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