Answer
$$A = \frac{1}{{20}}$$
Work Step by Step
$$\eqalign{
& y = {x^4} + \ln \left( {x + 10} \right),\,\,\,\,\,\,y = {x^3} + \ln \left( {x + 10} \right) \cr
& {\text{Find the points of intersection of the graphs}} \cr
& {x^4} + \ln \left( {x + 10} \right) = {x^3} + \ln \left( {x + 10} \right) \cr
& {x^4} = {x^3} \cr
& x = 0,\,\,x = 1 \cr
& {\text{Then with the critical value we have the intervals }}\left( {0,1} \right) \cr
& {\text{For the interval }}\left( {0,1} \right){\text{ we have that }}{x^3} + \ln \left( {x + 10} \right) > {x^4} + \ln \left( {x + 10} \right) \cr
& \cr
& {\text{The area between the curves is given by}}: \cr
& A = \int_0^1 {\left( {{x^3} + \ln \left( {x + 10} \right) - {x^4} + \ln \left( {x + 10} \right)} \right)} dx \cr
& A = \int_0^1 {\left( {{x^3} - {x^4}} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left( {\frac{1}{4}{x^4} - \frac{1}{5}{x^5}} \right)_0^1 \cr
& A = \left( {\frac{1}{4}{{\left( 1 \right)}^4} - \frac{1}{5}{{\left( 1 \right)}^5}} \right) - \left( {\frac{1}{4}{{\left( 0 \right)}^4} - \frac{1}{5}{{\left( 0 \right)}^5}} \right) \cr
& A = \frac{1}{4} - \frac{1}{5} - 0 \cr
& A = \frac{1}{{20}} \cr} $$
