Q. 50 - 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 $g from x=a to x=b.$.

$f\left(x\right)=2^x, g\left(x\right)=x^2, [a,b]=[0,3]$

Step-by-Step Solution

Answer

The exact area is $2.224$ .

Step 1. Given Information.

The given function and interval is$f\left(x\right)=2^x, g\left(x\right)=x^2, [a,b]=[0,3].$

Step 2. Graph.

The graph of the functions is,

Step 3. Required Area.

The exact area is,

$$\begin{gathered} {\int }_0^3\left|f\right(x)-g(x\left)\right|dx={\int }_0^2\left(f\right(x)-g(x\left)\right)dx+{\int }_2^3\left(g\right(x)-f(x\left)\right)dx \\ ={\int }_0^2\left(2^x-x^2\right)dx+{\int }_2^3\left(x^2-2^x\right)dx \\ ={\left[\frac{2^x}{\ln 2}-\frac{x^3}{3}\right]}_0^2+{\left[\frac{x^3}{3}-\frac{2^x}{\ln 2}\right]}_2^3 \\ =\left[\frac{4}{\ln 2}-\frac{8}{3}-\frac{1}{\ln 2}\right]+\left[\frac{27}{3}-\frac{8}{\ln 2}-\frac{8}{3}+\frac{4}{\ln 2}\right] \\ =\left(\frac{3}{\ln 2}-\frac{8}{3}\right)+\left(\frac{19}{3}-\frac{4}{\ln 2}\right) \\ =\frac{11}{3}-\frac{1}{\ln 2} \\ =2.224 \end{gathered}$$