# How do you find the integral of x ln x from (1,0)?

Jul 14, 2015

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

#### Explanation:

You can integrat by Parts to get:
$\int x \ln \left(x\right) \mathrm{dx} = {x}^{2} / 2 \ln \left(x\right) - \int {x}^{2} / 2 \cdot \frac{1}{x} \mathrm{dx} =$
$= {x}^{2} / 2 \ln \left(x\right) - \int \frac{x}{2} \mathrm{dx} = {x}^{2} / 2 \ln \left(x\right) - {x}^{2} / 4 {|}_{1}^{0}$$=$
$= 0 - \left(- \frac{1}{4}\right) = \frac{1}{4}$

Jul 16, 2015

You can do this slightly differently as well.

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

From the Fundamental Theorem of Calculus and basic properties of Integrals, where
${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = - {\int}_{b}^{a} f \left(x\right) \mathrm{dx}$

$= F \left(b\right) - F \left(a\right) = - \left(F \left(a\right) - F \left(b\right)\right)$

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

Using Integration by Parts, let:
$u = \ln x$
$\mathrm{du} = \frac{1}{x} \mathrm{dx}$
$\mathrm{dv} = x \mathrm{dx}$
$v = {x}^{2} / 2$

$u v - \int v \mathrm{du}$

= -[(x^2lnx)/2 - int x^2/2*1/xdx]|_(0)^(1

= -[(x^2lnx)/2 - 1/2int xdx]|_(0)^(1

$= \left[{x}^{2} / 4 - \frac{{x}^{2} \ln x}{2}\right] {|}_{0}^{1}$

$= \left[{1}^{2} / 4 - \cancel{\frac{{1}^{2} \ln 1}{2}}\right] - \left[{0}^{2} / 4 - \frac{{0}^{2} \ln 0}{2}\right]$

$= \frac{1}{4} + 0 \cdot \ln 0$

$= \frac{1}{4} + 0 \cdot \left(- \infty\right)$

Let's do the limit instead, on the right term, due to an indeterminate form, and then L'Hopital's Rule.

${\lim}_{x \to 0} \frac{{x}^{2} \ln x}{2} = {\lim}_{x \to 0} \ln \frac{x}{2 {x}^{- 2}}$

$= {\lim}_{x \to 0} \frac{\frac{1}{x}}{- \frac{4}{x} ^ 3} = {\lim}_{x \to 0} - {x}^{3} / \left(4 x\right)$

$= {\lim}_{x \to 0} - {x}^{2} / 4 = 0$

Thus, the answer is $\frac{1}{4}$.