# How do you find the integral of xln(x) dx from 0 to 1?

##### 3 Answers
Apr 14, 2015

$- \frac{1}{4}$

Solution

${\int}_{0}^{1} x \ln x \mathrm{dx}$

Integrating by parts

$\ln x . {x}^{2} / 2 {|}_{0}^{1} - {\int}_{0}^{1.} \left(\frac{1}{x} \left({x}^{2} / 2\right)\right) \mathrm{dx}$

$= \ln 1 {\left(1\right)}^{2} / 2 - \ln 0. {\left(0\right)}^{2} / 2 - \frac{1}{2} {\int}_{0}^{1} x \mathrm{dx}$

$= 0 \left(\frac{1}{2}\right) - 0 - {x}^{2} / 4 {|}_{0}^{1}$

$= 0 - 0 - {1}^{2} / 4 + {0}^{2} / 4$

$= 0 - \frac{1}{4} + 0$

$= - \frac{1}{4}$

Apr 14, 2015

Because $\ln \left(0\right)$ is not defined (does not exist), $0$ is not in the domain of the integrand.

This is an improper integral, so we use:

${\int}_{0}^{1} x \ln \left(x\right) \mathrm{dx} = {\lim}_{a \rightarrow 0} {\int}_{a}^{1} x \ln \left(x\right) \mathrm{dx}$

Use integration by parts (with $u = \ln x$ and $\mathrm{dv} = x$) to get:

lim_(a rarr 0) int _a^1 x ln(x) dx = lim_(a rarr 0) (1/2 x^2 lnx - 1/4 x^2]_a^1)

Recalling that $\ln \left(1\right) = 0$, we get:

${\int}_{0}^{1} x \ln \left(x\right) \mathrm{dx} = {\lim}_{a \rightarrow 0} \left(\left(0 - \left(\frac{1}{4}\right)\right) - {a}^{2} / 2 \ln a - \frac{1}{4} {a}^{2}\right)$

${\lim}_{a \rightarrow 0} \frac{1}{4} {a}^{2} = 0$,

But, ${\lim}_{a \rightarrow 0} \left({a}^{2} / 2 \ln a\right)$ has indeterminate form: $0 \cdot - \infty$.

Rewriting as: $\ln \frac{a}{\frac{2}{a} ^ 2}$ Which has indeterminate form: $\frac{- \infty}{\infty}$.

Applying l'Hopital's Rule gives us:

${\lim}_{a \rightarrow 0} \left({a}^{2} / 2 \ln a\right) = {\lim}_{a \rightarrow 0} \ln \frac{a}{\frac{2}{a} ^ 2} = {\lim}_{a \rightarrow 0} \frac{\frac{1}{a}}{\frac{- 4}{a} ^ 3} = {\lim}_{a \rightarrow 0} \left({a}^{2} / - 4\right) = 0$

So, we conclude:

${\int}_{0}^{1} x \ln \left(x\right) \mathrm{dx} = - \frac{1}{4} - 0 - 0 = - \frac{1}{4}$

Apr 15, 2015

The integration by parts of x ln(x) would be as follows:

ln(x) $\left({x}^{2} / 2\right)$ - $\int$$\frac{1}{x}$ $\left({x}^{2} / 2\right)$ dx

ln(x) $\left({x}^{2} / 2\right)$ -$\int$ $\frac{x}{2}$dx

ln(x) $\left({x}^{2} / 2\right)$ - ${x}^{2} / 4$

On taking limits from 0 to 1, the final answer would be $- \frac{1}{4}$