x} f(u)du[/tex]
    then
    h(g(x)) = [tex]\int_{a}^{g(x)} f(u)du[/tex]
    Now use the chain rule and the first part of the fundamental theorem of calculus
    [tex]\frac{d h(g(x))}{dx} = \frac{d g(x)}{dx} \frac{d h}{dx} = \frac{d g(x)}{dx} f(g(x))[/tex]

  5. #5
    Thanks to both of you for the help. Reminding me that c need not be between a and b made life way simpler.

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