Posted by Michael on Wednesday, April 18, 2012 at 1:43pm.
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)
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