Question

Estimate the integral ∫6,0 x^2dx by the midpoint estimate, n = 6

Answer 1

Splitting up the interval [0, 6] into 6 subintervals means we have

$[0,1]\cup[1,2]\cup[2,3]\cup\cdots\cup[5,6]$

and the respective midpoints are $\dfrac12,\dfrac32,\dfrac52,\ldots,\dfrac{11}2$. We can write these sequentially as ${x_i}^*=\dfrac{2i+1}2$ where $0\le i\le5$.

So the integral is approximately

$\displaystyle\int_0^6x^2\,\mathrm dx\approx\sum_{i=0}^5({x_i}^*)^2\Delta x_i=\frac{6-0}6\sum_{i=0}^5({x_i}^*)^2=\sum_{i=0}^5\left(\frac{2i+1}2\right)^2$

Recall that

$\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$
$\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2$
$\displaystyle\sum_{i=1}^n1=n$

so our sum becomes

$\displaystyle\sum_{i=0}^5\left(\frac{2i+1}2\right)^2=\sum_{i=0}^5\left(i^2+i+\frac14\right)$
$=\displaystyle\frac{5(6)(11)}6+\frac{5(6)}2+\frac54=\frac{143}2$