Q. 20 - Precalculus Enhanced with Graphing Utilities

Question

Calculus: Simpson’s Rule The figure shows the graph of $y=a{x}^2+bx+c$. Suppose that the points  $\left(-h,y_0\right),\left(0,y\right) and \left(h,y_2\right)$are on the graph. It can be shown that the area enclosed by the parabola, the $x$-axis, and the lines$x=-h,x=h$ is Area $=\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(2a{h}^2+6c\right).$ Show that this area may also be given by Area  $=\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(y_0+4y_1+y_2\right).$

Step-by-Step Solution

Answer

$y=a{x}^2+bx+c$It is proven that area$=\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(y_0+4y_1+y_2\right).$

Step 1. Given information

The given figure is,

The figure shows the graph $y=a{x}^2+bx+c.$ Suppose that the point $\left(-h,y_0\right),\left(0,y\right) and \left(h,y_2\right)$is on the graph and the area is an area$=\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(2a{h}^2+6c\right).$ We need to show that the area may be given by Area $=\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(y_0+4y_1+y_2\right).$

Step 2. Simplify

The figure is $y=a{x}^2+bx+c.$

For the point $\left(-h,y_0\right),$ the function is,

$y_0=a·{\left(-h\right)}^2+b\left(-h\right)+c.$

$y_0=h^2-bh+c\cdots \left(1\right).$

For the point $\left(0,y_1\right),$ the function is,

$y_1=c\cdots \left(2\right).$

For the point $\left(h,y_2\right)$ the function is,

$y_2=a{h}^2+bh+c\cdots \left(3\right).$

Add equations (1),(3) and include equation (2).

$y_0+y_2=2·a·h^2+2·y_1.$

$2·a·h^2=y_0+y_2-2y_1\cdots \left(4\right).$

Now include equation (4) in the equation of the area.

Area $=\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(2a{h}^2+6c\right).$

         $=\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(y_0+4y_1-2y_1+6y_1\right).$

          $=\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(y_0+4y_1+y_2\right).$