evaluate the following integral. \n\n∫₁ᵉ² x² ln(x) dx

Explanation:

Step1: Apply integration - by - parts

The formula for integration by parts is $\int u;dv=uv-\int v;du$. Let $u = \ln(x)$ and $dv=x^{2}dx$. Then $du=\frac{1}{x}dx$ and $v=\frac{1}{3}x^{3}$. So, $\int x^{2}\ln(x)dx=\frac{1}{3}x^{3}\ln(x)-\int\frac{1}{3}x^{3}\cdot\frac{1}{x}dx=\frac{1}{3}x^{3}\ln(x)-\frac{1}{3}\int x^{2}dx$.

Step2: Integrate $x^{2}$

We know that $\int x^{2}dx=\frac{1}{3}x^{3}+C$. So, $\int x^{2}\ln(x)dx=\frac{1}{3}x^{3}\ln(x)-\frac{1}{9}x^{3}+C$.

Step3: Evaluate the definite integral

$\int_{1}^{e^{2}}x^{2}\ln(x)dx=\left[\frac{1}{3}x^{3}\ln(x)-\frac{1}{9}x^{3}\right]_{1}^{e^{2}}$. First, substitute $x = e^{2}$: $\frac{1}{3}(e^{2})^{3}\ln(e^{2})-\frac{1}{9}(e^{2})^{3}=\frac{1}{3}e^{6}\cdot2-\frac{1}{9}e^{6}=\frac{2}{3}e^{6}-\frac{1}{9}e^{6}=\frac{6e^{6}-e^{6}}{9}=\frac{5}{9}e^{6}$. Then substitute $x = 1$: $\frac{1}{3}(1)^{3}\ln(1)-\frac{1}{9}(1)^{3}=0 - \frac{1}{9}=-\frac{1}{9}$. Finally, $\frac{5}{9}e^{6}-\left(-\frac{1}{9}\right)=\frac{5e^{6}+1}{9}$.

Answer:

$\frac{5e^{6}+1}{9}$