Answer
$$A = 10$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 2x - 14{\text{; }}\left[ {6,10} \right] \cr
& {\text{First check if the graph crosses the }}x - {\text{axis in the given interval}} \cr
& 2x - 14 = 0 \cr
& {\text{solve for }}x \cr
& 2x = 14 \cr
& x = 7 \cr
& {\text{the graph crosses the }}x{\text{ - axis at }}x = 7{\text{ and the given interval is }}\left[ {6,10} \right],{\text{ }} \cr
& f\left( 6 \right) = 2\left( 6 \right) - 14 = - 2,{\text{ }}f\left( x \right){\text{ is negative for }}\left( {6,7} \right) \cr
& {\text{then}} \cr
& {\text{the definite integral of the total area will be}}: \cr
& A = \left| {\int_6^7 {\left( {2x - 14} \right)} dx} \right| + \int_7^{10} {\left( {2x - 14} \right)} dx \cr
& A = \left| {\left( {{x^2} - 14x} \right)_6^7} \right| + \left( {{x^2} - 14x} \right)_7^{10} \cr
& A = \left| {\left( {{{\left( 7 \right)}^2} - 14\left( 7 \right)} \right) - \left( {{{\left( 6 \right)}^2} - 14\left( 6 \right)} \right)} \right| + \left( {{{\left( {10} \right)}^2} - 14\left( {10} \right)} \right) - \left( {{{\left( 7 \right)}^2} - 14\left( 7 \right)} \right) \cr
& {\text{simplifying}} \cr
& A = \left| { - 49 + 48} \right| - 40 + 49 \cr
& A = 1 - 40 + 49 \cr
& A = 10 \cr} $$
