Question Number 120628 by TANMAY PANACEA last updated on 01/Nov/20
$${selective}\:{intregals} \\ $$
Commented by TANMAY PANACEA last updated on 01/Nov/20
$${thanks} \\ $$$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{{b}} {f}\left({a}+{b}−{x}\right){dx} \\ $$
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
$$\underset{−\frac{\pi}{\mathrm{2}}} {\int}^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{{e}^{{sinx}} +\mathrm{1}}.{p}={I} \\ $$$$=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{{e}^{−{sinx}} +\mathrm{1}}={I} \\ $$$$\mathrm{2}{I}=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {dx} \\ $$$${I}=\frac{\pi}{\mathrm{2}} \\ $$
Commented by mathmax by abdo last updated on 01/Nov/20
![I =∫_0 ^(π/3) [(√3)tanx]dx changement (√3)tanx =t give tanx=(t/( (√3))) ⇒x=arctan((t/( (√3)))) I =∫_0 ^3 [t]×(1/( (√3)(1+(t^2 /3))))dt =(√3)∫_0 ^3 (([t])/(t^2 +3))dt= =(√3){∫_0 ^1 0dt +∫_1 ^2 (1/(t^2 +3))dt +∫_2 ^3 (2/(t^2 +3))dt} ∫_1 ^2 (dt/(t^2 +3)) =_(t=(√3)u) ∫_(1/( (√3))) ^(2/( (√3))) (((√3)du)/(3(1+u^2 ))) =((√3)/3)[arctanu]_(1/( (√3))) ^(2/( (√3))) =((√3)/3){arctan((2/( (√3))))−arctan((1/( (√3))))} ∫_2 ^3 ((2dt)/(t^2 +3)) =_(t=(√3)u) ∫_(2/( (√3))) ^(√3) ((2(√3)du)/(3(1+u^2 ))) =((2(√3))/3){arctan((√3))−arctan((2/( (√3))))} ⇒ I =arctan((2/( (√3))))−arctan((1/( (√3))))+2 arctan((√3))−2arctan((2/( (√3)))) =−arctan((2/( (√3))))−((π/2)−arctan((√3)))+2arctan((√3)) =3arctan((√3))−(π/2)−arctan((2/( (√3)))) we have arctan((√3))=(π/3) ⇒I =π−(π/2) −arctan((2/( (√3)))) I=(π/2)−arctan((2/( (√3))))(→c)](https://www.tinkutara.com/question/Q120655.png)
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \left[\sqrt{\mathrm{3}}\mathrm{tanx}\right]\mathrm{dx}\:\:\:\mathrm{changement}\:\sqrt{\mathrm{3}}\mathrm{tanx}\:=\mathrm{t}\:\mathrm{give}\:\mathrm{tanx}=\frac{\mathrm{t}}{\:\sqrt{\mathrm{3}}}\:\Rightarrow\mathrm{x}=\mathrm{arctan}\left(\frac{\mathrm{t}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{3}} \left[\mathrm{t}\right]×\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\left(\mathrm{1}+\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{3}}\right)}\mathrm{dt}\:=\sqrt{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{3}} \:\:\frac{\left[\mathrm{t}\right]}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{3}}\mathrm{dt}= \\ $$$$=\sqrt{\mathrm{3}}\left\{\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{0dt}\:+\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} +\mathrm{3}}\mathrm{dt}\:+\int_{\mathrm{2}} ^{\mathrm{3}} \:\frac{\mathrm{2}}{\mathrm{t}^{\mathrm{2}} +\mathrm{3}}\mathrm{dt}\right\} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} +\mathrm{3}}\:=_{\mathrm{t}=\sqrt{\mathrm{3}}\mathrm{u}} \:\:\:\:\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}} \:\:\:\:\frac{\sqrt{\mathrm{3}}\mathrm{du}}{\mathrm{3}\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left[\mathrm{arctanu}\right]_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left\{\mathrm{arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)−\mathrm{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right\} \\ $$$$\int_{\mathrm{2}} ^{\mathrm{3}} \:\:\frac{\mathrm{2dt}}{\mathrm{t}^{\mathrm{2}} +\mathrm{3}}\:=_{\mathrm{t}=\sqrt{\mathrm{3}}\mathrm{u}} \:\:\:\:\int_{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}} ^{\sqrt{\mathrm{3}}} \:\:\:\:\frac{\mathrm{2}\sqrt{\mathrm{3}}\mathrm{du}}{\mathrm{3}\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}\:=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\left\{\mathrm{arctan}\left(\sqrt{\mathrm{3}}\right)−\mathrm{arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)\right\}\:\Rightarrow \\ $$$$\mathrm{I}\:=\mathrm{arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)−\mathrm{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+\mathrm{2}\:\mathrm{arctan}\left(\sqrt{\mathrm{3}}\right)−\mathrm{2arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$=−\mathrm{arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)−\left(\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\left(\sqrt{\mathrm{3}}\right)\right)+\mathrm{2arctan}\left(\sqrt{\mathrm{3}}\right) \\ $$$$=\mathrm{3arctan}\left(\sqrt{\mathrm{3}}\right)−\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{arctan}\left(\sqrt{\mathrm{3}}\right)=\frac{\pi}{\mathrm{3}}\:\Rightarrow\mathrm{I}\:=\pi−\frac{\pi}{\mathrm{2}}\:−\mathrm{arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$\mathrm{I}=\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)\left(\rightarrow\mathrm{c}\right) \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 01/Nov/20
$$\mathrm{A}\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{sinx}} \:+\mathrm{1}}\:\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}=−\mathrm{t}\:\Rightarrow \\ $$$$\mathrm{A}\:=−\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{−\mathrm{dt}}{\mathrm{e}^{−\mathrm{sint}} +\mathrm{1}}\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{dx}}{\mathrm{e}^{−\mathrm{sinx}} +\mathrm{1}}\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{e}^{\mathrm{sinx}} }{\mathrm{1}+\mathrm{e}^{\mathrm{sinx}} }\mathrm{dx}\:\Rightarrow \\ $$$$\mathrm{2A}\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{e}^{\mathrm{sinx}} }+\frac{\mathrm{e}^{\mathrm{sinx}} }{\mathrm{1}+\mathrm{e}^{\mathrm{sinx}} }\right)\mathrm{dx}\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{dx}\:=\pi\:\Rightarrow\mathrm{A}\:=\frac{\pi}{\mathrm{2}} \\ $$
Commented by mathmax by abdo last updated on 01/Nov/20
$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}^{+} } \:\:\:\:\frac{\int_{\mathrm{1}} ^{\mathrm{x}} \mid\mathrm{t}−\mathrm{1}\mid\mathrm{dt}}{\mathrm{sin}\left(\mathrm{x}−\mathrm{1}\right)}\:=_{\mathrm{t}−\mathrm{1}=\mathrm{u}} \:\:\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}^{+} } \:\:\:\:\frac{\int_{\mathrm{0}} ^{\mathrm{x}−\mathrm{1}} \mid\mathrm{u}\mid\mathrm{du}}{\mathrm{sin}\left(\mathrm{x}−\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}^{+} } \:\:\:\:\frac{\mathrm{f}^{'} \left(\mathrm{x}\right)}{\mathrm{g}^{'} \left(\mathrm{x}\right)}\:=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}^{+} } \:\:\:\:\frac{\mathrm{x}−\mathrm{1}}{\left(\mathrm{x}−\mathrm{1}\right)\mathrm{cos}\left(\mathrm{x}−\mathrm{1}\right)}=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}^{+} } \:\:\:\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{x}−\mathrm{1}\right)} \\ $$$$=\mathrm{1} \\ $$
Commented by TANMAY PANACEA last updated on 01/Nov/20
$${thank}\:{you}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 01/Nov/20
![let f(x)=(1/n^3 ){[1^2 x]+[2^2 x]+.....[n^2 x]} =(1/n^3 )Σ_(k=1) ^n [k^2 x] we have [u]≤u <[u]+1 ⇒u−1 <[u]≤u ⇒k^2 x−1 <[k^2 x]≤k^2 x ⇒ Σ_(k=1) ^n (k^2 x−1)<Σ_(k=1) ^n [k^2 x]≤Σ_(k=1) ^n k^2 x ⇒ x Σ_(k=1) ^n k^2 −n <Σ_(k=1) ^n [k^2 x]≤x Σ_(k=1) ^n k^2 ⇒ (x/6)n(n+1)(2n+1)−n<Σ_(k=1) ^n [k^2 x]≤((xn(n+1)(2n+1))/6) ⇒ ((xn(n+1)(2n+1))/(6n^3 ))−(1/n^2 )<f(x)≤((xn(n+1)(2n+1))/6) we have lim_(n→+∞) (x/(6n^3 ))n(n+1)(2n+1) =xlim_(n→+∞) ((2n^3 )/(6n^3 )) =(x/3) ⇒ lim_(n→+∞) f(x) =(x/3)](https://www.tinkutara.com/question/Q120663.png)
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\left\{\left[\mathrm{1}^{\mathrm{2}} \mathrm{x}\right]+\left[\mathrm{2}^{\mathrm{2}} \mathrm{x}\right]+…..\left[\mathrm{n}^{\mathrm{2}} \mathrm{x}\right]\right\}\:=\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left[\mathrm{k}^{\mathrm{2}} \mathrm{x}\right]\:\mathrm{we}\:\mathrm{have} \\ $$$$\left[\mathrm{u}\right]\leqslant\mathrm{u}\:<\left[\mathrm{u}\right]+\mathrm{1}\:\Rightarrow\mathrm{u}−\mathrm{1}\:<\left[\mathrm{u}\right]\leqslant\mathrm{u}\:\Rightarrow\mathrm{k}^{\mathrm{2}} \mathrm{x}−\mathrm{1}\:<\left[\mathrm{k}^{\mathrm{2}} \mathrm{x}\right]\leqslant\mathrm{k}^{\mathrm{2}} \mathrm{x}\:\Rightarrow \\ $$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(\mathrm{k}^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)<\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\left[\mathrm{k}^{\mathrm{2}} \mathrm{x}\right]\leqslant\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{k}^{\mathrm{2}} \mathrm{x}\:\Rightarrow \\ $$$$\mathrm{x}\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{k}^{\mathrm{2}} −\mathrm{n}\:<\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left[\mathrm{k}^{\mathrm{2}} \mathrm{x}\right]\leqslant\mathrm{x}\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{k}^{\mathrm{2}} \:\Rightarrow \\ $$$$\frac{\mathrm{x}}{\mathrm{6}}\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)−\mathrm{n}<\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left[\mathrm{k}^{\mathrm{2}} \mathrm{x}\right]\leqslant\frac{\mathrm{xn}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{6}}\:\Rightarrow \\ $$$$\frac{\mathrm{xn}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{6n}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }<\mathrm{f}\left(\mathrm{x}\right)\leqslant\frac{\mathrm{xn}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{6}}\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\mathrm{x}}{\mathrm{6n}^{\mathrm{3}} }\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)\:=\mathrm{xlim}_{\mathrm{n}\rightarrow+\infty} \:\:\frac{\mathrm{2n}^{\mathrm{3}} }{\mathrm{6n}^{\mathrm{3}} }\:=\frac{\mathrm{x}}{\mathrm{3}}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{x}}{\mathrm{3}} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 01/Nov/20
$$\mathrm{Q}\:\mathrm{45}\:\mathrm{is}\:\mathrm{not}\:\mathrm{clear} \\ $$
Commented by TANMAY PANACEA last updated on 01/Nov/20
$${great} \\ $$
Commented by TANMAY PANACEA last updated on 01/Nov/20
$${ok}\:{sir}\:{let}\:{me}\:{see}\:{the}\:{source}\:{book} \\ $$
Commented by Bird last updated on 01/Nov/20
$${you}\:{are}\:{welcome} \\ $$
Commented by bobhans last updated on 01/Nov/20
$$\left(\mathrm{42}\right)\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\underset{\mathrm{1}} {\overset{\mathrm{x}} {\int}}\:\mid\mathrm{t}−\mathrm{1}\mid\:\mathrm{dt}}{\mathrm{sin}\:\left(\mathrm{x}−\mathrm{1}\right)}\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mid\mathrm{x}−\mathrm{1}\mid}{\mathrm{cos}\:\left(\mathrm{x}−\mathrm{1}\right)}\:=\:\mathrm{0} \\ $$