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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}
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