Chapter 7 - Applications of Integration - 7.1 Exercises - Page 443: 58

$A = \frac{{15}}{2}$

$$\eqalign{ & {\text{Let the points }}A\left( {0,0} \right),{\text{ }}B\left( {1,2} \right){\text{,}}C\left( {3, - 2} \right){\text{ and }}C\left( {1, - 3} \right) \cr & {\text{*Find the equation of the line between the points }}A{\text{ and }}B \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 0 = \frac{{2 - 0}}{{1 - 0}}\left( {x - 0} \right) \cr & y = 2x \cr & \cr & {\text{*Find the equation of the line between the points }}B{\text{ and }}C \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 2 = \frac{{2 + 2}}{{1 - 3}}\left( {x - 1} \right) \cr & y - 2 = - 2\left( {x - 1} \right) \cr & y = - 2x + 4 \cr & \cr & {\text{*Find the equation of the line between the points }}C{\text{ and }}D \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y + 3 = \frac{{ - 2 + 3}}{{3 - 1}}\left( {x - 1} \right) \cr & y + 3 = \frac{1}{2}\left( {x - 1} \right) \cr & y = \frac{1}{2}x - \frac{7}{2} \cr & \cr & {\text{*Find the equation of the line between the points }}A{\text{ and }}D \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y = \frac{{ - 3 - 0}}{{1 - 0}}x \cr & y = - 3x \cr & \cr & {\text{From the image shown below}}{\text{, we can define the area as}} \cr & A = \int_0^1 {\left( {2x + 3x} \right)dx + \int_1^3 {\left[ {\left( { - 2x + 4} \right) - \left( {\frac{1}{2}x - \frac{7}{2}} \right)} \right]} } dx \cr & A = \int_0^1 {5x} dx + \int_1^3 {\left( { - 2x + 4 - \frac{1}{2}x + \frac{7}{2}} \right)} dx \cr & A = \int_0^1 {5x} dx + \int_1^3 {\left( { - \frac{5}{2}x + \frac{{15}}{2}} \right)} dx \cr & {\text{Integrate and evaluate}} \cr & A = \left[ {\frac{5}{2}{x^2}} \right]_0^1 + \left[ { - \frac{5}{4}{x^2} + \frac{{15}}{2}x} \right]_1^3 \cr & A = \frac{5}{2} - \frac{5}{4}{\left( 3 \right)^2} + \frac{{15}}{2}\left( 3 \right) + \frac{5}{4}{\left( 1 \right)^2} - \frac{{15}}{2}\left( 1 \right) \cr & {\text{Simplifying}} \cr & A = \frac{{15}}{2} \cr} $$