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Tricky integral

  1. Jan 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Evaluate the integral to find the area.
    2. Relevant equations
    3. The attempt at a solution

    IMG_2118.jpg
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    So I know how to find an anti-derivative for the most part. Here it's tricky because my equation has an exponent AKA square root. I tried to use the chain rule with anti-differentiation. My result was 0. I'm not sure how to deal with that exponent AKA square root.
     
  2. jcsd
  3. Jan 14, 2016 #2

    blue_leaf77

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    Never heard of that. It doesn't seem to work, does it? Try to use the substitution method with the help of trigonometric identity, remember that ##a^2\sin^2\theta + a^2\cos^2\theta = a^2##.
    Note: exponent is not the same as square root.
     
  4. Jan 15, 2016 #3

    HallsofIvy

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    The "chain rule with integration" is "substitution"- but you are doing it wrong.
    If you had [itex]\int (ax- b)^n dx[/itex] then you could let u= ax- b so that du= adx and dx= du/a. Replacing ax- b by u and dx by du/a, the integral becomes [itex]\int u^n (du/a)[/itex] and because a is a constant we can take it out and get [itex]\frac{1}{a}\int u^n du[/tex].

    But if we have [itex]\int (cos(x)- b)^n dx[/itex] and try to do the same thing, we run into a problem. Letting u= cos(x)- b, we have du= -sin(x)dx so that dx= -du/cos(x). Putting those into the integral, [itex]\int u^n (du/cos(x))[/itex]. But now we cannot take "cos(x)" outside the integral because it is NOT a constant! And we cannot integrate cos(x) with respect to u.

    The "chain rule" with differentiation says that if f(x)= g(u(x)) for some differentiable function u, then df/dx= (dg/du)(du/dx). That is, we multiply together the derivatives of g with respect to u and u with respect to x. But doing the "opposite", substituting in the integral for u(x), [itex]\int g(u(x))dx[/itex] we cannot just do the integration with respect to u of g and then multiply by the integral of g. We have to have something line [itex]\int g(u(x))u'(x)dx[/itex]. u'dx= du so the integral becomes [itex]\int g(u)du[/itex]. But the derivative, u', has to already be in the integral. We cannot just multiply as we can with differentiation.
     
  5. Jan 15, 2016 #4

    Mark44

    Staff: Mentor

    You have errors on every line of the work you have done. Here is your work, reproduced using LaTeX, with comments.
    ##\int_0^{\sqrt{19}} \sqrt{19 - h^2} dh##
    ##= (19 - h^2)^{1/2}## No. All you have done is to remove the integration symbol.
    ##= \frac{2(19 - h^2)^{1/2}}{3} \cdot (19h - \left. \frac{h^3}{3}) \right|_0^{\sqrt{19}}## See comment below.
    ##= 0## Not a reasonable answer. The integral represents the area of a quarter-circle, whose area is not zero.

    From above:
    No, this isn't how it works. To get ##19h - \frac{h^3} 3## you integrated ##19 - h^2## but ignored the fact that ##19 - h^2## is raised to the 1/2 power. You can verify that your integration is incorrect by differentiating your antiderivative. For example, ##\int x^3 dx = \frac {x^4} 4 + C##. If I differentiate ##\frac {x^4} 4 + C##, I get ##x^3##, the integrand I started with.
     
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