Question

$\int _0^{\pi }\sin \left(x\right)dx \::value$

Answer 1

$\int _0^{\pi }\sin \left(x\right)dx=2$

$Steps$

$\mathrm{Use\:the\:common\:integral}:\quad \int \sin \left(x\right)dx=-\cos \left(x\right)$

$=\left[-\cos \left(x\right)\right]^{\pi }_0$

$\mathrm{Compute\:the\:boundaries}:\quad \left[-\cos \left(x\right)\right]^{\pi }_0=2$

$\int _a^bf\left(x\right)dx=F\left(b\right)-F\left(a\right)=\lim _{x\to \:b-}\left(F\left(x\right)\right)-\lim _{x\to \:a+}\left(F\left(x\right)\right)$

$\lim _{x\to \:0+}\left(-\cos \left(x\right)\right)=-1$

$\lim _{x\to \:\pi -}\left(-\cos \left(x\right)\right)=1$

$=1-\left(-1\right)$

$=2$

Answer 2

$\mathrm{Use\:the\:common\:integral}:\quad \int \sin \left(x\right)dx=-\cos \left(x\right)$

$=\left[-\cos \left(x\right)\right]^{\pi }_0$

$\mathrm{Use\:the\:following\:trivial\:identity}:\quad \cos \left(0\right)=1$

$\lim _{x\to \:\pi -}\left(-\cos \left(x\right)\right)=1$

$\lim _{x\to \:0+}\left(-\cos \left(x\right)\right)=-1$

$=1-\left(-1\right)$

= 2