What is the integral from 0 to 4 of lnx dx?

Question

Emily Ammerman · Accepted Answer

For  #int_0^4lnx*dx#, it converges by a sum of #4ln4-4#. Let's introduce the idea of improper integrals. Remember that integrals are based on the sums of the individual terms as shown in the Sigma notation below: #int_a^bf(x)dx = lim_(n to oo)sum_(i=0)^nf(x_i)Deltax_i#, for #Deltax =(b-a)/n#,
(From https://tutor.hix.ai) where #a# is the first term, #b# is the last term, and #n# is the number of "parts." Think of it as going from a to b by splitting individual rectangles of area and adding them up.  According to the function #lnx#, if you insert the first term #0# for x, you get no solution since the line on the graph below continues going down to  #-oo#, making #x<=0# undefined:
 graph{lnx [-7.87, 12.13, -4.13, 5.875]} 
Using the Type 2 definition of improper integrals, let's instead make #0# as a #t# limit variable for the integral as #t to 0# from the right (#0^+#): (1) #lim_(t to 0^+)[int_t^4lnx*dx]# To integrate #lnx#, integration by parts comes in handy if you set it as #1*lnx# using the equation: (2) #intuv'=uv-intvu'# (the prime indicates the derivative). Let #u=lnx#, and #v'=1*dx#. By differentiating the #u# and integrating the #v'#, #u'=1/x*dx# and #v=x#. Substituting the #u# and #v# variables for Equation 2, #int (lnx*1)dx=xlnx-int 1*dx#. This allows us to easily integrate #lnx#, only now we need to evaluate its definite integral from #t to 4#: (3) #int_t^4lnx*dx=[lnx(x)~|_t^4-int_t^4 1*dx=(4ln4-tlnt)-(4-t)# Now we can do the math using Eqs. 1 and 3 and split the limit: #lim_(t to 0^+)[int_t^4lnx*dx] = lim_(t to 0^+)[4ln4-tlnt-4+t]# =#lim_(t to 0^+)(-tlnt)+lim_(t to 0^+)(4ln4-4+t)# =#lim_(t to 0^+)(-tlnt)+4ln4-4# Notice though that #lim_(t to 0^+)(-tlnt)# gives us a #-0*oo# indeterminate form. 
Rewriting it would allow us to use l'Hôpital's Rule to find the limit: #lim_(t to 0^+)(-lnt/(1/t))=>-oo/oo=>#indeterminate form. So by LHR,
#lim_(t to 0^+)(-lnt/(1/t))=lim_(t to 0^+)((-1/cancel(t))/(-1/t^cancel(2)))=lim_(t to 0^+)(t)=0# Thus, #int_0^4lnx*dx=4ln4-4#. Hopefully it makes sense with such a long text!

Scarlett Allison · Answer

undefined The integral from 0 to 4 of ln(x) dx is equal to 4 ln(4) - 4.