## Calculus: Early Transcendentals (2nd Edition)

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