Answer
$$A = \frac{{81\ln 3 - 20}}{4}$$
Work Step by Step
$$\eqalign{
& y = {x^3}\ln x,{\text{ }}y = 0,{\text{ }}x = 1,{\text{ }}x = 3 \cr
& {\text{From the graph we can define the area as}} \cr
& A = \int_1^3 {{x^3}\ln x} dx \cr
& {\text{Integrating by the formula of exercise 69}} \cr
& \int {{x^n}\ln x} dx = \frac{{{x^{n + 1}}}}{{{{\left( {n + 1} \right)}^2}}}\left[ { - 1 + \left( {n + 1} \right)\ln x} \right] + C \cr
& A = \int_1^3 {{x^3}\ln x} dx \cr
& n = 3 \cr
& A = \left( {\frac{{{x^{3 + 1}}}}{{{{\left( {3 + 1} \right)}^2}}}\left[ { - 1 + \left( {3 + 1} \right)\ln x} \right]} \right)_1^3 \cr
& A = \left( {\frac{{{x^4}}}{{16}}\left[ {4\ln x - 1} \right]} \right)_1^3 \cr
& {\text{Evaluating}} \cr
& A = \left( {\frac{{{3^4}}}{{16}}\left[ {4\ln 3 - 1} \right]} \right) - \left( {\frac{{{1^4}}}{{16}}\left[ {4\ln 1 - 1} \right]} \right) \cr
& A = \left( {\frac{{81}}{{16}}\left[ {4\ln 3 - 1} \right]} \right) - \left( {\frac{{{1^4}}}{{16}}\left[ { - 1} \right]} \right) \cr
& A = \frac{{81}}{{16}}\left[ {4\ln 3 - 1} \right] + \frac{1}{{16}} \cr
& A = \frac{{324\ln 3}}{{16}} - \frac{{81}}{{16}} + \frac{1}{{16}} \cr
& A = \frac{{324\ln 3 - 80}}{{16}} \cr
& A = \frac{{81\ln 3 - 20}}{4} \cr} $$
