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Derivative of a difficult integral

  1. Mar 4, 2013 #1
    Spivak, 14.1.iii: Derivative of F, where
    gif.gif

    I have the FTC. The answer is also given,

    gif.gif

    But I don't know how to find it.

    First I want to try a function decomposition:

    gif.gif

    and from the Chain Rule:

    gif.gif


    But now I'm stuck. For I do not know an expression for p' or q'. And worse, r' doesn't show up in the given solution. Thanks in advance for your time :).
     
  2. jcsd
  3. Mar 4, 2013 #2

    LCKurtz

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    If you let ##f(y) = \int_{8}^{y}\frac{1}{1+t^{2}+\sin^{2}t}dt## so ##F(x) =
    \int_{15}^xf(y)dy## what would ##F'(x)## be in terms of ##f##?
     
  4. Mar 4, 2013 #3
    Thanks for your reply. When you put it that way, we have [itex]F'(x) = f(y) = \int_{8}^{y} \frac{1}{1+t^{2}+sin^{2}t}dt[/itex]

    But the given answer has [itex]\int_{8}^{x} \frac{1}{1+t^{2}+sin^{2}t}dt[/itex]. How can we just change the integrand variable like that?
     
  5. Mar 4, 2013 #4

    LCKurtz

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    No it doesn't!! That y is a dummy variable and F is a function of x. And its derivative must be a function of x.
     
  6. Mar 4, 2013 #5

    Ray Vickson

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    The y inside the y-integral is "integrated over", so goes away. For example, you can think of ##F(x) = \int_8^x f(y) \, dy## as a limit of the form
    [tex] \lim_{n \to \infty} \sum_{i=1}^{n} \frac{x-8}{n} f\left(8 + \frac{i (x-8)}{n}\right),[/tex]
    and there is no 'y' at all in this expression. In other words,
    [tex] F(x) = \int_8^x f(\text{anything}) \; d\: \text{anything},[/tex]
    and it does not matter what name you give to "anything" (except that it should not be x, f or d).
     
  7. Mar 4, 2013 #6
    Thanks. I think I get it. Because we're integrating over ##dy##, that means we're plugging in ##x## for y when we integrate.
     
  8. Mar 4, 2013 #7
    Just remember.... [itex]\int _{ a }^{ b }{ f(x)dx } =F(b)-F(a)[/itex] so let's call [itex]\int _{ 8 }^{ y }{ g(t)dt } =G(y)-G(8)[/itex] where [itex] g(t)=\frac { 1 }{ 1+ t^{2}+sin^{2}t } [/itex] then [itex]\int _{ 15 }^{ x }{ \int _{ 8 }^{ y }{ g(t)dt } } =\int _{ 15 }^{ x }{ G(y)-G(8){dy} } [/itex] since the last term is a constant, we get [itex]F(x)=\int _{ 15 }^{ x }{ \int _{ 8 }^{ y }{ g(t)dt } } =\left[ H(x)-H(15) \right] -G(8)\left( x-15 \right) =H(x)-G(8)x+15G(8)-H(15) [/itex] Now derive and see what constants drop out :-)
     
  9. Mar 5, 2013 #8
    Ok guys, I struggled through that one and I think I understand it now. To test my knowledge, I've attempted another one, which isn't solved.

    ##F(x) = \sin \left ( \int_{0}^{x} \sin \left ( \int_{0}^{y} \sin^{3} t dt \right ) dy \right )##

    So now I will try another composition. I will use capital letters for functions defined by integrals, and lowercase for functions that aren't.

    ## f(x) = \sin x \\
    G(x) = \int_{0}^{x} \sin y dy\\
    H(y) = \int_{0}^{y} \sin^{3}t dt \\
    ##
    then,

    ##F(x) = f \circ G \circ H##

    By the chain rule,

    ## F'(x) = f'(G \circ H)\cdot G'(H) \cdot H' ##

    ## f'(x) = \cos x ##

    By the FTC,

    ## G'(x) = \sin x \\
    H'(y) = \sin^{3} y \\
    ##

    Composing,

    ##F'(x) = \cos(\int_{0}^{x} \sin \left ( \int_{0}^{y} \sin^{3}t dt \right )dy \cdot \sin \left (\int_{0}^{x} \sin^{3} t dt \right ) \cdot sin^{3} x##

    How'd I do?
     
  10. Mar 5, 2013 #9

    Dick

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    You are missing a big closing parentheses on the first cos( part. But it looks ok to me.
     
  11. Mar 5, 2013 #10

    LCKurtz

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    I don't think it is quite correct. If you call$$
    f(y) = \sin(\int_0^y \sin^3(t)dt)$$ then$$
    F(x) = \sin(\int_0^xf(y)dy)$$Then$$
    F'(x) = cos(\int_0^xf(y)dy)\cdot \frac d {dx}\int_0^xf(y)dy=
    cos(\int_0^xf(y)dy)\cdot f(x)
    =cos(\int_0^xf(y)dy)\cdot \sin(\int_0^x \sin^3(t)dt)$$ $$
    =cos(\int_0^x\sin(\int_0^y \sin^3(t)dt)dy)\cdot \sin(\int_0^x \sin^3(t)dt)
    $$I think you have an extra ##\sin^3(x)##.
     
  12. Mar 5, 2013 #11

    Dick

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    Yeah, you are right. Ooops.
     
  13. Mar 6, 2013 #12
    Though I am convinced by Kurtz's answer, I am having a hard time seeing where I went wrong in my own reasoning.

    Is my function composition wrong? Does ##F(x) ≠ (f \circ G \circ H)(x)##?
     
  14. Mar 6, 2013 #13

    LCKurtz

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    You have ##G(x) = \int_0^x \sin y\, dy##, but lets call that dummy variable of integration ##u## so ##G(x) = \int_0^x \sin u\, du##. Also, you have ##H(y) = \int_0^y \sin^3 t\, dt##. Then$$
    G(H(y)) =\int_0^{H(y)} \sin u\, du=\int_0^{\int_0^y \sin^3 t\, dt} \sin u\, du$$I don't think that is what you want.
     
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