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
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