Calculus: Early Transcendentals, Chapter 5, 5.3, Section 5.3, Problem 34

You need to evaluate the definite integral using the fundamental theorem of calculus, such that:
int_a^b f(u) du = F(b) - F(a)
int_0^3 (2sin x - e^x) dx =int_0^3 2sin x dx - int_0^3 e^x dx
int_0^3 2sin x dx = 2(-cos x)|_0^3 = -2cos 3 + 2cos 0 = 2 - 2cos 3
int_0^3 e^x dx = e^x|_0^3 = e^3 - e^0 = e^3 - 1
Gathering the results yields:
int_0^3 (2sin x - e^x) dx = 2 - 2cos 3 - e^3 + 1 = 3 - 2cos 3 - e^3
Hence, evaluating the definite integral, yields int_0^3 (2sin x - e^x) dx = 3 - 2cos 3 - e^3.