Q. 49 - 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 $x=a to x=b.$

$f\left(x\right)=x^3, g\left(x\right)=(x-2{)}^3, [a,b]=[-1,2]$

Step-by-Step Solution

Answer

The exact area is $\frac{145}{12}$.

Step 1. Given Information.

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

Step 2. Graph of the function.

The graph of the functions is,

Step 3. Required area.

The exact area is,

$$\begin{gathered} {\int }_{-1}^2\left|f\right(x)-g(x\left)\right|dx={\int }_{-1}^1\left(g\right(x)-f(x\left)\right)dx+{\int }_1^2\left(f\right(x)-g(x\left)\right)dx \\ ={\int }_{-1}^1\left((x-2{)}^2-x^3\right)dx+{\int }_1^2\left(x^3-(x-2{)}^2\right)dx \\ =\left[{\int }_{-1}^1{x}^2-4x+4-x^{3}dx\right]+\left[{\int }_1^2{x}^3-x^2+4x-4dx\right] \\ ={\left[\frac{x^3}{3}-4\frac{x^2}{2}+4x-\frac{x^4}{4}\right]}_{-1}^1+{\left[\frac{x^4}{4}-\frac{x^3}{3}+4\frac{x^2}{2}-4x\right]}_1^2 \\ =\left[\left(\frac{1}{3}-2+4-\frac{1}{4}\right)-\left(\frac{-1}{3}-2-4-\frac{1}{4}\right)\right]+\left[\left(4-\frac{8}{3}+8-8\right)-\left(\frac{1}{4}-\frac{1}{3}+2-4\right)\right] \\ =\frac{26}{3}+\frac{41}{12} \\ =\frac{145}{12} \end{gathered}$$