Evaluate the integral of ln x with respect to x from 0 to 1.

Step-by-step solution

1. 1

   Notice that the integral is improper before computing anything. $\ln x\to-\infty$ as $x\to0^+$, so the integrand is unbounded at the lower endpoint. The integral must be defined as a limit:

   $\int_0^1\ln x\,dx=\lim_{a\to0^+}\int_a^1\ln x\,dx$

   Skipping this and plugging in $0$ directly produces the undefined expression $0\cdot\ln 0$.

2. 2

   Find the antiderivative by parts. Take $u=\ln x$ and $dv=dx$, so $du=\tfrac{dx}{x}$ and $v=x$:

   $\int\ln x\,dx=x\ln x-\int x\cdot\frac1x\,dx=x\ln x-x+C$

   Choosing $dv=dx$ is what makes this work: it converts $\ln x$ into $1/x$, which then cancels against $v=x$.

3. 3

   Evaluate on $[a,1]$.

   $\int_a^1\ln x\,dx=\big[x\ln x-x\big]_a^1=(0-1)-(a\ln a-a)=-1-a\ln a+a$

4. 4

   Take the limit at the singular endpoint. The only delicate term is $a\ln a$. Writing it as $\dfrac{\ln a}{1/a}$ and applying L'Hôpital gives $\dfrac{1/a}{-1/a^2}=-a\to0$. So

   $\lim_{a\to0^+}\left(-1-a\ln a+a\right)=-1$

   The integral converges even though the integrand blows up — the singularity is logarithmic, hence integrable.

5. 5

   Sanity-check the sign and size. On $(0,1)$ we have $\ln x<0$, so a negative answer is expected. The mean value of $\ln x$ over $[0,1]$ is therefore $-1$, matching $e^{-1}\approx0.368$ as the point where $\ln x$ equals its average — consistent with the graph ✓.