Split up the interval [0, 2] into 4 subintervals, so that
[tex][0,2]=\left[0,\dfrac12\right]\cup\left[\dfrac12,1\right]\cup\left[1,\dfrac32\right]\cup\left[\dfrac32,2\right][/tex]
Each subinterval has width [tex]\dfrac{2-0}4=\dfrac12[/tex]. The area of the trapezoid constructed on each subinterval is [tex]\dfrac{f(x_i)+f(x_{i+1})}4[/tex], i.e. the average of the values of [tex]x^2[/tex] at both endpoints of the subinterval times 1/2 over each subinterval [tex][x_i,x_{i+1}][/tex].
So,
[tex]\displaystyle\int_0^2x^2\,\mathrm dx\approx\dfrac{0^2+\left(\frac12\right)^2}4+\dfrac{\left(\frac12\right)^2+1^2}4+\dfrac{1^2+\left(\frac32\right)^2}4+\dfrac{\left(\frac32\right)^2+2^2}4[/tex]
[tex]=\displaystyle\sum_{i=1}^4\frac{\left(\frac{i-1}2\right)^2+\left(\frac i2\right)^2}4=\frac{11}4[/tex]
