Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.1 Area Between Two Curves - Exercises Set 5.1 - Page 353: 6

$$\eqalign{ & \left( {\text{a}} \right)A = 9 \cr & \left( {\text{b}} \right)A = 9 \cr} $$

$$\eqalign{ & \left( {\text{a}} \right){\text{Integrating with respect to }}x \cr & {y^2} = 4x \Rightarrow y = \sqrt {4x} = 2\sqrt x \cr & y = 2x - 4 \cr & 2\sqrt x \geqslant 2x - 4{\text{ on the interval 1}} \leqslant x \leqslant {\text{4}} \cr & {\text{The area is given by}} \cr & A = 2\int_0^1 {2\sqrt x dx} + \int_1^4 {\left( {2\sqrt x - \left( {2x - 4} \right)} \right)} dx \cr & A = 4\int_0^1 {{x^{1/2}}dx} + \int_1^4 {\left( {2{x^{1/2}} - 2x + 4} \right)} dx \cr & {\text{Integrate}} \cr & A = 4\left[ {\frac{2}{3}{x^{3/2}}} \right]_0^1 + \left[ {\frac{4}{3}{x^{3/2}} - {x^2} + 4x} \right]_1^4 \cr & A = \frac{8}{3}\left[ {{{\left( 1 \right)}^{3/2}}} \right] + \left[ {\frac{4}{3}{{\left( 4 \right)}^{3/2}} - {{\left( 4 \right)}^2} + 4\left( 4 \right)} \right] - \left[ {\frac{4}{3}{{\left( 1 \right)}^{3/2}} - {{\left( 1 \right)}^2} + 4\left( 1 \right)} \right] \cr & {\text{Simplifying}} \cr & A = \frac{8}{3} + \frac{{32}}{3} - \frac{{13}}{3} \cr & A = 9 \cr & \cr & \left( {\text{b}} \right){\text{Integrating with respect to }}y \cr & {y^2} = 2x \Rightarrow x = \frac{1}{4}{y^2} \cr & y = 2x - 4 \Rightarrow x = \frac{{y + 4}}{2} \cr & \frac{{y + 4}}{2} \geqslant \frac{1}{4}{y^2}y{\text{ on the interval }} - 2 \leqslant y \leqslant {\text{4}} \cr & {\text{The area is given by}} \cr & A = \int_{ - 2}^4 {\left( {\frac{1}{2}y + 2 - \frac{1}{4}{y^2}} \right)} dy \cr & A = \left[ {\frac{1}{4}{y^2} + 2y - \frac{1}{{12}}{y^3}} \right]_{ - 2}^4 \cr & A = \left[ {\frac{1}{4}{{\left( 4 \right)}^2} + 2\left( 4 \right) - \frac{1}{{12}}{{\left( 4 \right)}^3}} \right] - \left[ {\frac{1}{4}{{\left( { - 2} \right)}^2} + 2\left( { - 2} \right) - \frac{1}{{12}}{{\left( { - 2} \right)}^3}} \right] \cr & A = \frac{{20}}{3} + \frac{7}{3} \cr & A = 9 \cr} $$