The evaluation theorem states that if we have a function, then we find its derivative (in this case) g' and then integrate it we get the original function in the form g(b)-g(a) where b and a are the bounds on the integral so basically when you integrate g' from 0 to 2pi you get the function g in the form g(b)-g(a) where a is 0 and b is 2 pi that is:

g(2pi)-g(0)=pi (given in the question) also given in question is g(0) =4 simply take 4 to the other side and solve for g(2pi) which gives us pi+4 as the final Answer.

Hope that helps

This looks like an application of the fundamental theorem of calculus. Since you know the integral of the derivative of g(t), you have some information about g(t) itself. In particular, you know:

integral (from 0 to 2pi) of g ' (t) dt = pi

implies

g(2pi) - g(0) = pi

(to see why this is, think about how you evaluate integrals in the usual case)

and since you know g(0) = 4, then you have:

g(2pi) - 4 = pi

so

g(2pi) = pi + 4

hope this helps!