Question

For each pair of functions f and g and interval $[a,b]$in Exercises 41–52, use definite integrals and the Fundamental Theorem of Calculus to find the exact area of the region between the graphs of $f$and $g$from $x=a\mathrm{to}x=b$

$f\left(x\right)=x^2-x-1$and $g\left(x\right)=5-x^2$,$\left[-2,3\right]$

Short answer

the area of the region between $f\left(x\right)=x^2-x-1$and $g\left(x\right)=5-x^2$$38.75\mathrm{sq}.\mathrm{unit}$

Step-by-step solution

Step 1. Given Information

Two curves $f\left(x\right)=x^2-x-1$and $g\left(x\right)=5-x^2$are given.

The graph and region between the curve is follow:

We need to find the exact are of region bounded by $f\left(x\right)=x^2-x-1$and $5-x^2$in interval $\left[-2,3\right]$i.e. the shaded region shown below:

Step 1. formula used

we know that area between two curve $f\left(x\right)\mathrm{and}g\left(x\right)$on some interval $\left[a,b\right]$is

$\mathrm{Area}={\int }_a^b\left|f\left(x\right)-g\left(x\right)\right|dx$

Here $a=-2\mathrm{and}b=3$

$f\left(x\right)=x^2-x-1$and$g\left(x\right)=5-x^2$

Step 2. Solution

From the graph we see that in interval $\left[-2,-1,5\right]$,$f\left(x\right)=x^2-x-1$is above $g\left(x\right)=5-x^2$

In interval $\left[-1.5,2\right]$, $g\left(x\right)=5-x^2$is above $f\left(x\right)=x^2-x-1$

And in interval $\left[2,3\right]$, $f\left(x\right)=x^2-x-1$is above $g\left(x\right)=5-x^2$

There Area of the region can be expanded piecewise as follow:

${\int }_{-2}^3\left|\left(x^2-x-2\right)-\left(5-x^2\right)\right|dx={\int }_{-2}^{1.5}\left[\left(x^2-x-2\right)-\left(5-x^2\right)\right]dx+{\int }_{-1.5}^2\left[\left(5-x^2\right)-\left(x^2-x-2\right)\right]dx+{\int }_2^3\left[\left(x^2-x-2\right)-\left(5-x^2\right)\right]dx$

Solving further we get,

$$\begin{gathered} {\int }_{-2}^3\left|\left(x^2-x-2\right)-\left(5-x^2\right)\right|dx={\int }_{-2}^{1.5}\left[\left(x^2-x-2\right)-\left(5-x^2\right)\right]dx+{\int }_{-1.5}^2\left[\left(5-x^2\right)-\left(x^2-x-2\right)\right]dx+{\int }_2^3\left[\left(x^2-x-2\right)-\left(5-x^2\right)\right]dx \\ {\int }_{-2}^3\left|\left(x^2-x-2\right)-\left(5-x^2\right)\right|dx={\left[\frac{2}{3}{x}^3-\frac{x^2}{2}-7x\right]}_{-2}^{1.5}+{\left[-\frac{2}{3}{x}^3+\frac{x^2}{2}+7x\right]}_{-1.5}^2+{\left[\frac{2}{3}{x}^3-\frac{x^2}{2}-7x\right]}_2^3 \\ {\int }_{-2}^3\left|\left(x^2-x-2\right)-\left(5-x^2\right)\right|dx=\frac{2}{3}{(-1.5)}^3-\frac{{(-1.5)}^2}{2}-7(-1.5)-\frac{2}{3}{\left(2\right)}^3+\frac{{\left(2\right)}^2}{2}+7\left(2\right)-\frac{2}{3}{2}^3+\frac{2^2}{2}+7.2+\frac{2}{3}{(-1.5)}^3-\frac{{(-1.5)}^2}{2}-7(-1.5)+\frac{2}{3}.3^3-\frac{3^2}{2}-7.3-\frac{2}{3}.2^3+\frac{2^2}{2}+7.2 \\ {\int }_{-2}^3\left|\left(x^2-x-2\right)-\left(5-x^2\right)\right|dx=2.\frac{2}{3}{(-1.5)}^3-2.\frac{{(-1.5)}^2}{2}-2.7(-1.5)-3.\frac{2}{3}{\left(2\right)}^3+3.\frac{{\left(2\right)}^2}{2}+3.7\left(2\right)+\frac{2}{3}.3^3-\frac{3^2}{2}-7.3 \\ {\int }_{-2}^3\left|\left(x^2-x-2\right)-\left(5-x^2\right)\right|dx=-4.5-2.25+21-16+6+42+18-4.5-21 \\ {\int }_{-2}^3\left|\left(x^2-x-2\right)-\left(5-x^2\right)\right|dx=-2.25+41=38.75 \end{gathered}$$

Step 4. Conclusion

Therefore the exact area of the region between the graph of$f=x^2-x-1\mathrm{and}g=5-x^2$from$a=-2\mathrm{to}b=3$is$38.75\mathrm{sq}.\mathrm{unit}$$38.75\mathrm{sq}.\mathrm{unit}$