Concept:If f(x) and g(x) are continuous on [a, b] and g(x) < f(x) for all x in [a, b], then we have the following formula.Area = \(\displaystyle \int_b^a \left[ f(x)?g(x) \right] dx\)Calculation:Given:The line y = x lies above the curve y = x3 in the first quadrant and y = x3 lies above the line y = x in the third quadrant.Solve the curves to obtain the point of intersection y = x3, y = x ? x3 = x.Equating the values of y, we get,? x = x3? x3 - x = 0 ? x(x2 - 1) = 0 ? x (x - 1) (x + 1) = 0 ? x = {0, ± 1}Required area in the first quadrant, A = \(\displaystyle \int^x_0(x -x^3)dx\)? A = \(\displaystyle \int^1_0 x \ dx -\int^1_0 x^3 \ dx\)? A = \(\displaystyle \left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1\)? A = \(\displaystyle \left(\frac{1}{2}-0\right)- \left(\frac{1}{4}-0\right)\)? A = \(\displaystyle\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\)? The area included in the first quadrant between the curves y = x and y = x3 is \(\displaystyle\frac{1}{4}\) square units.