Mathematics

# Solve : $\dfrac{\pi}{128} < \underset{\pi/4}{\overset{\pi/2}{\int}} (\sin \, x)^{10} dx < \dfrac{\pi}{4}$

##### SOLUTION
If $m$ and $M$ be global minima and global maxima of $f(x)$ in $[a,b]$  then
$m(b-a)\leq \int_{a}^{b}f(x)dx\leq M(b-a)$

$M=$ maximum of $\sin^{10} x=\sin^{10} \dfrac{\pi }{2}=1$

$m=$  minimum of  $\sin^{10} x=\sin^{10} \dfrac{\pi }{4}=\left ( \cfrac{1}{\sqrt{2}} \right )^{10}=\cfrac{1}{32}$

Hence,
$\dfrac{1}{32}\left ( \dfrac{\pi }{4} \right )\leq \int_{\frac{\pi }{4} }^{\frac{\pi }{2} }\sin^{10} x\: dx\leq 1\left ( \dfrac{\pi }{4} \right )$

$\left ( \dfrac{\pi }{128} \right )\leq \int_{\frac{\pi }{4} }^{\frac{\pi }{2} }\sin^{10} x\: dx\leq \left ( \dfrac{\pi }{4} \right )$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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