Question

Area under a Curve The area under the graph of $y=\frac{1}{1+x^2}$and above the $x$-axis between $x=aandx=b$is given by

${\tan }^{-1}b-{\tan }^{-1}a$

(a) Find the exact area under the graph of $y=\frac{1}{1+x^2}$and above the $x$-axis between $x=0andx=\sqrt{3}.$

(b) Find the exact area under the graph of $y=\frac{1}{1+x^2}$and above the $x$-axis between$x=-\frac{\sqrt{3}}{3}andx=1$.

Short answer

The exact Area of the curve is:

$$\begin{gathered} \left(a\right)\frac{\mathrm{\pi }}{3} \\ \left(b\right)\frac{5\mathrm{\pi }}{12}. \end{gathered}$$

Step-by-step solution

Step 1. Given Information.

The given function is

${\tan }^{-1}b-{\tan }^{-1}a$.

Part (a) Step 1. Calculation of area.

The given values are $a=0,b=\sqrt{3}$

Then, we get

$\begin{array}{rcl}{\tan }^{-1}b-{\tan }^{-1}a& =& {\tan }^{-1}\left(\sqrt{3}\right)-{\tan }^{-1}\left(0\right)\\ & =& \frac{\mathrm{\pi }}{3}-0\\ & =& \frac{\mathrm{\pi }}{3}.\end{array}$

Part (b) Step 1. Calculation of area.

The given values are $b=1,a=-\frac{\sqrt{3}}{3}$.

The values we get,

$\begin{array}{rcl}{\tan }^{-1}b-{\tan }^{-1}a& =& -{\tan }^{-1}\left(-\frac{\sqrt{3}}{3}\right)+{\tan }^{-1}\left(1\right)\\ & =& -\left(-\frac{\mathrm{\pi }}{6}\right)-\frac{\mathrm{\pi }}{4}\\ & =& \frac{\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{4}\\ & =& \frac{5\mathrm{\pi }}{12}.\end{array}$.