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

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#

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#