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Calculus: trig functions anti-derivatives

  1. Oct 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Let g'(x)=-17x16* sin(Ax9) - 9Ax25*cos(Ax9)

    g(1)=143/9
    Where A is a real number such that tanA=1/sqrt(80), 0<A<pi/2

    Find g(0)

    3. The attempt at a solution
    I was able to get the antiderivative of g'(x), so that:
    g(x) = -x17*sin(Ax9)

    I got A=6.3794 degrees.

    BUT, I don't get g(1)=143/9

    What am I doing wrong? Did I assume wrong that the g'(x) was created by the product rule?
     
  2. jcsd
  3. Oct 16, 2008 #2

    Mark44

    Staff: Mentor

    Your antiderivative looks fine. Did you remember to add a constant? IOW, you should have
    g(x) = -x17*sin(Ax9) + C

    You're sort of given A, and you're given that g(1) = 143/9, so you can find C. If you did all that, you can still run into problems calculating things with a calculator, such as when it's in radian mode but you're working with degrees.
     
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