definite integration(2)

0 = x^3 – x = x(x – 1)(x+1) => x = -1, 0 or 1y = x^3 – x intersects with the x-axis (y=0) at(-1, 0), (0, 0) and (1, 0)The portion of the curve from (-1, 0) and (0, 0) is above the x-axisThe portion of the curve from (0, 0) and (1, 0) is below the x-axisThe area of these parts are ∫ (x^3 – x)dx [from -1 to 0] + ∫ (-(x^3 – x)dx [from 0 to 1]= [x^4/4 – x^2/2] from -1 to 0 + [-x^4 /4 +x^2/2] from 0 to 1= 1/4 + 1/4 = 1/2