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

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Answer 1 · 2015-07-14T17:36:00

I found: #int_1^0xln(x)dx=1/4#

You can integrat by Parts to get:
#intxln(x)dx=x^2/2ln(x)-intx^2/2*1/xdx=#
#=x^2/2ln(x)-intx/2dx=x^2/2ln(x)-x^2/4|_1^0##=#
#=0-(-1/4)=1/4#

Answer 2 · 2015-07-16T06:52:53

You can do this slightly differently as well.

#int_(1)^0 xlnxdx#

From the Fundamental Theorem of Calculus and basic properties of Integrals, where
#int_a^b f(x)dx = -int_b^a f(x)dx#

# = F(b) - F(a) = -(F(a) - F(b))#

#=> -int_(0)^1 xlnxdx#

Using Integration by Parts, let:
#u = lnx#
#du = 1/xdx#
#dv = xdx#
#v = x^2/2#

#uv - intvdu#

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

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

#= [x^2/4 - (x^2lnx)/2]|_(0)^(1)#

#= [1^2/4 - cancel((1^2ln1)/2)] - [0^2/4 - (0^2ln0)/2]#

#= 1/4 + 0*ln0#

#= 1/4 + 0*(-oo)#

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

#lim_(x->0) (x^2lnx)/2 = lim_(x->0) lnx/(2x^(-2))#

#= lim_(x->0) (1/x)/(-4/x^3) = lim_(x->0) -x^3/(4x)#

#= lim_(x->0) -x^2/4 = 0#

Thus, the answer is #1/4#.