Calculus
posted by Michael .
Find the area of the region enclosed by the given curves:
y=e^6x, y=2sin(x), x=0, x=pi/2

what's the problem?
e^6x > 2sinx, so the area is just
∫[0,pi/2](e^6x  2sinx) dx
= 1/6 e^6x + 2cosx [0,pi/2]
= (1/6e^3pi + 0)  (1/6 + 2)]
= 1/6(e^3pi  13)