Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises - Page 443: 41

$A = \frac{5}{2}$

$$\eqalign{ & \left( {0,0} \right),{\text{ }}\left( {3,1} \right),{\text{ }}\left( {1,2} \right) \cr & {\text{Find the equation between the points }}\left( {0,0} \right){\text{ and }}\left( {1,2} \right) \cr & m = \frac{{2 - 0}}{{1 - 0}} \cr & m = 2 \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 0 = 2\left( {x - 0} \right) \cr & y = 2x \cr & {\text{Find the equation between the points }}\left( {0,0} \right){\text{ and }}\left( {3,1} \right) \cr & m = \frac{{1 - 0}}{{3 - 0}} \cr & m = \frac{1}{3} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 0 = \frac{1}{3}\left( {x - 0} \right) \cr & y = \frac{1}{3}x \cr & {\text{Find the equation between the points }}\left( {1,2} \right){\text{ and }}\left( {3,1} \right) \cr & m = \frac{{1 - 2}}{{3 - 1}} \cr & m = - \frac{1}{2} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 2 = - \frac{1}{2}\left( {x - 1} \right) \cr & y - 2 = - \frac{1}{2}x + \frac{1}{2} \cr & y = - \frac{1}{2}x + \frac{5}{2} \cr & \cr & {\text{The area is given by}} \cr & A = \int_0^1 {\left( {2x - \frac{1}{3}x} \right)} dx + \int_1^3 {\left[ {\left( { - \frac{1}{2}x + \frac{5}{2}} \right) - \frac{1}{3}x} \right]} dx \cr & A = \frac{5}{3}\int_0^1 x dx + \int_1^3 {\left( { - \frac{5}{6}x + \frac{5}{2}} \right)} dx \cr & {\text{Integrate}} \cr & A = \frac{5}{3}\left[ {\frac{{{x^2}}}{2}} \right]_0^1 + \left[ { - \frac{5}{{12}}{x^2} + \frac{5}{2}x} \right]_1^3 \cr & {\text{Evaluate}} \cr & A = \frac{5}{3}\left[ {\frac{{{{\left( 1 \right)}^2}}}{2} - 0} \right] + \left[ { - \frac{5}{{12}}{{\left( 3 \right)}^2} + \frac{5}{2}\left( 3 \right)} \right] - \left[ { - \frac{5}{{12}}{{\left( 1 \right)}^2} + \frac{5}{2}\left( 1 \right)} \right] \cr & A = \frac{5}{6} + \frac{{15}}{4} - \frac{{25}}{{12}} \cr & A = \frac{5}{2} \cr} $$