Chapter 3.4, Problem 24AYU

Use the result obtained in Problem 20 to find the area enclosed by $f\left(x\right)=-x^2+x+4$, the $x\text{-axis}$, and the lines $x=-1$ and $x=\text{ 1}$.

To determine

To calculate: The area enclosed by the function $f(x)=-x^2+x+4$ the $x\text{-axis}$, and the lines $x=-1$ and $x=1$.

Answer to Problem 24AYU

The area enclosed by the parabola $f(x)=-x^2+x+4$ the $x\text{-axis}$ and the lines $x=-1$ and $x=1$ is

$Area=\frac{22}{3}$

Explanation of Solution

Given:

The given function is $f(x)=-x^2+x+4$ and the lines are $x=-1$ and $x=1$.

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 $a=-1,b=1,c=4$

We have $h=1$ as the distance between the line $x=-1$ and the $y\text{-axis}$ and the line $x=1$ and the $y\text{-axis}$ is 1.

Thus, the area is

$\begin{array}{l}Area=\frac{1}{3}\left(2(-1){(1)}^2+6(4)\right)\\ =\frac{22}{3}\end{array}$