Wednesday

July 30, 2014

July 30, 2014

Posted by **quita** on Wednesday, May 2, 2007 at 4:15pm.

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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