How do you evaluate the integral #int lnxdx# from #[0,1]#?

Question

Samantha Adkison · Accepted Answer

#int_0^1ln(x)\ dx=-1# First we need to work out the antiderivative. To do this, we will use integration by parts. I will let #f=ln(x)# and #g'=1#. We know: #f'=1/x# #g=x# So we can rewrite the integral like so: #int\ ln(x)\ dx=xln(x)-int\ x*1/x\ dx=xln(x)-int\ 1\ dx# #=xln(x)-x# Now we know the antiderivative, we can plug in the limits of integration to compute the definite integral: #int_0^1 ln(x)\ dx=[xln(x)-x]_0^1=1ln(1)-1-(0ln(0)-0)# This gives us a problem, since #ln(0)# isn't defined, but we can get around this by taking the limit as the value approaches #0#: #lim_(x->0)xln(x)=0# This is quite clearly equal to #0# (because #0# multiplied by anything is #0#), so we can now carry on evaluating our definite integral: #ln(1)-1-(0)=0-1-0=-1# So, the answer is: #int_0^1ln(x)\ dx=-1#

William Abdalla · Answer

undefined To evaluate the integral of ln(x) from 0 to 1, we use integration by parts. Let u = ln(x) and dv = dx. Then, du = (1/x) dx and v = x.

Applying the integration by parts formula ∫u dv = uv - ∫v du, we have:

∫ln(x) dx = x ln(x) - ∫x (1/x) dx
          = x ln(x) - ∫dx
          = x ln(x) - x + C

To evaluate the definite integral from 0 to 1, we substitute the limits of integration:

∫[0,1] ln(x) dx = [1 ln(1) - 1] - [0 ln(0) - 0]
                = [0 - 1] - [0 - 0]
                = -1 - 0
                = -1