What is the area bounded by y=x^2 and y=x^3-2x?

Answer 1

Euuw, sketch that
where do they hit?

x^2 = x^3 - 2x
x^3 -x^2 -2x = 0
x(x^2 -x -2) = 0

x (x-2)(x+1) = 0
so crashes at x = -1, 0 and 2
between -1 and 0, the cubic is on top
between 0 and 2, the parabola is on top

so do your integral from -1 to 0 of the cubic minus the parabola
then add the integral from 0 to 2 of the parabola minus the cubic.

Answer 2

so it should be 5/12+8/3=37/12?

Answer 3

must be done in 2 parts

total area
= ∫ (x^3 - 2x - x^2) dx from -1 to 0 + ∫(x^2 - x^3 + 2x) dx from 0 to 2
= [x^4/4 - x^2 - x^3/3]from -1 to 0 + [ x^3/3 - x^4/4 + x^2] from 0 to 2
= (0 - (1/4 - 1 - 1/3) + (8/3 - 4 + 4 - 0)
= 1 - 1/4 + 1/3 + 8/3
= 4 - 1/4
= 15/4

check my arithmetic

Answer 4

yes, 37/12 I get

Answer 5

(0 - (1/4 - 1 + 1/3) note plus 1/3
1 -1/4 -1/3
12/12 -3/12 - 4/12
5/12

then
5/12 +8/3
5/12 + 32/12
37/12