Chapter 3.4, Problem 23AYU

Use the result obtained in Problem 20 to find the area enclosed by $f\left(x\right)=x^2+3x+5$, the $x\text{-axis}$, and the lines $x=-4$ and $x=\text{ 4}$.

To determine

To calculate: The area enclosed by the function $f(x)=x^2+3x+5$, the $x\text{-axis}$ , and the lines $x=-4$ and $x=4$.

Answer to Problem 23AYU

The area enclosed by the parabola $f(x)=x^2+3x+5$ the $x\text{-axis}$ and the lines $x=-4$ and $x=4$ is

$Area=\frac{248}{3}$

Explanation of Solution

Given:

The given function is $f(x)=x^2+3x+5$ and the lines are $x=-4$ and $x=4$.

Formula used:

The area enclosed by the parabola $y=a{x}^2+bx+c$, the $x\text{-axis}$, and the lines $x=h$ and $x=-h$ is

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

Calculation:

In the given equation, we have $a=1,b=3,c=5$.

We have $h=4$ as the distance between the line $x=-4$ and the $y\text{-axis}$ and the line $x=4$ and the $y\text{-axis}$ is 4.

Thus, the area is

$\begin{array}{l}Area=\frac{4}{3}\left(2(1){(4)}^2+6(5)\right)\\ = \frac{248}{3}\end{array}$