Partition the interval [0,1] as
\(\left[0,\dfrac15\right]\cup\left[\dfrac15,\dfrac25\right]\cup\left[\dfrac25,\dfrac35\right]\cup\left[\dfrac35,\dfrac45\right]\cup\left[\dfrac45,5\right]\)
The midpoints of the intervals are, respectively \(\dfrac1{10},\dfrac3{10},\dfrac12,\dfrac7{10},\dfrac9{10}\) - these are your sample points.
The integral is approximated by
\(\displaystyle\int_0^1x^2\,\mathrm dx\approx\sum_{n=1}^5f(x_n)\Delta x\)
where \(\Delta x\) is the difference between the partition endpoints, i.e. \(\Delta x=\dfrac{1-0}5=\dfrac15\), and \(x_n\) is the midpoint of the \(n\)th partition. You have
\(\displaystyle\sum_{n=1}^5f(x_n)\Delta x=\frac15\left(\left(\frac1{10}\right)^2+\left(\frac3{10}\right)^2+\left(\frac12\right)^2+\left(\frac7{10}\right)^2+\left(\frac9{10}\right)^2\right)=\dfrac{33}{100}\)
For comparison, the actual value of the integral is \(\dfrac13\), so the approximation is valid to two decimal places.
