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I don't understand how to differentiate integrals!

  1. Jan 30, 2013 #1
    I am in Calculus 2 and we're just reviewing calc 1. Can someone break down the concept of differentiating definite integrals for me? I am mostly struggling on the trig functions. The problem I am stuck on is
    ∫[itex]^{}x[/itex][itex]_{}0[/itex] cos(t[itex]^{}2[/itex]) dt
     
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  3. Jan 30, 2013 #2

    Simon Bridge

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    You exploit the fact that the integration is the inverse of differentiation, so:

    $$\int \frac{d}{dt}y(t)dt = y(t) + c$$ and $$\frac{d}{dt}\int y(t)dt = y(t) + c$$... for indefinite integrals. For definite integrals, the next step is to apply the limits.

    But it looks like you want something a but different ... the chain rule for integrating $$\int x_0\cos(t^2)dt$$ ... is that correct?

    If you want to evaluate: $$\int f(g(x))g^\prime(x).dx$$ then you can put ##u=g(x)## then solve to find out what dx is. When you make the substitution, you end up evaluating the integral of f(u)du which may be easier to do. That what you are after?
     
  4. Jan 31, 2013 #3

    Mark44

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    I think this is what you were trying to write:
    $$ \int_0^x cos(t^2)dt$$

    Click what I wrote to see my LaTeX script, or click Quote on my post.

    I don't think so. I believe that 0 and x were the lower and upper limits of integration.
     
  5. Jan 31, 2013 #4

    Simon Bridge

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    Yeah that makes sense... we'll soon see :)

    That particular example looks a tad rough for level 2 - have they just learned Fresnel integrals or something?
     
  6. Jan 31, 2013 #5
    All you need to do is exploit the fundamental theorem of calculus:

    [tex]\int_{g(x)}^{h(x)} f(t) dt = F(h(x))-F(g(x))[/tex]
    So differentiating via the chain rule gives:

    [tex]\frac{d}{dx} \int_{g(x)}^{h(x)} f(t) dt = f(h(x))h'(x)-f(g(x))g'(x)[/tex]
    In the OP's example g(x)=0 and h(x)=x, so:

    [tex]\frac{d}{dx} \int_{0}^{x} cos(t^2) dt = cos(x^2)-0[/tex]
     
  7. Jan 31, 2013 #6
    Yes! This is what I was trying to write - sorry, new to the site :)
     
  8. Jan 31, 2013 #7
    Ah thank you! Life makes sense again haha
     
  9. Jan 31, 2013 #8
    Actually I am unclear with your last step there. So f(h(x))h'(x) equals x, do you just plug x in for t? What if it equaled x2? Would it end up being cos x4? Hypothetically speaking.
     
  10. Jan 31, 2013 #9
    If you were integrating from 0 to x2, you would get:

    [tex]2xcos(x^4)[/tex]
     
  11. Jan 31, 2013 #10

    HallsofIvy

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    The "fundamental theorem of calculus" (part of it) says the if [itex]F(x)= \int f(x) dx[/itex], then dF/dx= f(x). That says that [itex]d/dx \int_a^x f(t)dt= f(x)[/itex]. (The other part says that if f(x)= dF/dx, then [itex]F(x)= \int f(x)dx+ C[/itex] for some constant C.)

    "Leibniz's formula" is more general:
    [tex]\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t)dt=f(x, \beta(x))\frac{d\beta(x)}{dx}- f(x, \alpha(x))\frac{d\alpha(x)}{dx}+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f(x, t)}{\partial x}dt[/tex].

    The "fundamental theorem" is clearly a special case of "Leibniz's formula" with [itex]\alpha(x)= 0[/itex], so that [itex]d\alpha/dx= 0[/itex], [itex]\beta(x)= x[/itex] so that [itex]d\beta/dx= 1[/itex] and f(t) not a function of x so that [itex]\partial f/\partial x= 0[/itex]:
    [tex]\frac{d}{dx}\int_0^x f(t)dt= (1)f(x)- (0)f(0)+ \int_0^x 0 dt= f(x)[/tex].

    To do the initial example, [itex]d/dx\int_0^x cos(t^2)dt[/itex], you just need the "fundamental theorem", [itex]d/dx\int_0^x cos(t^2)dt= cos(x^2)[/itex].

    To do the example elfmotat gives, [itex]d/dx \int_0^{x^2} cos(t^2)dt[/itex] take [itex]\beta(x)= x^2[/itex], so that [itex]d\beta/dx= 2x[/itex], and [itex]\alpha(x)= 0[/itex]. [itex]f(x, t)= cos(t^2)[/itex] which does NOT depend on x so that [itex]\partial f/\partial x= 0[/itex]. Leibniz' forumula becomes
    [tex]2x cos((x^2)^2)- 0 cos(0^2)+ \int_0^{x^2} 0 dt= 2x cos(x^4)[/tex].
     
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