UNKNOWN Questions

Question Number 83225 by 09658867628 last updated on 28/Feb/20

$$\:\underset{\:\mathrm{0}} {\overset{\mathrm{100}} {\int}}\:\mathrm{sin}\:\left({x}−\left[{x}\right]\right)\pi\:{dx}\:= \\$$

Commented by mathmax by abdo last updated on 29/Feb/20

$${A}\:=\sum_{{n}=\mathrm{0}} ^{\mathrm{99}} \int_{{n}} ^{{n}+\mathrm{1}} {sin}\left(\left({x}−{n}\right)\pi\right){dx} \\$$$$=\sum_{{n}=\mathrm{0}} ^{\mathrm{99}} \:\int_{{n}} ^{{n}+\mathrm{1}} {sin}\left(\pi{x}−{n}\pi\right){dx}\:=\sum_{{n}=\mathrm{0}} ^{\mathrm{99}} \:\int_{{n}} ^{{n}+\mathrm{1}} \left(−\mathrm{1}\right)^{{n}} {sin}\left(\pi{x}\right){dx} \\$$$$=\sum_{{n}=\mathrm{0}} ^{\mathrm{99}} \left(−\mathrm{1}\right)^{{n}} \:\:\left[−\frac{\mathrm{1}}{\pi}{cos}\left(\pi{x}\right)\right]_{{n}} ^{{n}+\mathrm{1}} \\$$$$=\frac{\mathrm{1}}{\pi}\sum_{{n}=\mathrm{0}} ^{\mathrm{99}} \left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left\{{cos}\left({n}+\mathrm{1}\right)\pi\:−{cos}\left({n}\pi\right)\right\} \\$$$$=\frac{\mathrm{1}}{\pi}\sum_{{n}=\mathrm{0}} ^{\mathrm{99}} \left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left\{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} −\left(−\mathrm{1}\right)^{{n}} \right\} \\$$$$=−\frac{\mathrm{2}}{\pi}\sum_{{n}=\mathrm{0}} ^{\mathrm{99}} \left(−\mathrm{1}\right)^{{n}} ×\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:=\frac{\mathrm{2}}{\pi}\sum_{{n}=\mathrm{0}} ^{\mathrm{99}} \left(\mathrm{1}\right) \\$$$$=\frac{\mathrm{200}}{\pi} \\$$