Chapter 6.5, Problem 39E

To determine

$T_n=\frac{LRA{M}_n+RRA{M}_n}{2}$

Explanation of Solution

Given information:

The value is

  $T_n=\frac{LRA{M}_n+RRA{M}_n}{2}$

Formula used:

  $LRA{M}_n=h\left(y_0+y_1+\mathrm{.....}+y_{n-1}\right)$

Calculation:

LRAM means left-hand endpoint rectangular approximation method. The name suggests the choice we made when determining the heights of the approximating rectangles: we evaluated the function at the left-hand endpoint of each subinterval.

  $LRA{M}_n=h\left(y_0+y_1+\mathrm{.....}+y_{n-1}\right)$

RRAM means right-hand endpoint rectangular approximation method. The name suggests the choice we made when determining the heights of the approximating rectangles: we evaluated the function at the right-hand endpoint of each subinterval.

  $RRA{M}_n=h\left(y_1+y_2+\mathrm{.....}+y_{n-1}+y_n\right)$

In the trapezoidal rule, if [a, b] is partitioned into n subintervals of equal length $h=\frac{b-a}{n}$ , the graph of f 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,

  $\begin{array}{l}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=h\left(\frac{y_0+y_1}{2}\right)+h\left(\frac{y_1+y_2}{2}\right)+\mathrm{....}+h\left(\frac{y_{n-1}+y_n}{2}\right)\\ =h\left(\frac{y_0}{2}+y_1+y_2+\mathrm{....}+y_{n-1}+\frac{y_n}{2}\right)\\ =\frac{h}{2}\left(y_0+2y_1+\mathrm{.....}+2y_{n-1}+y_n\right)\end{array}$

Where,

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

Therefore,

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

Conclusion:

Hence it is proved that $T_n=\frac{h\left[LRA{M}_n+RRA{M}_n\right]}{2}$