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October 20, 2014

October 20, 2014

Posted by **sam** on Tuesday, June 3, 2008 at 9:53pm.

Then F(5)=?

I am confused on how I would find the antiderivative, after I do that do I plug in five?

- math -
**drwls**, Tuesday, June 3, 2008 at 10:19pmYou might want to use a table of integrals for the first term. Most people prefer to call the "antiderivative" the (indefinite) integral.

In your case the integral is

F(t) = 2 tan t -2t^3 + C

where C is an arbitary constant.

Since F(0) = 0, C = 0

Now solve for F(5)

- math -
**sam**, Tuesday, June 3, 2008 at 10:32pmwhen i plug this answer in I get -500 is this correct?

- math -
**drwls**, Tuesday, June 3, 2008 at 11:02pmI get 2 tan 5 - 2*125 = -249.8

The "5" is understood to be in radians. The -2t^3 term dominates

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