| Direction: The mean value of the continuous functions f(x) in the interval [a, b} is given by the formula |
| Mean value \[=\frac{\int\limits_{a}^{b}{f(x)\,dx}}{b-a}\] |
A) \[\frac{1}{3}\]
B) \[\frac{1}{2}\]
C) 0
D) 1
Correct Answer: A
Solution :
Put \[x=\,\sin \,\theta \] \[I=\int_{0}^{\pi /2}{\frac{\cos \,\theta }{\cos \,\theta \,(\sqrt{\cos \,\theta +\sqrt{\sin \,\theta }{{)}^{4}}}}d\theta }\] \[=\int_{0}^{\pi /2}{\frac{1}{{{(\sqrt{\cos \,\theta +}\sqrt{\sin \,\theta })}^{4}}}d\theta }\] \[=\int_{0}^{\pi /2}{\frac{1}{{{\cos }^{2}}\theta {{(1+\sqrt{\tan \,\theta })}^{4}}}d\theta }\] \[=\int_{0}^{\pi /2}{\frac{{{\sec }^{2}}\theta }{{{(1+\sqrt{\tan \,\theta })}^{4}}}d\theta }\] Let \[1+\sqrt{\tan \,\theta }=z\] \[\Rightarrow \] \[\frac{1}{2\,\sqrt{\tan \,\theta }}{{\sec }^{2}}\theta \,d\theta =dz\] \[I=\,\int_{1}^{\infty }{\,\,\frac{2(z-1)}{{{z}^{4}}}dz}\] \[=2\,\int_{1}^{\infty }{({{z}^{-3}}-{{z}^{-4}})dz}\] \[=2\,\left[ \frac{1}{-2{{z}^{2}}}+\frac{1}{3{{z}^{3}}} \right]_{1}^{\infty }\] \[=2\,\left[ 0-\left( -\frac{1}{2}+\frac{1}{3} \right) \right]\] \[=\frac{1}{3}\]
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