# Calculus 2

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Sketch the region enclosed by y=e^2x, y=e^6x, and x=1. Decide whether to integrate with respect to x or y. Then find the area of the region.

• Calculus 2 -

since e^6x is greater than e^2x on [0,1], I'd integrate with respect to x.

Int(e^6x - e^2x dx)[0,1]
= 1/6 e^6x - 1/2 e^2x [0,1]
= (1/6 e^6 - 1/2 e^2) - (1/6 - 1/2)
= 1/6 e^2 (e^4 - 3) + 1/3
= 63.8

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