Calculus

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Find the area of the region enclosed by the given curves:

y=e^6x, y=2sin(x), x=0, x=pi/2

  • Calculus -

    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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