Question Number 47651 by prof Abdo imad last updated on 12/Nov/18
![calculate A_n =∫_0 ^1 sin([nx])e^(−2x) dx with n integr natural .](Q47651.png)
$${calculate}\:\:{A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{sin}\left(\left[{nx}\right]\right){e}^{−\mathrm{2}{x}} {dx}\:{with}\:{n} \\ $$$${integr}\:{natural}\:. \\ $$
Commented by maxmathsup by imad last updated on 13/Nov/18
![we have A_n =Σ_(k=0) ^(n−1) ∫_(k/n) ^((k+1)/n) sin([nx])e^(−2x) dx =Σ_(k=0) ^(n−1) ∫_(k/n) ^((k+1)/n) sin(k)e^(−2x) dx =Σ_(k=0) ^n sin(k)∫_(k/n) ^((k+1)/n) e^(−2x) dx =−(1/2)Σ_(k=0) ^(n−1) sin(k)( e^(−2((k+1)/n)) −e^((−2k)/n) ) =(1/2) Σ_(k=0) ^(n−1) e^((−2k)/n) −(1/2)Σ_(k=0) ^(n−1) e^(−2((k+1)/n)) sin(k) =((1−e^(−(2/n)) )/2) Σ_(k=0) ^(n−1) e^((−2k)/n) sin(k) but Σ_(k=0) ^(n−) e^((−2k)/n) sink =Im(Σ_(k=0) ^(n−1) e^(ik−(2/n)k) ) and Σ_(k=0) ^(n−1) e^(ik−(2/n)k) =Σ_(k=0) ^(n−1) e^((i−(2/n))k) =((1−(e^((i−(2/n))) )^n )/(1−e^(i−(2/n)) )) =((1−e^(−2) e^(in) )/(1−e^(−(2/n)) e^i )) =((1−e^(−2) cosn −ie^(−2) sin(n))/(1−e^(−(2/n)) cos(1)−i e^(−(2/n)) sin(1))) =(((1−e^(−2) cos(n)−i e^(−2) sinn)(1−e^(−(2/n)) cos(1)+i e^(−(2/n)) sin(1)))/((1−e^(−2) cos(1))^2 +e^(−(4/n)) sin^2 (1))) =(((1−e^(−2) cos(n))(1−e^(−(2/n)) cos(1))+i(1−e^(−2) cos(n))e^(−(2/n)) sin(1)−ie^(−2) sin(n)(1−e^(−(2/n)) cos(1))+e^(4/n) sin(n)sin(1))/((1−e^(−2) cos(1))^2 +e^(−(4/n)) sin^2 (1))) so the value of A_n is clear determined after extracting Im(Σ....).](Q47719.png)
$${we}\:{have}\:{A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{\frac{{k}}{{n}}} ^{\frac{{k}+\mathrm{1}}{{n}}} \:{sin}\left(\left[{nx}\right]\right){e}^{−\mathrm{2}{x}} {dx} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{\frac{{k}}{{n}}} ^{\frac{{k}+\mathrm{1}}{{n}}} \:{sin}\left({k}\right){e}^{−\mathrm{2}{x}} {dx}\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{sin}\left({k}\right)\int_{\frac{{k}}{{n}}} ^{\frac{{k}+\mathrm{1}}{{n}}} \:{e}^{−\mathrm{2}{x}} {dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{sin}\left({k}\right)\left(\:{e}^{−\mathrm{2}\frac{{k}+\mathrm{1}}{{n}}} −{e}^{\frac{−\mathrm{2}{k}}{{n}}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{\frac{−\mathrm{2}{k}}{{n}}} \:−\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{−\mathrm{2}\frac{{k}+\mathrm{1}}{{n}}} \:{sin}\left({k}\right) \\ $$$$=\frac{\mathrm{1}−{e}^{−\frac{\mathrm{2}}{{n}}} }{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{\frac{−\mathrm{2}{k}}{{n}}} \:{sin}\left({k}\right)\:{but}\:\sum_{{k}=\mathrm{0}} ^{{n}−} \:{e}^{\frac{−\mathrm{2}{k}}{{n}}} \:{sink}\:={Im}\left(\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{{ik}−\frac{\mathrm{2}}{{n}}{k}} \right)\:{and} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{{ik}−\frac{\mathrm{2}}{{n}}{k}} \:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{\left({i}−\frac{\mathrm{2}}{{n}}\right){k}} \:=\frac{\mathrm{1}−\left({e}^{\left({i}−\frac{\mathrm{2}}{{n}}\right)} \right)^{{n}} }{\mathrm{1}−{e}^{{i}−\frac{\mathrm{2}}{{n}}} }\:=\frac{\mathrm{1}−{e}^{−\mathrm{2}} \:{e}^{{in}} }{\mathrm{1}−{e}^{−\frac{\mathrm{2}}{{n}}} \:{e}^{{i}} } \\ $$$$=\frac{\mathrm{1}−{e}^{−\mathrm{2}} {cosn}\:−{ie}^{−\mathrm{2}} {sin}\left({n}\right)}{\mathrm{1}−{e}^{−\frac{\mathrm{2}}{{n}}} {cos}\left(\mathrm{1}\right)−{i}\:{e}^{−\frac{\mathrm{2}}{{n}}} {sin}\left(\mathrm{1}\right)} \\ $$$$=\frac{\left(\mathrm{1}−{e}^{−\mathrm{2}} {cos}\left({n}\right)−{i}\:{e}^{−\mathrm{2}} {sinn}\right)\left(\mathrm{1}−{e}^{−\frac{\mathrm{2}}{{n}}} {cos}\left(\mathrm{1}\right)+{i}\:{e}^{−\frac{\mathrm{2}}{{n}}} {sin}\left(\mathrm{1}\right)\right)}{\left(\mathrm{1}−{e}^{−\mathrm{2}} {cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:+{e}^{−\frac{\mathrm{4}}{{n}}} {sin}^{\mathrm{2}} \left(\mathrm{1}\right)} \\ $$$$=\frac{\left(\mathrm{1}−{e}^{−\mathrm{2}} {cos}\left({n}\right)\right)\left(\mathrm{1}−{e}^{−\frac{\mathrm{2}}{{n}}} {cos}\left(\mathrm{1}\right)\right)+{i}\left(\mathrm{1}−{e}^{−\mathrm{2}} {cos}\left({n}\right)\right){e}^{−\frac{\mathrm{2}}{{n}}} {sin}\left(\mathrm{1}\right)−{ie}^{−\mathrm{2}} {sin}\left({n}\right)\left(\mathrm{1}−{e}^{−\frac{\mathrm{2}}{{n}}} {cos}\left(\mathrm{1}\right)\right)+{e}^{\frac{\mathrm{4}}{{n}}} {sin}\left({n}\right){sin}\left(\mathrm{1}\right)}{\left(\mathrm{1}−{e}^{−\mathrm{2}} {cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:+{e}^{−\frac{\mathrm{4}}{{n}}} \:{sin}^{\mathrm{2}} \left(\mathrm{1}\right)} \\ $$$${so}\:{the}\:{value}\:{of}\:\:{A}_{{n}} \:{is}\:{clear}\:{determined}\:{after}\:{extracting}\:{Im}\left(\Sigma....\right). \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 13/Nov/18
![∫_0 ^(1/n) sin([nx])e^(−2x) dx+∫_(1/n) ^1 sin([nx])e^(−2x) dx ∫_0 ^(1/n) sin(0)e^(−2x) dx+sin(1)∫_(1/n) ^1 e^(−2x) dx sin(1)×∣(e^(−2x) /(−2))∣_(1/n) ^1 sin(1)×(((e^(−2) −e^((−2)/n) )/(−2)))](Q47670.png)
$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} {sin}\left(\left[{nx}\right]\right){e}^{−\mathrm{2}{x}} {dx}+\int_{\frac{\mathrm{1}}{{n}}} ^{\mathrm{1}} {sin}\left(\left[{nx}\right]\right){e}^{−\mathrm{2}{x}} {dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{n}}} {sin}\left(\mathrm{0}\right){e}^{−\mathrm{2}{x}} {dx}+{sin}\left(\mathrm{1}\right)\int_{\frac{\mathrm{1}}{{n}}} ^{\mathrm{1}} {e}^{−\mathrm{2}{x}} {dx} \\ $$$${sin}\left(\mathrm{1}\right)×\mid\frac{{e}^{−\mathrm{2}{x}} }{−\mathrm{2}}\mid_{\frac{\mathrm{1}}{{n}}} ^{\mathrm{1}} \\ $$$${sin}\left(\mathrm{1}\right)×\left(\frac{{e}^{−\mathrm{2}} −{e}^{\frac{−\mathrm{2}}{{n}}} }{−\mathrm{2}}\right) \\ $$