Answer
$\dfrac{2e^3+1}{9} \approx 4.574$
Work Step by Step
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
Here, $u=\ln x$ and $dv=x^2 dx \implies v=\dfrac{x^3}{3}$
$\displaystyle \int_1^{e} x^2 \ln x \ dx=\ln x (\dfrac{x^3}{3})-\int_1^{e} (\dfrac{x^3}{3}) (\dfrac{1}{x}) \ dx$
or, $=\ln x (\dfrac{x^3}{3})-\dfrac{1}{3} \int_1^e x^2 \ dx$
or, $=[ \ln x (\dfrac{x^3}{3})-\dfrac{x^3}{9}]_1^e$
or, $=[ \ln e (\dfrac{e^3}{3})-\dfrac{e^3}{9}]-[ \ln (1) (\dfrac{(1)^3}{3})-\dfrac{(1)^3}{9}]$
Therefore, the required area is: $Area=\dfrac{2e^3+1}{9} \approx 4.574$
