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October 30, 2014

October 30, 2014

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

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

- Calculus -
**Steve**, Wednesday, April 18, 2012 at 4:37pmwhat'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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