Question

Using 4 equal-width intervals, show that the trapezoidal rule is the average of the upper and lower sum estimates for the integral from 0 to 2 of x squared, dx .

Answer 1

Split up the interval [0, 2] into 4 subintervals, so that

$[0,2]=\left[0,\dfrac12\right]\cup\left[\dfrac12,1\right]\cup\left[1,\dfrac32\right]\cup\left[\dfrac32,2\right]$

Each subinterval has width $\dfrac{2-0}4=\dfrac12$. The area of the trapezoid constructed on each subinterval is $\dfrac{f(x_i)+f(x_{i+1})}4$, i.e. the average of the values of $x^2$ at both endpoints of the subinterval times 1/2 over each subinterval $[x_i,x_{i+1}]$.

So,

$\displaystyle\int_0^2x^2\,\mathrm dx\approx\dfrac{0^2+\left(\frac12\right)^2}4+\dfrac{\left(\frac12\right)^2+1^2}4+\dfrac{1^2+\left(\frac32\right)^2}4+\dfrac{\left(\frac32\right)^2+2^2}4$

$=\displaystyle\sum_{i=1}^4\frac{\left(\frac{i-1}2\right)^2+\left(\frac i2\right)^2}4=\frac{11}4$