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Integration using the fundamental theorem of calculus

  1. Nov 7, 2013 #1
    Hello PF.

    1. The problem statement, all variables and given/known data
    Find a function g such that

    [itex]\int_0^{x^2} \ tg(t) \, \mathrm{d}t = x^2+x[/itex]

    2. Relevant equations

    From the fundamental theorem of calculus:

    [itex]f(x) = \frac{d}{dx}\int_a^x \ f(t) \, \mathrm{d}t[/itex]

    3. The attempt at a solution

    After taking the derivative of both sides of the equation:

    [itex]\frac{d}{dx}\int_0^{x^2} \ tg(t) \, \mathrm{d}t = 2x+1[/itex]

    Thus, from the fundamental theorem and chain rule,

    [itex](2x)(x^2)g(x^2) = 2x+1[/itex]

    [itex]2x^3g(x^2) = 2x+1[/itex]

    [itex]g(x^2) = \frac{1}{x^2} + \frac{1}{2x^3}[/itex]


    However, I know that this answer is wrong because the definite integral of this function (when substituted into the original equation) is not equal to x + x^2. I'm having difficulty identifying my error. Could someone please point me in the right direction? :)
     
  2. jcsd
  3. Nov 7, 2013 #2
    What did you put back into your integral for checking. You solved only down to g(##x^2##). Did you put that back in or g(x)? I got the right answer when I put in g(x).
     
  4. Nov 7, 2013 #3
    I put g(x^2) back into the integral. How would I simplify g(x^2) to get g(x)? Would I square every expression for x in the equation (i.e x^2 and x^3)?
     
  5. Nov 7, 2013 #4
    To break it into two steps: Let u = ##x^2## and figure out what g(u) is. Then since it doesn't matter what symbol you use the function g(u) is the same as g(x).

    More directly, just stick an x in everywhere you have an ##x^2##.
     
  6. Nov 7, 2013 #5
    Aha! Then U = sqr(x) and that can be substituted in the equation. Thanks brmath!
     
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