Sunday

December 21, 2014

December 21, 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

**Answer this Question**

**Related Questions**

Calc - Problems, once again. 1. Compute the average value of: f(x} = x/(x+3) ...

calculus - 1. Find the area of the region bounded by f(x)=x^2 +6x+9 and g(x)=5(x...

math - Sketch the region enclosed by the given curves. Decide whether to ...

Math- Calc - Sketch the region enclosed by the given curves. Decide whether to ...

calc check - <<y=(1/A)*integral from a to b of: (1/2)[f(x)]^2 dx >> ...

Calc - Find the area of the region bounded by y=x^2 and y = -(x-4)^2 +4 and the ...

calculus - 1. Let S be the region in the first quadrant bounded by the graphs of...

Calculus - Find the area of the region bounded by the line y=3x and y= x^3 + 2x^...

Calc - finding area bounded by curve - Find the area bounded by x=cubed root of ...

Calculus (Area Between Curves) - Find the area of the region bounded by the ...