Question Number 92772 by Power last updated on 09/May/20
Commented by mathmax by abdo last updated on 09/May/20
![A =∫_0 ^6 [x] sin(((πx)/6))dx ⇒ A =Σ_(k=0) ^5 ∫_k ^(k+1) k sin(((πx)/6))dx =Σ_(k=0) ^5 k ∫_k ^(k+1) sin(((πx)/6))dx =Σ_(k=0) ^5 k [−(6/π)cos(((πx)/6))]_k ^(k+1) =−(6/π) Σ_(k=0) ^5 k{ cos(((π(k+1))/6))−cos(((kπ)/6))} =−(6/π)( cos((π/3))−cos((π/6))+cos((π/2))−cos((π/3))+cos(((2π)/3))−cos((π/2)) +cos(((5π)/6))−cos(((2π)/3)) +cos(π)−cos(((5π)/6))) =−(6/π){−((√3)/2)−(1/2) +(1/2) −1} =(6/π)(1+((√3)/2)) =(6/(2π))(1+(√3))](https://www.tinkutara.com/question/Q92911.png)
$${A}\:=\int_{\mathrm{0}} ^{\mathrm{6}} \:\left[{x}\right]\:{sin}\left(\frac{\pi{x}}{\mathrm{6}}\right){dx}\:\Rightarrow\:{A}\:=\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} \:\int_{{k}} ^{{k}+\mathrm{1}} \:{k}\:{sin}\left(\frac{\pi{x}}{\mathrm{6}}\right){dx} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} \:{k}\:\int_{{k}} ^{{k}+\mathrm{1}} \:{sin}\left(\frac{\pi{x}}{\mathrm{6}}\right){dx} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} \:{k}\:\left[−\frac{\mathrm{6}}{\pi}{cos}\left(\frac{\pi{x}}{\mathrm{6}}\right)\right]_{{k}} ^{{k}+\mathrm{1}} \\ $$$$=−\frac{\mathrm{6}}{\pi}\:\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} {k}\left\{\:{cos}\left(\frac{\pi\left({k}+\mathrm{1}\right)}{\mathrm{6}}\right)−{cos}\left(\frac{{k}\pi}{\mathrm{6}}\right)\right\} \\ $$$$=−\frac{\mathrm{6}}{\pi}\left(\:\:{cos}\left(\frac{\pi}{\mathrm{3}}\right)−{cos}\left(\frac{\pi}{\mathrm{6}}\right)+{cos}\left(\frac{\pi}{\mathrm{2}}\right)−{cos}\left(\frac{\pi}{\mathrm{3}}\right)+{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)−{cos}\left(\frac{\pi}{\mathrm{2}}\right)\right. \\ $$$$\left.+{cos}\left(\frac{\mathrm{5}\pi}{\mathrm{6}}\right)−{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\:+{cos}\left(\pi\right)−{cos}\left(\frac{\mathrm{5}\pi}{\mathrm{6}}\right)\right) \\ $$$$=−\frac{\mathrm{6}}{\pi}\left\{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:−\mathrm{1}\right\}\:=\frac{\mathrm{6}}{\pi}\left(\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:=\frac{\mathrm{6}}{\mathrm{2}\pi}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right) \\ $$