Question

Please help
Find the area of the region bounded by:
y=x^3-3x, y=x using integration

Answer 1

I believe that the bounds are -2, 2 and that the equation looks like this\[\int\limits_{-2}^{0}[x^3-4x]dx+\int\limits_{0}^{2}[x^3+4x]dx\]

Answer 2

area = the integral from a to b of f(x)=g(x)
to find a and b you need to find where they intersect, where f(x)=g(x) so set the two equations equal to each other, and solve for x, you get that they intersect at 0 and 2. so that is your interval, no you need to integrate x-(x^3-3x) from 0 to 2
the integral is (1/2)*x^2-(1/4)*x^4+(3/2)*x^2. Now plug in the upper value, 2, then subtract the value when you plug in the lower value, 0.

Answer 3

^then double the answer

Answer 4

so the area is 2*[(.5*2^2)-((1/4)*2^4)+(3/2)*(2^2)]

Answer 5

so the answer is 16?

Answer 6

that's not what I got

Answer 7

I got 8