Q. 47 - 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 to x = b.

width="330" height="23" style="max-width: none; vertical-align: -5px;" $f\left(x\right)=x-1, g\left(x\right)=x^2-2x-1, [a,b]=[-1,3]$

Step-by-Step Solution

Answer

The exact area is $\frac{19}{3}$.

Step 1. Given Information.

The given function and interval is $f\left(x\right)=x-1, g\left(x\right)=x^2-2x-1, [a,b]=[-1,3]$.

Step 2. Graph of the function.

The graph of the functions is,

Step 3. Required area.

The exact area will be,

$$\begin{gathered} {\int }_{-1}^3\left|f\right(x)-g(x\left)\right|dx={\int }_{-1}^0\left(g\right(x)-f(x\left)\right)dx+{\int }_0^3\left(f\right(x)-g(x\left)\right)dx \\ ={\int }_{-1}^0{x}^2-3xdx+{\int }_0^3-x^2+3xdx \\ ={\left[\frac{x^3}{3}-3\frac{x^2}{2}\right]}_{-1}^0+{\left[\frac{-x^3}{3}+3\frac{x^2}{2}\right]}_0^3 \\ =\left[0-\left(\frac{(-1{)}^3}{3}-3\frac{(-1{)}^2}{2}\right)\right]+\left[\frac{-(3{)}^3}{3}+3\frac{(3{)}^2}{2}-0\right] \\ =\frac{1}{3}+\frac{3}{2}-\frac{27}{3}+\frac{27}{2} \\ =\frac{-26}{3}+15 \\ =\frac{19}{3} \end{gathered}$$