Calculus: trig functions anti-derivatives

  • Thread starter Melawrghk
  • Start date
  • #1
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Homework Statement


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)

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?
 

Answers and Replies

  • #2
35,204
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Homework Statement


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)

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?

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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