Answer
$$A = 26$$
Work Step by Step
$$\eqalign{
& x = 0,\,\,\,\,\,x = 6,\,\,\,\,\,y = 5x,\,\,\,\,\,\,y = 3x + 10 \cr
& {\text{Find the points of intersection of the graphs}} \cr
& 5x = 3x + 10 \cr
& 2x = 10 \cr
& x = 5 \cr
& {\text{Then with the given points and the critical value }}x = 5{\text{ we have the intervals}} \cr
& \left( {0,5} \right){\text{ and }}\left( {5,6} \right) \cr
& {\text{For the interval }}\left( {0,5} \right){\text{ we have that }}3x + 10 \geqslant 5x{\text{ and for the interval }}\left( {5,6} \right) \cr
& 5x \geqslant 3x + 10 \cr
& {\text{The area is given by}}: \cr
& A = \int_0^5 {\left( {3x + 10 - 5x} \right)} dx + \int_5^{10} {\left( {5x - 3x - 10} \right)} dx \cr
& A = \int_0^5 {\left( {10 - 2x} \right)} dx + \int_5^{10} {\left( {2x - 10} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left( {10x - {x^2}} \right)_0^5 + \left( {{x^2} - 10x} \right)_5^6 \cr
& A = \left( {10\left( 5 \right) - {{\left( 5 \right)}^2}} \right) - \left( {10\left( 0 \right) - {{\left( 0 \right)}^2}} \right) + \left( {{{\left( 6 \right)}^2} - 10\left( 6 \right)} \right) - \left( {{{\left( 5 \right)}^2} - 10\left( 5 \right)} \right) \cr
& A = 25 - 24 + 25 \cr
& A = 26 \cr} $$
