tgt

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