Question and Answers Forum

All Questions      Topic List

Atomic Structure Questions

Previous in All Question      Next in All Question      

Previous in Atomic Structure      Next in Atomic Structure      

Question Number 125015 by Khalmohmmad last updated on 07/Dec/20

Answered by Dwaipayan Shikari last updated on 07/Dec/20

∫_(π/6) ^(π/3) ((sinx)/x)dx=Si((π/3))−Si((π/6))=0.4697

$$\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \frac{{sinx}}{{x}}{dx}={Si}\left(\frac{\pi}{\mathrm{3}}\right)−{Si}\left(\frac{\pi}{\mathrm{6}}\right)=\mathrm{0}.\mathrm{4697} \\ $$

Answered by mathmax by abdo last updated on 07/Dec/20

I =∫_(π/6) ^(π/3)  ((sinx)/x)dx  let determine spproximste vslue of I  we hsve  x−(x^3 /6)≤sinx≤x ⇒1−(x^2 /6)≤((sinx)/x)≤1  ⇒  ∫_(π/6) ^(π/3) (1−(x^2 /6))dx ≤∫_(π/6) ^(π/3)  ((sinx)/x)dx≤(π/3)−(π/6)(=(π/6))  but  ∫_(π/6) ^(π/3) (1−(x^2 /6))dx =[x−(1/(18))x^3 ]_(π/6) ^(π/3) =(π/3)−(1/(18))×(π^3 /(27))−(π/6)+(1/(18))×(π^3 /6^3 )  =(π/6)−(π^3 /(18.27))+(π^3 /(18.216)) ⇒(π/6)+((1/(18.216))−(1/(18.27)))π^3 ≤I≤(π/6)    so v_0 =(π/6)+((1/(36.216))−(1/(36.27)))π^3   is a approximate value of this  integral0

$$\mathrm{I}\:=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{\mathrm{sinx}}{\mathrm{x}}\mathrm{dx}\:\:\mathrm{let}\:\mathrm{determine}\:\mathrm{spproximste}\:\mathrm{vslue}\:\mathrm{of}\:\mathrm{I} \\ $$$$\mathrm{we}\:\mathrm{hsve}\:\:\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\leqslant\mathrm{sinx}\leqslant\mathrm{x}\:\Rightarrow\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{6}}\leqslant\frac{\mathrm{sinx}}{\mathrm{x}}\leqslant\mathrm{1}\:\:\Rightarrow \\ $$$$\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{6}}\right)\mathrm{dx}\:\leqslant\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{\mathrm{sinx}}{\mathrm{x}}\mathrm{dx}\leqslant\frac{\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{6}}\left(=\frac{\pi}{\mathrm{6}}\right)\:\:\mathrm{but} \\ $$$$\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{6}}\right)\mathrm{dx}\:=\left[\mathrm{x}−\frac{\mathrm{1}}{\mathrm{18}}\mathrm{x}^{\mathrm{3}} \right]_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} =\frac{\pi}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{18}}×\frac{\pi^{\mathrm{3}} }{\mathrm{27}}−\frac{\pi}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{18}}×\frac{\pi^{\mathrm{3}} }{\mathrm{6}^{\mathrm{3}} } \\ $$$$=\frac{\pi}{\mathrm{6}}−\frac{\pi^{\mathrm{3}} }{\mathrm{18}.\mathrm{27}}+\frac{\pi^{\mathrm{3}} }{\mathrm{18}.\mathrm{216}}\:\Rightarrow\frac{\pi}{\mathrm{6}}+\left(\frac{\mathrm{1}}{\mathrm{18}.\mathrm{216}}−\frac{\mathrm{1}}{\mathrm{18}.\mathrm{27}}\right)\pi^{\mathrm{3}} \leqslant\mathrm{I}\leqslant\frac{\pi}{\mathrm{6}} \\ $$$$ \\ $$$$\mathrm{so}\:\mathrm{v}_{\mathrm{0}} =\frac{\pi}{\mathrm{6}}+\left(\frac{\mathrm{1}}{\mathrm{36}.\mathrm{216}}−\frac{\mathrm{1}}{\mathrm{36}.\mathrm{27}}\right)\pi^{\mathrm{3}} \:\:\mathrm{is}\:\mathrm{a}\:\mathrm{approximate}\:\mathrm{value}\:\mathrm{of}\:\mathrm{this} \\ $$$$\mathrm{integral0} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com