Find the area under the graph of \(f( x) =x\ln x\) from \(1\) to \(2.\)
Then \[ \begin{equation*} du=\frac{1}{x} dx\qquad \hbox{and}\qquad v=\int x dx=\frac{x^{2}}{2} \end{equation*} \]
and \[ \begin{eqnarray*} A=\int_{1}^{2}x\ln x\,dx&=&\left[ \frac{x^{2}}{2}\ln x\right] _{1}^{2}-\int_{1}^{2}\frac{x^{2}}{2}\left( \frac{1}{x} dx\right) =2\ln 2-\dfrac{1}{2}\int_{1}^{2}x\,dx\\[5pt] &=&2\ln 2- \dfrac{1}{2}\left[ \frac{x^{2}}{2}\right] _{1}^{2}=2\ln 2-\dfrac{3}{4} \end{eqnarray*} \]