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$x=1$.

Short answer

The area is$7.33$.

Step-by-step solution

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

$=\frac{1}{3}(2+4·4+4)$

$=\frac{22}{3}$

$=7.33$