Answer
\[ = - \frac{1}{4}\]
Work Step by Step
\[\begin{gathered}
\int_0^1 {x\ln x\,dx} \hfill \\
\hfill \\
use\,\,the\,\,Formula\,\,for\,\,in\,tegration\,\,by\,\,parts \hfill \\
\hfill \\
\int_{}^{} {udv = uv - \int_{}^{} {vdu} } \hfill \\
\hfill \\
set \hfill \\
du = \ln x\,\,\,\,\,\,\,then\,\,\,\,\,\,du = \frac{1}{x}dx \hfill \\
dv = xdx\,\,\,\,\,\,\,then\,\,\,v = \frac{{{x^2}}}{2} \hfill \\
\hfill \\
\int_{}^{} {x\ln x\,dx} = \left[ {\frac{{{x^2}}}{2}\ln x - \int_{}^{} {\frac{x}{2}dx} } \right] \hfill \\
\hfill \\
integrating \hfill \\
\hfill \\
\int_0^1 {x\ln x\,dx} \, = \,\,\left[ {\frac{{{x^2}}}{2}\ln x - \frac{{{x^2}}}{4}} \right] \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\mathop {\lim }\limits_{a \to 0} \int_a^1 {x\ln xdx} = \mathop {\lim }\limits_{a \to 0} \,\,\left[ {\frac{{{x^2}}}{2}\ln x - \frac{{{x^2}}}{4}} \right]_a^1 \hfill \\
\hfill \\
use\,\,the\,\,ftc \hfill \\
\hfill \\
= \mathop {\lim }\limits_{a \to 0} \,\,\left[ { - \frac{1}{4} - \frac{{{a^2}}}{2}\ln a + \frac{{{a^2}}}{4}} \right] \hfill \\
\hfill \\
evaluate\,\,the\,\,limit \hfill \\
\hfill \\
= - \frac{1}{4} \hfill \\
\hfill \\
\end{gathered} \]
