91Ó°ÊÓ

To determine the points of intersection between the curves y = x^2 and y = x^3, we will set them equal to each other. \begin{align*} x^2 &= x^3 \\ 0 &= x^3 - x^2 \\ x^2(x - 1) &= 0 \end{align*} The curves intersect at x = 0 and x = 1.

For the points between the intersections, we can plug in a test value, such as \(x = 0.5\), to see which curve is above the other. \begin{align*} y_1 &= 0.5^2 = 0.25 \\ y_2 &= 0.5^3 = 0.125 \\ \end{align*} Since \(y_1 > y_2\), the curve y = x^2 is above y = x^3 between x = 0 and x = 1.

To find the area between the two curves, we will evaluate the definite integral from x = -1 to x = 0 for the leftmost vertical boundary line, and between x = 0 and x = 2 for the rightmost vertical boundary line. We will integrate the difference between the two curves to find the total area enclosed. For x = -1 to x = 0, \begin{align*} A_1 &= \int_{-1}^{0} (x^3 - x^2) dx \\ &= \left[\frac{1}{4}x^4 - \frac{1}{3}x^3 \right]_{-1}^0 \\ &= \frac{1}{12} \end{align*} For x = 0 to x = 1, \begin{align*} A_2 &= \int_{0}^{1} (x^2 - x^3) dx \\ &= \left[\frac{1}{3}x^3 - \frac{1}{4}x^4 \right]_{0}^1 \\ &= \frac{1}{12} \end{align*} For x = 1 to x = 2, \begin{align*} A_3 &= \int_{1}^{2} (x^3 - x^2) dx \\ &= \left[\frac{1}{4}x^4 - \frac{1}{3}x^3 \right]_{1}^2 \\ &= \frac{4}{3} \end{align*}

Finally, we will add the area of all intervals to find the total area of the region R. $$ A = A_1 + A_2 + A_3 = \frac{1}{12} + \frac{1}{12} + \frac{4}{3} = \frac{17}{6} $$ The area of the enclosed region R is \(\frac{17}{6}\).