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Question Number 120628 by TANMAY PANACEA last updated on 01/Nov/20

$$selectiveintregals$$

Commented by TANMAY PANACEA last updated on 01/Nov/20

$$thanks$$$${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$$

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by Dwaipayan Shikari last updated on 01/Nov/20

$${\underset{-\frac{\pi }{2}}{\int }}^{\frac{\pi }{2}}\frac{dx}{e^{sinx}+1}.p=I$$$$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{dx}{e^{-sinx}+1}=I$$$$2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}dx$$$$I=\frac{\pi }{2}$$

Commented by mathmax by abdo last updated on 01/Nov/20

$$\mathrm{I}={\int }_0^{\frac{\pi }{3}}\left[\sqrt{3}\mathrm{tanx}\right]\mathrm{dx}\mathrm{changement}\sqrt{3}\mathrm{tanx}=\mathrm{t}\mathrm{give}\mathrm{tanx}=\frac{\mathrm{t}}{\sqrt{3}}\Rightarrow \mathrm{x}=\arctan \left(\frac{\mathrm{t}}{\sqrt{3}}\right)$$$$\mathrm{I}={\int }_0^3\left[\mathrm{t}\right]\times \frac{1}{\sqrt{3}(1+\frac{{\mathrm{t}}^2}{3})}\mathrm{dt}=\sqrt{3}{\int }_0^3\frac{\left[\mathrm{t}\right]}{{\mathrm{t}}^2+3}\mathrm{dt}=$$$$=\sqrt{3}\{{\int }_0^{1}0\mathrm{dt}+{\int }_1^2\frac{1}{{\mathrm{t}}^2+3}\mathrm{dt}+{\int }_2^3\frac{2}{{\mathrm{t}}^2+3}\mathrm{dt}\}$$$${\int }_1^2\frac{\mathrm{dt}}{{\mathrm{t}}^2+3}{=}_{\mathrm{t}=\sqrt{3}\mathrm{u}}{\int }_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}\frac{\sqrt{3}\mathrm{du}}{3(1+{\mathrm{u}}^2)}=\frac{\sqrt{3}}{3}{\left[\mathrm{arctanu}\right]}_{\frac{1}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}}$$$$=\frac{\sqrt{3}}{3}\{\arctan \left(\frac{2}{\sqrt{3}}\right)-\arctan \left(\frac{1}{\sqrt{3}}\right)\}$$$${\int }_2^3\frac{2\mathrm{dt}}{{\mathrm{t}}^2+3}{=}_{\mathrm{t}=\sqrt{3}\mathrm{u}}{\int }_{\frac{2}{\sqrt{3}}}^{\sqrt{3}}\frac{2\sqrt{3}\mathrm{du}}{3(1+{\mathrm{u}}^2)}=\frac{2\sqrt{3}}{3}\{\arctan \left(\sqrt{3}\right)-\arctan \left(\frac{2}{\sqrt{3}}\right)\}\Rightarrow$$$$\mathrm{I}=\arctan \left(\frac{2}{\sqrt{3}}\right)-\arctan \left(\frac{1}{\sqrt{3}}\right)+2\arctan \left(\sqrt{3}\right)-2\arctan \left(\frac{2}{\sqrt{3}}\right)$$$$=-\arctan \left(\frac{2}{\sqrt{3}}\right)-(\frac{\pi }{2}-\arctan \left(\sqrt{3}\right))+2\arctan \left(\sqrt{3}\right)$$$$=3\arctan \left(\sqrt{3}\right)-\frac{\pi }{2}-\arctan \left(\frac{2}{\sqrt{3}}\right)$$$$\mathrm{we}\mathrm{have}\arctan \left(\sqrt{3}\right)=\frac{\pi }{3}\Rightarrow \mathrm{I}=\pi -\frac{\pi }{2}-\arctan \left(\frac{2}{\sqrt{3}}\right)$$$$\mathrm{I}=\frac{\pi }{2}-\arctan \left(\frac{2}{\sqrt{3}}\right)(\to \mathrm{c})$$$$$$

Commented by mathmax by abdo last updated on 01/Nov/20

$$\mathrm{A}={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\mathrm{dx}}{{\mathrm{e}}^{\mathrm{sinx}}+1}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}=-\mathrm{t}\Rightarrow$$$$\mathrm{A}=-{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{-\mathrm{dt}}{{\mathrm{e}}^{-\mathrm{sint}}+1}={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\mathrm{dx}}{{\mathrm{e}}^{-\mathrm{sinx}}+1}={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{\mathrm{e}}^{\mathrm{sinx}}}{1+{\mathrm{e}}^{\mathrm{sinx}}}\mathrm{dx}\Rightarrow$$$$2\mathrm{A}={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(\frac{1}{1+{\mathrm{e}}^{\mathrm{sinx}}}+\frac{{\mathrm{e}}^{\mathrm{sinx}}}{1+{\mathrm{e}}^{\mathrm{sinx}}})\mathrm{dx}={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\mathrm{dx}=\pi \Rightarrow \mathrm{A}=\frac{\pi }{2}$$

Commented by mathmax by abdo last updated on 01/Nov/20

$${\lim }_{\mathrm{x}\to 1^{+}}\frac{{\int }_1^{\mathrm{x}}\mid \mathrm{t}-1\mid \mathrm{dt}}{\sin (\mathrm{x}-1)}{=}_{\mathrm{t}-1=\mathrm{u}}{\lim }_{\mathrm{x}\to 1^{+}}\frac{{\int }_0^{\mathrm{x}-1}\mid \mathrm{u}\mid \mathrm{du}}{\sin (\mathrm{x}-1)}$$$$={\lim }_{\mathrm{x}\to 1^{+}}\frac{{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)}{{\mathrm{g}}^{{}^{\prime }}\left(\mathrm{x}\right)}={\lim }_{\mathrm{x}\to 1^{+}}\frac{\mathrm{x}-1}{(\mathrm{x}-1)\cos (\mathrm{x}-1)}={\lim }_{\mathrm{x}\to 1^{+}}\frac{1}{\cos (\mathrm{x}-1)}$$$$=1$$

Commented by TANMAY PANACEA last updated on 01/Nov/20

$$thankyousir$$

Commented by mathmax by abdo last updated on 01/Nov/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{n}}^3}\{\left[1^2\mathrm{x}\right]+\left[2^2\mathrm{x}\right]+.....\left[{\mathrm{n}}^2\mathrm{x}\right]\}=\frac{1}{{\mathrm{n}}^3}\sum _{\mathrm{k}=1}^{\mathrm{n}}\left[{\mathrm{k}}^2\mathrm{x}\right]\mathrm{we}\mathrm{have}$$$$\left[\mathrm{u}\right]⩽\mathrm{u}<\left[\mathrm{u}\right]+1\Rightarrow \mathrm{u}-1<\left[\mathrm{u}\right]⩽\mathrm{u}\Rightarrow {\mathrm{k}}^2\mathrm{x}-1<\left[{\mathrm{k}}^2\mathrm{x}\right]⩽{\mathrm{k}}^2\mathrm{x}\Rightarrow$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}({\mathrm{k}}^2\mathrm{x}-1)<\sum _{\mathrm{k}=1}^{\mathrm{n}}\left[{\mathrm{k}}^2\mathrm{x}\right]⩽\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{k}}^2\mathrm{x}\Rightarrow$$$$\mathrm{x}\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{k}}^2-\mathrm{n}<\sum _{\mathrm{k}=1}^{\mathrm{n}}\left[{\mathrm{k}}^2\mathrm{x}\right]⩽\mathrm{x}\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{k}}^2\Rightarrow$$$$\frac{\mathrm{x}}{6}\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)-\mathrm{n}<\sum _{\mathrm{k}=1}^{\mathrm{n}}\left[{\mathrm{k}}^2\mathrm{x}\right]⩽\frac{\mathrm{xn}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}\Rightarrow$$$$\frac{\mathrm{xn}(\mathrm{n}+1)(2\mathrm{n}+1)}{{6\mathrm{n}}^3}-\frac{1}{{\mathrm{n}}^2}<\mathrm{f}\left(\mathrm{x}\right)⩽\frac{\mathrm{xn}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}\mathrm{we}\mathrm{have}$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{\mathrm{x}}{{6\mathrm{n}}^3}\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)={\mathrm{xlim}}_{\mathrm{n}\to +\mathrm{\infty }}\frac{{2\mathrm{n}}^3}{{6\mathrm{n}}^3}=\frac{\mathrm{x}}{3}\Rightarrow$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{3}$$$$$$

Commented by mathmax by abdo last updated on 01/Nov/20

$$\mathrm{Q}45\mathrm{is}\mathrm{not}\mathrm{clear}$$

Commented by TANMAY PANACEA last updated on 01/Nov/20

$$great$$

Commented by TANMAY PANACEA last updated on 01/Nov/20

$$oksirletmeseethesourcebook$$

Commented by Bird last updated on 01/Nov/20

$$youarewelcome$$

Commented by bobhans last updated on 01/Nov/20

$$\left(42\right)\underset{x\to 1}{\lim }\frac{\underset{1}{\stackrel{\mathrm{x}}{\int }}\mid \mathrm{t}-1\mid \mathrm{dt}}{\sin (\mathrm{x}-1)}=\underset{x\to 1}{\lim }\frac{\mid \mathrm{x}-1\mid }{\cos (\mathrm{x}-1)}=0$$

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