Calculus help?

Question

..........π
f(x)=∫(1+cost)dt , f'(x)= ?
..........x

Answer 1

you are having trouble because there is a minus sign?
I cannot see.... that is a f '(x) right?

by the chain rule:

d/dx f(x)= d/dx ∫(1+cost)dt = (1+cosx)*d/dx(pi) - (1+cosx)*d/dx(x)

- (1+cosx)

Answer 2

You want f'(x), you should use the Fundamental Theorem of Calculus (Part 1). All you have to do is evaluate the function inside the integral at the endpoints, and multiply by the derivative of the endpoints.

In this case:

(1+cos(1))*(1)' -(1+cosx)*(x)' = -(1+cosx)

See that wasn't so bad!

Answer 3

Well the antiderivative of 1+cos(t) is
t+sin(t)

so the answer is

PI+sin(pi) - (x+sin(x))

But sin(pi) is zero, so f(x) = pi-x-sinx