Answer
$$A = \frac{1}{6}$$
Work Step by Step
$$\eqalign{
& {\text{The region is given by }}y = {\left( {x - 1} \right)^2}{\text{ and }}y = 7x - 19 \cr
& {\text{Find the intersection points, let }}y = y \cr
& {\left( {x - 1} \right)^2} = 7x - 19 \cr
& {x^2} - 2x + 1 = 7x - 19 \cr
& {x^2} - 9x + 20 = 0 \cr
& \left( {x - 5} \right)\left( {x - 4} \right) = 0 \cr
& {x_1} = 5,\,\,\,{x_2} = 4 \cr
& {\text{Where }}7x - 19 \geqslant {\left( {x - 1} \right)^2}{\text{ on the interval }}\left[ {4,5} \right] \cr
& \cr
& {\text{The area between the curves is given by}} \cr
& A = \int_4^5 {\left[ {7x - 19 - {{\left( {x - 1} \right)}^2}} \right]} dx \cr
& A = \int_4^5 {\left( {7x - 19 - {x^2} + 2x - 1} \right)} dx \cr
& A = \int_4^5 {\left( {9x - {x^2} - 20} \right)} dx \cr
& {\text{Integrate}} \cr
& A = \left[ {\frac{{9{x^2}}}{2} - \frac{{{x^3}}}{3} - 20x} \right]_4^5 \cr
& A = \left[ {\frac{{9{{\left( 5 \right)}^2}}}{2} - \frac{{{{\left( 5 \right)}^3}}}{3} - 20\left( 5 \right)} \right] - \left[ {\frac{{9{{\left( 4 \right)}^2}}}{2} - \frac{{{{\left( 4 \right)}^3}}}{3} - 20\left( 4 \right)} \right] \cr
& A = - \frac{{175}}{6} + \frac{{88}}{3} \cr
& A = \frac{1}{6} \cr} $$
