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I-0-pi-x-pi-x-sin-x-dx-




Question Number 1085 by 123456 last updated on 10/Jun/15
I=∫_0 ^π ((x(π−x))/(sin x))dx
$$\mathrm{I}=\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{x}\left(\pi−{x}\right)}{\mathrm{sin}\:{x}}{dx} \\ $$
Commented by 123456 last updated on 10/Jun/15
f(x)=((x(π−x))/(sin x))  f(0^+ )=^? π  f(π^− )=^? π  ∃ξ∈(0,π),∀x∈(0,π),f(ξ)≥f(x)  ∃ζ∈(0,π),∀x∈(0,π),f(ζ)≤f(x)  πf(ζ)≤^? ∫_0 ^π f(x)dx≤^? πf(ξ)
$${f}\left({x}\right)=\frac{{x}\left(\pi−{x}\right)}{\mathrm{sin}\:{x}} \\ $$$${f}\left(\mathrm{0}^{+} \right)\overset{?} {=}\pi \\ $$$${f}\left(\pi^{−} \right)\overset{?} {=}\pi \\ $$$$\exists\xi\in\left(\mathrm{0},\pi\right),\forall{x}\in\left(\mathrm{0},\pi\right),{f}\left(\xi\right)\geqslant{f}\left({x}\right) \\ $$$$\exists\zeta\in\left(\mathrm{0},\pi\right),\forall{x}\in\left(\mathrm{0},\pi\right),{f}\left(\zeta\right)\leqslant{f}\left({x}\right) \\ $$$$\pi{f}\left(\zeta\right)\overset{?} {\leqslant}\underset{\mathrm{0}} {\overset{\pi} {\int}}{f}\left({x}\right){dx}\overset{?} {\leqslant}\pi{f}\left(\xi\right) \\ $$
Commented by 123456 last updated on 10/Jun/15
f((π/2))=^? (π^2 /(2(√2)))  π^2 <^? ∫_0 ^π f(x)dx<^? (π^3 /(2(√2)))
$${f}\left(\frac{\pi}{\mathrm{2}}\right)\overset{?} {=}\frac{\pi^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\pi^{\mathrm{2}} \overset{?} {<}\underset{\mathrm{0}} {\overset{\pi} {\int}}{f}\left({x}\right){dx}\overset{?} {<}\frac{\pi^{\mathrm{3}} }{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$
Commented by prakash jain last updated on 10/Jun/15
lim_(x→0) ((x(π−x))/(sin x))=π  lim_(x→π) ((x(π−x))/(sin x))=π
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\left(\pi−{x}\right)}{\mathrm{sin}\:{x}}=\pi \\ $$$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{{x}\left(\pi−{x}\right)}{\mathrm{sin}\:{x}}=\pi \\ $$

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