Chapter 6.5, Problem 40E

To determine

$S_{2n}=\frac{MRA{M}_n+2T_{2n}}{3}$

Explanation of Solution

Given information:

  $S_{2n}=\frac{MRA{M}_n+2T_{2n}}{3}$

Formula used:

  $S_n=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+\mathrm{....}+2y_{n-2}+4y_{n-1}+y_n\right)$

Calculation:

To approximate ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ , use

  $S_n=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+\mathrm{....}+2y_{n-2}+4y_{n-1}+y_n\right)$

Where [a, b] is partitioned into an even number n of subintervals of equal length $h=\frac{b-a}{n}$

Hence,

  $S_{2n}=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+\mathrm{....}+2y_{2n-2}+4y_{2n-1}+y_{2n}\right)$

In the trapezoidal rule, if [a, b] is partitioned into n subintervals of equal length $h=\frac{b-a}{n}$ , the graph off on [a, b] can be approximated by a straight line segment at each subinterval. The region between the curve and the x-axis is then approximated by the trapezoids, the area of each trapezoid being the length of its horizontal “altitude" times the average of two vertical “bases".

That is,

  $\begin{array}{l}{T}_n=h\left(\frac{y_0+y_1}{2}\right)+h\left(\frac{y_1+y_2}{2}\right)+\mathrm{....}+h\left(\frac{y_{n-1}+y_n}{2}\right)\\ =h\left(\frac{y_0}{2}+y_1+y_2+\mathrm{....}+y_{n-1}+\frac{y_n}{2}\right)\\ =\frac{h}{2}\left(y_0+2y_1+\mathrm{.....}+2y_{n-1}+y_n\right)\end{array}$

Where,

  $\begin{array}{l}{y}_0=f\left(a\right)\\ y_1=f\left(x_1\right)\\ .\\ .\\ y_{n-1}=f\left(x_{n-1}\right)\\ y_n=f\left(b\right)\end{array}$

Hence,

  $T_{2n}=\frac{h}{2}\left(y_0+2y_1+2y_2+\mathrm{....}+2y_{2n-1}+y_{2n}\right)$

MRAM means midpoint rectangular approximation method. The name suggests the choice we made when determining the heights of the approximating rectangles, we evaluated the function at the midpoint of each subinterval. Subintervals are taken as $\left[x_0,x_2\right]\mathrm{.......}\left[x_{2n-2},x_n\right]$

  $MRA{M}_n=h\left(y_1+y_3+\mathrm{...}+y_{2n-1}\right)$

Now

  $\begin{array}{l}{S}_{2n}=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+\mathrm{.....}+2y_{2n-2}+4y_{2n-1}+y_{2n}\right)\\ =\frac{h}{3}\left(y_0+2y_1+2y_1+2y_2+2y_3+2y_3+\mathrm{....}+2y_{2n-2}+2y_{2n-1}+2y_{2n-1}+y_{2n}\right)\\ =\frac{h\left(y_0+2y_1+2y_2+\mathrm{.....}+2y_{2n-1}+y_{2n}\right)+2h\left(y_1+y_3+\mathrm{....}+y_{2n-1}\right)}{3}\\ S_{2n}=\frac{2T_{2n}+MRA{M}_n}{3}\end{array}$

Conclusion:

Hence it is proved that $S_{2n}=\frac{2T_{2n}+MRA{M}_n}{3}$