Chapter 6, Problem 53RE

(a)

To determine

Area of trapezoid plus twice the area of the rectangle equals $h\left(y_1+4y_2+y_3\right)$ .

Explanation of Solution

Given information:

Figure shows an interval of length 2h with a trapezoid, a midpoint rectangle, and a parabolic region on it:

  

First

  $\text{Area}\text{of}\text{trapezoid}=\frac{1}{2}\cdot 2h\cdot \left(y_1+y_3\right)$

Then

  $\text{Area}\text{of}\text{rectangle}=2h\cdot y_2$

Now,

Sum up the area of trapezoid and twice the area of rectangle:

  $\begin{array}{l}\text{Area}\text{of}\text{trapezoid}\text{+}2\left(\text{Area}\text{of}\text{rectangle}\right)=\frac{1}{2}\cdot 2h\cdot \left(y_1+y_3\right)+2\left(2h\cdot y_2\right)\\ =h\left(y_1+y_3\right)+4h{y}_2\\ =h\left(y_1+4y_2+y_3\right)\end{array}$

(b)

To determine

Prove the expression using result in Part (a).

Explanation of Solution

Given information:

Figure shows an interval of length 2h with a trapezoid, a midpoint rectangle, and a parabolic region on it:

  

Setup the equation using Simpson’s rule:

  $\begin{array}{l}{S}_{2n}=\frac{h}{3}\left[y_0+4y_1+2y_2+4y_3+2y_4+\mathrm{...}+2y_{2n-2}+4y_{2n-1}+y_{2n}\right]\\ =\frac{1}{3}\left[h\left(y_0+4y_1+y_2\right)+h\left(y_2+4y_3+y_4\right)+\mathrm{...}+h\left(y_{2n-2}+4y_{2n-1}+y_{2n}\right)\right]\end{array}$

According to Part (a) result,

  $h\left(y_{2i-2}+4y_{2i-1}+y_{2i}\right)=2\left(\text{Area}\text{of}\text{the}\text{ith}\text{of}n\text{rectangles}\right)+\text{Area}\text{of}\text{the}\text{ith}\text{of}n\text{trapezoids}$

Thus,

  $\begin{array}{l}{S}_{2n}=\frac{1}{3}\left(2{\text{MRAM}}_n+T_n\right)\\ =\frac{2{\text{MRAM}}_n+T_n}{3}\end{array}$