Calculus with Applications (10th Edition)

$$A = 26$$
\eqalign{ & f\left( x \right) = 4x - 32{\text{; }}\left[ {5,10} \right] \cr & {\text{First check if the graph crosses the }}x - {\text{axis in the given interval}} \cr & 4x - 32 = 0 \cr & {\text{solve for }}x \cr & 4x = 32 \cr & x = 8 \cr & {\text{the graph crosses the }}x{\text{ - axis at }}x = 8{\text{ and the given interval is }}\left[ {5,10} \right],{\text{ }} \cr & f\left( 7 \right) = 4\left( 7 \right) - 32 = - 4,{\text{ }}f\left( x \right){\text{ is negative for }}\left( {5,8} \right) \cr & {\text{then}} \cr & {\text{the definite integral of the total area will be}}: \cr & A = \left| {\int_5^8 {\left( {4x - 32} \right)} dx} \right| + \int_8^{10} {\left( {4x - 32} \right)} dx \cr & A = \left| {\left( {2{x^2} - 32x} \right)_5^8} \right| + \left( {2{x^2} - 32x} \right)_8^{10} \cr & A = \left| {\left( {2{{\left( 8 \right)}^2} - 32\left( 8 \right)} \right) - \left( {2{{\left( 5 \right)}^2} - 32\left( 5 \right)} \right)} \right| + \left( {2{{\left( {10} \right)}^2} - 32\left( {10} \right)} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\, - \left( {2{{\left( 8 \right)}^2} - 32\left( 8 \right)} \right) \cr & {\text{simplifying}} \cr & A = \left| { - 128 + 110} \right| - 120 + 128 \cr & A = 18 - 120 + 128 \cr & A = 26 \cr}