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

3 Answers
Apr 14, 2015

-1/414

Solution

int_0^1xlnxdx10xlnxdx

Integrating by parts

lnx.x^2/2|_0^1-int_0^1.(1/x(x^2/2))dxlnx.x221010.(1x(x22))dx

=ln1(1)^2/2-ln0.(0)^2/2-1/2int_0^1xdx=ln1(1)22ln0.(0)221210xdx

=0(1/2)-0-x^2/4|_0^1=0(12)0x2410

=0-0-1^2/4+0^2/4=00124+024

=0-1/4+0=014+0

=-1/4=14

Apr 14, 2015

Because ln(0)ln(0) is not defined (does not exist), 00 is not in the domain of the integrand.

This is an improper integral, so we use:

int _0^1 x ln(x) dx = lim_(a rarr 0) int _a^1 x ln(x) dx

Use integration by parts (with u = lnx and 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(1)=0, we get:

int _0^1 x ln(x) dx = lim_(a rarr 0) ((0-(1/4))- a^2/2 lna - 1/4 a^2)

lim_(a rarr 0) 1/4 a^2 = 0,

But, lim_(a rarr 0) ( a^2/2 lna ) has indeterminate form: 0 * -oo.

Rewriting as: lna/(2/a^2) Which has indeterminate form: (-oo)/oo.

Applying l'Hopital's Rule gives us:

lim_(a rarr 0) ( a^2/2 lna ) = lim_(a rarr 0) lna/(2/a^2) = lim_(a rarr 0) (1/a)/((-4)/a^3) = lim_(a rarr 0)(a^2 / -4) = 0

So, we conclude:

int _0^1 x ln(x) dx = -1/4 - 0 - 0 = -1/4

Apr 15, 2015

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

ln(x) (x^2/2) - int1/x (x^2/2) dx

ln(x) (x^2/2) -int x/2dx

ln(x) (x^2/2) - x^2/4

On taking limits from 0 to 1, the final answer would be -1/4