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Math Help - limits of integration are variables

  1. #1
    Member pflo's Avatar
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    limits of integration are variables

    How would I find the derivatives of these functions?

    1.  f(x) = \int_{x}^{1} \sqrt{t+sin(t)} dt

    2.  y = \int_{\sqrt{x}}^{x} \frac{cos\theta}{\theta} d\theta
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    MHF Contributor Siron's Avatar
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    Re: limits of integration are variables

    In general:
    D\left(\int f(x)dx\right)=D[F(x)+C]=f(x)
    But because they're definite integrals it can be useful to calculate them.
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    Member pflo's Avatar
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    Re: limits of integration are variables

    So is this true then?

    1.  f(x) = \int_{x}^{1} \sqrt{t+sin(t)} dt
    Let g(t)=\sqrt{t+sin(t) so that f(x) = \int_{x}^{1} g(t) dt
    Then D\left( f(x) \right) = D\left( \int_{x}^{1} g(t) dt \right) = D\left( G(1) - G(x) \right) = D\left( G(1) \right) - D\left( G(x) \right) = D\left( -G(x) \right) = -g(x)

    2.  y = \int_{\sqrt{x}}^{x} \frac{cos\theta}{\theta} d\theta
    Let  y = f(x) and g(\theta)=\sqrt{\theta+sin(\theta) so that f(x) = \int_{x}^{1} g(\theta) d\theta
    Then D\left( f(x) \right) = D\left( \int_{x}^{1} g(\theta) d\theta \right) = D\left( G(x) - G(\sqrt{x}) \right) = g(x) - g(\sqrt{x})

    Seems correct, but it seems too simple. I suppose these problems could be just to illustrate how integration and differentiation are inverse processes.
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    Re: limits of integration are variables

    Do not make this hard.
    D_x \left( {\int_0^x {f(t)dt} } \right)=f(x)

    Because {\int_x^0 {f(t)dt}=-{\int_0^x {f(t)dt} we have D_x \left( {\int_x^0 {f(t)dt} } \right)=-f(x)

    D_x \left( {\int_{g(x)}^{h(x)} {f(t)dt} } \right) = f \circ h(x)h'(x) - f \circ g(x)g'(x)
    Last edited by Plato; August 24th 2011 at 08:39 AM.
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  5. #5
    Member pflo's Avatar
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    Re: limits of integration are variables

    Right. Thanks. "KISS = Keep It Simple, Stupid" I'll try to remember that.

    I'm going with these answers then:

    1.  D(f(x)) = -\sqrt{x+sin(x)}

    2.  D(y) = \frac{cos(x)-\sqrt{x} \, cos\sqrt{x}}{x}

    (unless someone says "no, don't" - anyone?)
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    Member pflo's Avatar
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    Re: limits of integration are variables

    Quote Originally Posted by Plato View Post
    D_x \left( {\int_{g(x)}^{h(x)} {f(t)dt} } \right) = f \circ h(x)h'(x) - f \circ g(x)g'(x)
    Since Plato edited his response to include the above, it appears at least one of my answers is wrong.

    2.  y = \int_{\sqrt{x}}^{x} \frac{cos\theta}{\theta} d\theta
    Let  y = f(x) and t(\theta)=\frac{cos\theta}{\theta} so that f(x) = \int_{\sqrt{x}}^{x} t(\theta) d\theta

    Also, if g(x) = \sqrt{x} and h(x) = x then this would conform with Plato's formula and the derivatives would be h'(x) = 1, g'(x)=\frac{1}{2\sqrt{x}}.

    The integral would NOT be what I posted earlier:  D(y) = \frac{cos(x)-\sqrt{x} \, cos\sqrt{x}}{x} .

    Instead, this is what I'm thinking:

    D\left( f(x) \right) = D\left( \int_{g(x)}^{h(x)} t(\theta) d\theta \right) = t(x)*1 - t(\sqrt{x})*\frac{1}{2\sqrt{x}}
    = \frac{cos(x)}{x}*1 - \frac{cos(\sqrt{x})}{\sqrt{x}}*\frac{1}{2\sqrt{x}}
    = \frac{2cos(x)-cos(\sqrt{x})}{2x}

    Because the derivative of x is 1, I'm sticking with my original results for the first problem, which was  D(f(x)) = -\sqrt{x+sin(x)}

    Once again, can anyone confirm or deny these answers? Thanks.
    Last edited by pflo; August 24th 2011 at 11:27 PM. Reason: added original problem 1 answer
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    Re: limits of integration are variables

    More generally, "Liebniz's rule":
    \frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t)dt= \frac{d\beta(x)}{dx}f(x,\beta(x))- \frac{d\alpha(x)}{dx}f(x, \alpha(x)) + \int_{\alpha(x)}^{\beta(x)}\frac{\partial f}{\partial x} dt
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