Q. 22 - Precalculus Enhanced with Graphing Utilities

Question

Use the result obtained in Problem $20$ to find the area enclosed by $f\left(x\right)=2x^2+8$, the $x-axis$ and the lines $x=-2$ and $x=2$.

Step-by-Step Solution

Answer

The area is $42.67$.

Step 1. Given information

It is given that $f\left(x\right)=2x^2+8$, we need to use the result obtained in problem $20$ to determine the area enclosed by $f\left(x\right)$, the $x-axis$ and the lines $x=-2$ and $x=2$.

Step 2. Simplify

$f\left(x\right)=2x^2+8$

For $x=-2$,

$y_0=2·{(-2)}^2+8$

    $=16$

For $x=0$,

$y_1=2·{\left(0\right)}^2+8$

    $=8$

For $x=2$,

$y_2=2·{\left(2\right)}^2+8$

    $=16$

So, Area $=\frac{2}{3}(y_0+4y_1+y_2)$

               $=\frac{2}{3}(16+4·8+16)$

               $=\frac{128}{3}$

               $=42.67$