Question: The function $f(x)=x^3$ has the derivative $3x^2$ which is equal to $0$ at $x=0$ but the function $f(x)=x^3$ has neither maxima nor minima at that point. How is this possible?

Answer: 
Graph $y=\ln x$. The area under the curve from $x=1$ to $x=x\_0$ is an anti-derivative of $\ln x\_0$.

Make the rectangle with base on the $x$-axis from $x=0$ to $x=x\_0$ and height equal to $\ln x\_0$. The area you want is $x\_0 \ln x\_0$ minus the area above the curve and inside the rectangle. That is:

$$\int\_1^{x\_0} \ln x \; dx = x\_0\ln x\_0 - \int\_0^{\ln x\_0} e^y \; dy.$$
$$\dfrac{d(x^m\ln x)}{dx}=x^{m-1}+mx^{m-1}\ln x$$

$$\implies x^m\ln x=\int x^{m-1}\ dx+m\int(x^{m-1}\ln x)dx$$

Set $m=1$