Question

Integral 0 sampai 1/2phi 3(3sin2x-cosx) dx=

Answer 1

Π/2
∫   9sin 2x -  3cos x) dx
0.
.
= [- 9/2 cos 2x - 3 sin x ] (π/2 ...0)
= -9/2 {cos π- cos 0}  - 3(sin π/2 - sin 0)
= -9/2(-1 - 1) - 3 (1 - 0)
= -9/2(-2)  - 3(1)
= 9 - 3
= 6