Q. 24 - Precalculus Enhanced with Graphing Utilities

Question

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

Step-by-Step Solution

Answer

The area is $7.33$.

Step 1. Given information

It is given that $f\left(x\right)=-x^2+x+4$, 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 $x=-1$ and $x=1$.

Step 2. Simplify

$f\left(x\right)=-x^2+x+4$

For $x=-1$,

$y_0=-{(-1)}^2+(-1)+4$

    $=2$

For $x=0$,

$y_1=-{\left(0\right)}^2+0+4$

    $=4$

For $x=1$,

$y_2=-{\left(1\right)}^2+1+4$

    $=4$

So, Area $=\frac{1}{3}(y_0+4y_1+y_2)$

               $=\frac{1}{3}(2+4·4+4)$

               $=\frac{22}{3}$

               $=7.33$