Chapter 3.4, Problem 20AYU

To determine

To Prove:

The area that is given by Area $=\frac{h}{3}\left(y_0+4y_1+y_2\right)$ .

Explanation of Solution

Given Information:

  $y=a{x}^2+bx+c$

Area $=\frac{h}{3}\left(2a{h}^2+6c\right)$

Area $=\frac{h}{3}\left(y_0+4y_1+y_2\right)$

  

The given equation of parabola, $3$ points through which it passes and a formula for the area enclosed by the parabola, the $x$ -axis and $2$ lines:

  $\begin{array}{l}y=a{x}^2+bx+c\\ \left(-h,y_0\right)\\ \left(0,y_1\right)\\ \left(h,y_2\right)\\ A=\frac{h}{3}\left(2a{h}^2+6c\right)\\ x=h\\ x=-h\end{array}$

Now prove that the area may also be given by the formula,

Area $=\frac{h}{3}\left(y_0+4y_1+y_2\right)$

Now compute the $y_0,y_1,y_2$ using the $3$ points coordinates and the fact that the $3$ points belong to the parabola is,

  $\begin{array}{l}{y}_0=a{\left(-h\right)}^2+b\left(-h\right)+c\\ y_0=a{h}^2-bh+c\\ y_1=0^2+b.0+c\\ y_1=c\\ y_2=a{h}^2+bh+c\end{array}$

Now compute $A_1=\frac{h}{3}\left(y_0+4y_1+y_2\right)$

  $\begin{array}{l}{A}_1=\frac{h}{3}\left(y_0+4y_1+y_2\right)\\ A_1=\frac{h}{3}\left(a{h}^2-bh+c+4c+a{h}^2+bh+c\right)\\ A_1=\frac{h}{3}\left(2a{h}^2+6c\right)\end{array}$

So, get the same formula.

  $A_1=A=\frac{h}{3}\left(2a{h}^2+6c\right)$