Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.1 Area Between Two Curves - Exercises Set 5.1 - Page 353: 7

$$A = \frac{{49}}{{192}}$$

$$\eqalign{ & {\text{From the graph we can note that the area is given by}} \cr & A = \int_{1/4}^1 {\left( {\sqrt x - {x^2}} \right)} dx \cr & {\text{Integrate}} \cr & A = \left[ {\frac{2}{3}{x^{3/2}} - \frac{1}{3}{x^3}} \right]_{1/4}^1 \cr & {\text{Evaluate}} \cr & A = \left[ {\frac{2}{3}{{\left( 1 \right)}^{3/2}} - \frac{1}{3}{{\left( 1 \right)}^3}} \right] - \left[ {\frac{2}{3}{{\left( {\frac{1}{4}} \right)}^{3/2}} - \frac{1}{3}{{\left( {\frac{1}{4}} \right)}^3}} \right] \cr & {\text{Simplify}} \cr & A = \frac{1}{3} - \frac{5}{{64}} \cr & A = \frac{{49}}{{192}} \cr} $$