Chapter 3.4, Problem 22AYU

To determine

Find the area enclosed by $f\left(x\right)=2x^2+8$ , the $x-$ axis, and the lines $x=-2$ and $x=2$ .

Answer to Problem 22AYU

The area is $A=\frac{128}{3}$ .

Explanation of Solution

Given:

  $f\left(x\right)=2x^2+8$ .

Lines are $x=-2$ and $x=2$ .

In exercise 20 as proved that the area may given by the formula: $A=\frac{h}{3}\left(y_0+4y_1+y_2\right)$

Calculation:

Now as having the area enclosed by a parabola with given equation, the $x$ -axis and $2$ lines,

  $\begin{array}{l}f\left(x\right)=2x^2+8\\ x=-2\\ x=2\end{array}$

In exercise 20 as proved that the area may give by the formula:

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

Now as identified the value of $h$ from the given equations of the $2$ lines and the values of $a,b,c$ from the parabolas equations.

Then, compute $y_0,y_1,y_2$ using the fact that they satisfy the equations:

  $\begin{array}{l}f\left(-2\right)=y_0\\ f\left(0\right)=y_1\\ f\left(2\right)=y_2\end{array}$

  $\begin{array}{l}h=2,a=2,b=0,c=8\\ y_0=f\left(-h\right)=f\left(-2\right)=a{\left(-2\right)}^2+b\left(-2\right)+c\\ y_0=2.4+8\\ y_0=16\\ \\ y_1=f\left(0\right)=a{.0}^2+b.0+c\\ y_1=c\\ y_1=8\\ \\ y_2=f\left(h\right)=f\left(2\right)=a{.2}^2+b.2+c\\ y_2=2.4+8\\ y_2=16\end{array}$

Now compute:

  $\begin{array}{l}A=\frac{h}{3}\left(y_0+4y_1+y_2\right)\\ A=\frac{2}{3}\left(16+4.8+16\right)\\ A=\frac{2}{3}.64\\ A=\frac{128}{3}\end{array}$

Conclusion:

Hence, the area is $A=\frac{128}{3}$ .