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Question Number 88033 by M±th+et£s last updated on 07/Apr/20

$$find{\int }_0^1\frac{sin\left(x\right)}{x}dx$$

Commented by mathmax by abdo last updated on 08/Apr/20

$$approximatevaluewehavesinx=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)!}{x}^{2n+1}\Rightarrow$$$$sinx=x-\frac{x^3}{3!}+o\left(x^3\right)\Rightarrow \mathrm{\forall }x\in [0,1]x-\frac{x^3}{6}⩽sinx⩽x\Rightarrow$$$$1-\frac{x^2}{6}⩽\frac{sinx}{x}⩽1\Rightarrow {\int }_0^1(1-\frac{x^2}{6})dx⩽{\int }_0^1\frac{sinx}{x}dx⩽1$$$$wehave{\int }_0^1(1-\frac{x^2}{6})dx={[x-\frac{1}{18}{x}^3]}_0^1=1-\frac{1}{18}=\frac{17}{18}\Rightarrow$$$$\frac{17}{18}⩽{\int }_0^1\frac{sinx}{x}dx⩽1wecantake{v}_0=\frac{1}{2}+\frac{17}{46}asaapprimate$$$$valueforthisintegral$$$$I\sim 0,5+0,369\Rightarrow I\sim 0,869$$

Commented by mathmax by abdo last updated on 08/Apr/20

$$forgiveerroroftypo{v}_0=\frac{1}{2}+\frac{17}{36}\Rightarrow I\sim 0,5+0,472\Rightarrow$$$$I\sim 0,972$$

Commented by M±th+et£s last updated on 08/Apr/20

$$thankyousir$$

Commented by abdomathmax last updated on 08/Apr/20

$$youarewelcome$$

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