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