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$x=2$.

Short answer

The area is$42.67$.

Step-by-step solution

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

$=\frac{2}{3}(16+4·8+16)$

$=\frac{128}{3}$

$=42.67$