Q. 45 - Calculus

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 $x=a and x=b$.

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

Step-by-Step Solution

Answer

The exact area is $15$.

Step 1. Given Information.

The given functions and interval is $f\left(x\right)=x^2,g\left(x\right)=x+2,[-3,3]$.

Step 2. Required graph.

The graph of the functions is,

Step 3. Required area.

The exact area will be,

$$\begin{gathered} {\int }_{-3}^3\left|f\right(x)-g(x\left)\right|dx={\int }_{-3}^{-1}\left(f\right(x)-g(x\left)\right)dx+{\int }_{-1}^2\left(g\right(x)-f(x\left)\right)dx+{\int }_2^3\left(f\right(x)-g(x\left)\right)dx \\ ={\int }_{-3}^{-1}\left(x^2-x-2\right)dx+{\int }_{-1}^2\left(x+2-x^2\right)dx+{\int }_2^3\left(x^2-x-2\right)dx \\ ={\left[\frac{x^3}{3}-\frac{x^2}{2}-2x\right]}_{-3}^{-1}+{\left[2x+\frac{x^2}{2}-\frac{x^3}{3}\right]}_{-1}^2+{\left[\frac{x^3}{3}-\frac{x^2}{2}-2x\right]}_2^3 \\ =\frac{26}{3}+\frac{9}{2}+\frac{11}{6} \\ =15 \end{gathered}$$