Q. 51 - 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 $a to x=b.$

$f\left(x\right)=\frac{2}{1+x^2};g\left(x\right)=1;[a,b]=[0,\sqrt{3}]$

Step-by-Step Solution

Answer

The exact area is$\sqrt{3}-2+\frac{\pi }{3}.$

Step 1. Given Information.

The given function and interval is$f\left(x\right)=\frac{2}{1+x^2};g\left(x\right)=1;[a,b]=[0,\sqrt{3}]$.

Step 2. Graph.

The graph of the functions is,

Step 3. Required area.

The exact area will be,

$$\begin{gathered} ={\int }_0^1\left[\frac{2}{1+x^2}-1\right]dx+{\int }_1^{\sqrt{3}}\left[1-\frac{2}{1+x^2}\right]dx \\ ={\left[2{\tan }^{-1}x-x\right]}_0^1+{\left[x-2{\tan }^{-1}x\right]}_1^{\sqrt{3}} \\ A=\left[2{\tan }^{-1}\left(1\right)-1-2{\tan }^{-1}\left(0\right)\right]+\left[\sqrt{3}-1-\left(2{\tan }^{-1}\sqrt{3}-2{\tan }^{-1}1\right)\right] \\ =2\left(\frac{\pi }{4}\right)-1-0+\sqrt{3}-1-\frac{2\pi }{3}+\frac{2\pi }{4} \\ =\frac{\pi }{2}-2+\sqrt{3}-\frac{2\pi }{3}+\frac{2\pi }{4} \\ =\sqrt{3}-2+\frac{\pi }{3} \end{gathered}$$