Let R be the region in the first quadrant bounded by the cur

Solution

The points they intersect can be found by setting them equal to each other: x^3 = 2x-x^2 => x^2 = 2 - x => x^2 + x - 2 = 0 => (x+2)(x-1) = 0 so x = -2 or x = 1. Since we divided both sides by x in the second step, we must take into account when x = 0. Since we want the first quadrant, we just take the integral between 0 and 1. To find the area of two functions, you take the integral of the higher function subtracted by the lower one. To find out which is the greater one, just plug in a point between the bounds and whichever gives you the greater value is the greater function. Taking x=.5, the first function is .125 and the second is .75. So integral between 0 and 1 of (2x-x^2) - x^3 dx = x^2 - (1/3)x^3 - (1/4) x^4 from 0 to 1 = (1 - (1/3) - (1/4)) - (0 - 0 - 0) = 2/3 - 1/4 = 5/12