ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 3- Worked Out Examples

Monday, 4 August 2014

CHAPTER 3- Worked Out Examples

Example: 1

Find the area under the curve $y = \dfrac{1}{{{x^2} + 1}}$ on its entire domain, i.e., evaluate $\int\limits_{ - \infty }^{ + \infty } {\dfrac{1}{{{x^2} + 1}}} \,\,dx$

Solution: 1

The graph for $y = \dfrac{1}{{{x^2} + 1}}$ is sketched below:

Although the graph extends to infinity on both sides, the area under the curve will still be finite, as we’ll now see:

	$\int\limits_{ - \infty }^{ + \infty } {\dfrac{1}{{{x^2} + 1}}} \,\,\,dx\,\, = \,\,\,\left. {{{\tan }^{ - 1}}x} \right\|_{ - \infty }^{ + \infty }$
	$= {\tan ^{ - 1}}\left( { + \infty } \right) - {\tan ^{ - 1}}\left( { - \infty } \right)$

What do we make of ${\tan ^{ - 1}}\left( { + \infty } \right)$ ? We can take it to mean $\mathop {\lim }\limits_{k \to \infty } \,\,{\tan ^{ - 1}}\left( k \right)$ which is $\dfrac{\pi }{2}$ . Similarly, ${\tan ^{ - 1}}\left( { - \infty } \right)$ would equal $\dfrac{{ - \pi }}{2}$ . Thus, the required area is $\pi$ .

Example: 2

Evaluate the area bounded between from $x = 0$ to $x = 1$ .

Solution: 2

The given curves are sketched in the region of interest below.

The required area is

	$A = \int\limits_0^1 {\left( {x - {x^3}} \right)dx}$
	$= \left. {\left( {\dfrac{{{x^2}}}{2} - \dfrac{{{x^4}}}{2}} \right)} \right\|_0^1$
	$= \left( {\dfrac{1}{2} - \dfrac{1}{4}} \right) - \left( {0 - 0} \right)$
	$= \dfrac{1}{4}$

Example: 3

Find the mean value of $f\left( x \right) = {\cos ^2}x\,$ on $\left[ {0,\dfrac{\pi }{2}} \right]$

Solution: 3

Let ${f_{av}}$ be the required mean value.

	$\Rightarrow \,\,\, \,{f_{av}}\left( {\dfrac{\pi }{2} - 0} \right) = \int\limits_0^{\pi /2} {{{\cos }^2}x\,dx}$
	$= \dfrac{1}{2}\,\int\limits_0^{\pi /2} {\left( {1 + \cos 2x} \right)dx}$
	$= \dfrac{1}{2}\left. {\left( {x + \dfrac{{\sin 2x}}{2}} \right)} \right\|_0^{\pi /2}$
	$= \dfrac{\pi }{4}$
	$\Rightarrow \,\,\,\, {f_{av}} = \dfrac{1}{2}$

Monday, 4 August 2014

CHAPTER 3- Worked Out Examples

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