Question

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

Answer 1

`-1/4`

Solution

`int_0^1xlnxdx`

Integrating by parts

`lnx.x^2/2|_0^1-int_0^1.(1/x(x^2/2))dx`

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

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

`=0-0-1^2/4+0^2/4`

`=0-1/4+0`

`=-1/4`

Answer 2

Because `ln(0)` 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(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`

Answer 3

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

ln(x) `(x^2/2)` - `int` `1/x` `(x^2/2)` dx

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

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

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