You 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)
Consider the function f(t)=2sec^2(t)–6t^2 . Let F(t) be the antiderivative of f(t) with F(0)=0 .
Then F(5)=?
I am confused on how I would find the antiderivative, after I do that do I plug in five?
3 answers
when i plug this answer in I get -500 is this correct?
I get 2 tan 5 - 2*125 = -249.8
The "5" is understood to be in radians. The -2t^3 term dominates
The "5" is understood to be in radians. The -2t^3 term dominates
1 answer