# calc

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find the area of the region bounded by
y=4x, y=x^3, x=0 and x=2

Check this on paper, but in my mind I see the differential area (x^3-3x)dx

integrate that from 0 to 2

y=4x and y=x^3 intersect when x=-2,0 and 2
the vertical boundaries are x=0 and x=2, so
the area is the integral of 4x-x^3 from 0 to 2
which is [2x^2 - (1/4)x^4] from 0 to 2
=8-4 - 0
=4

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