Calculus 2
posted by Rebecca .
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 
Steve
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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