Chapter 7.1, Problem 77AYU

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

See the figure.

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

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

To determine

To calculate: The exact area under the graph of $y=1/\left(1+x^2\right)$ and above the $x\text{-axis}$,

a. between $x=0$ and $x=\surd 3$.

Answer to Problem 77AYU

Solution:

a. exact area under the graph of $y=1/\left(1+x^2\right)$ and above the x-axis between $x=0$ and $x=\surd 3$.

$= {\text{ tan}}^{-1}\left(\surd 3\right) -{\text{ tan}}^{-1}\left(0\right) = \pi /3–0=\pi /3$

Explanation of Solution

Given:

The exact area under the graph of $y=1/\left(1+x^2\right)$ and above the $x\text{-axis}$ between $x=a$ and $x=b$ is given by ${\text{tan}}^{-1}b -{\text{ tan}}^{-1}a$,

Graph $=y=1/\left(1+x^2\right)$.

Area above the $x\text{-axis}$ between $x=a$ and $x=b$ is given by ${\text{tan}}^{-1}b -{\text{ tan}}^{-1}a\text{ -----1}$

a. To calculate area above the $x\text{-axis}$ between $x=0$ and $x=\surd 3$.

Here $a=0$ and $b=\surd 3$.

Putting values of $a$ and $b$ in 1 we get,

exact area under the graph of $y=1/\left(1+x^2\right)$ and above the $x\text{-axis}$ between $x=0$ and $x=\surd 3$.

$= {\text{ tan}}^{-1}\left(\surd 3\right) -{\text{ tan}}^{-1}0= \pi /3–0=\pi /3$

To determine

To calculate: The exact area under the graph of $y=1/\left(1+x^2\right)$ and above the $x\text{-axis}$,

b. between $x=-\surd 3/3$ and $x=1$.

Answer to Problem 77AYU

Solution:

b. exact area under the graph of $y=1/\left(1+x^2\right)$ and above the $x\text{-axis}$ between $x=-\surd 3/3$ and $x=1$.

$={\text{ tan}}^{-1}1 -{\text{ tan}}^{-1}\left(-\surd 3/3\right)= \pi /4–\left(-\pi /6\right) =\left(4\pi +6\pi \right)/24=10\pi /24=5\pi /12$

Explanation of Solution

Given:

The exact area under the graph of $y=1/\left(1+x^2\right)$ and above the $x\text{-axis}$ between $x=a$ and $x=b$ is given by ${\text{tan}}^{-1}b -{\text{ tan}}^{-1}a$,

Graph $=y=1/\left(1+x^2\right)$.

Area above the $x\text{-axis}$ between $x=a$ and $x=b$ is given by ${\text{tan}}^{-1}b -{\text{ tan}}^{-1}a\text{ -----1}$

b. To calculate area above the $x\text{-axis}$ between $x=-\surd 3/3$ and $x=1$.

Here $a=-\surd 3/3$ and $b=1$.

Putting values of $a$ and $b$ in 1 we get,

exact area under the graph of $y=1/\left(1+x^2\right)$ and above the $x\text{-axis}$ between $x=-\surd 3/3$ and $x=1$.

$={\text{ tan}}^{-1}1 -{\text{ tan}}^{-1}\left(-\surd 3/3\right)= \pi /4–\left(-\pi /6\right) =\left(4\pi +6\pi \right)/24=10\pi /24=5\pi /12$