Question Number 83225 by 09658867628 last updated on 28/Feb/20
![∫_( 0) ^(100) sin (x−[x])π dx =](https://www.tinkutara.com/question/Q83225.png)
$$\:\underset{\:\mathrm{0}} {\overset{\mathrm{100}} {\int}}\:\mathrm{sin}\:\left({x}−\left[{x}\right]\right)\pi\:{dx}\:= \\ $$
Commented by mathmax by abdo last updated on 29/Feb/20
![A =Σ_(n=0) ^(99) ∫_n ^(n+1) sin((x−n)π)dx =Σ_(n=0) ^(99) ∫_n ^(n+1) sin(πx−nπ)dx =Σ_(n=0) ^(99) ∫_n ^(n+1) (−1)^n sin(πx)dx =Σ_(n=0) ^(99) (−1)^n [−(1/π)cos(πx)]_n ^(n+1) =(1/π)Σ_(n=0) ^(99) (−1)^(n+1) {cos(n+1)π −cos(nπ)} =(1/π)Σ_(n=0) ^(99) (−1)^(n+1) {(−1)^(n+1) −(−1)^n } =−(2/π)Σ_(n=0) ^(99) (−1)^n ×(−1)^(n+1) =(2/π)Σ_(n=0) ^(99) (1) =((200)/π)](https://www.tinkutara.com/question/Q83247.png)
$${A}\:=\sum_{{n}=\mathrm{0}} ^{\mathrm{99}} \int_{{n}} ^{{n}+\mathrm{1}} {sin}\left(\left({x}−{n}\right)\pi\right){dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\mathrm{99}} \:\int_{{n}} ^{{n}+\mathrm{1}} {sin}\left(\pi{x}−{n}\pi\right){dx}\:=\sum_{{n}=\mathrm{0}} ^{\mathrm{99}} \:\int_{{n}} ^{{n}+\mathrm{1}} \left(−\mathrm{1}\right)^{{n}} {sin}\left(\pi{x}\right){dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\mathrm{99}} \left(−\mathrm{1}\right)^{{n}} \:\:\left[−\frac{\mathrm{1}}{\pi}{cos}\left(\pi{x}\right)\right]_{{n}} ^{{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\pi}\sum_{{n}=\mathrm{0}} ^{\mathrm{99}} \left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left\{{cos}\left({n}+\mathrm{1}\right)\pi\:−{cos}\left({n}\pi\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\pi}\sum_{{n}=\mathrm{0}} ^{\mathrm{99}} \left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left\{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} −\left(−\mathrm{1}\right)^{{n}} \right\} \\ $$$$=−\frac{\mathrm{2}}{\pi}\sum_{{n}=\mathrm{0}} ^{\mathrm{99}} \left(−\mathrm{1}\right)^{{n}} ×\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:=\frac{\mathrm{2}}{\pi}\sum_{{n}=\mathrm{0}} ^{\mathrm{99}} \left(\mathrm{1}\right) \\ $$$$=\frac{\mathrm{200}}{\pi} \\ $$