We have
#f(x) = xln(x)#
Its derivative can be found with the product rule:
#f'(x) = d/dx{x}ln(x)+xd/dx{ln(x)}#
#=1*ln(x)+x*1/x#
#=ln(x)+1#
Now that we have #f'(x)# we can use this to evaluate the integral:
#int_e^(2e)ln(x)dx#
Now we use a simple trick of adding #1# and subtracting #1# like this:
#=int_e^(2e)ln(x)+1-1dx#
We can now break this integral down like so:
#=int_e^(2e)ln(x)+1dx+int_e^(2e)-1dx#
#=int_e^(2e)ln(x)+1dx-int_e^(2e)1dx#
We already have the answer to the first integral from the first part of the question: the expression inside the integral is what we found for #f'(x)# so the integral will simply be #f(x)#. The second integral will of course just be #-x#. So evaluating the integral and its limits yiels:
#=[xln(x)]_e^(2e)-[x]_e^(2e)#
#={(2e)ln(2e)}-{eln(e)}-({2e}-{e})#
#2eln(2e)-e-2e+e#
#2e(ln(2e)-1)=2e(ln2+lne-1)#
#2eln(2)~~3.76834#
