Question

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

Answer 1

Answer:

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

Step-by-step explanation:

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

Compute the indefinite integral:

$\displaystyle\int \sin \left(x\right)dx$

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

$=-\cos \left(x\right)$

Add a constant to the solution

$=-\cos \left(x\right)+C$

Compute the boundaries:

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

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

$\mathrm{Plug\:in\:the\:value}\:x=0$

$=-\cos \left(0\right)$

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

$=-1$

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

$\mathrm{Plug\:in\:the\:value}\:x=\pi$

$=-\cos \left(\pi \right)$

$\mathrm{Simplify\:}-\cos \left(\pi \right)$

$-\cos \left(\pi \right)$

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

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

$\mathrm{Apply\:rule}\:-\left(-a\right)=a$

$=1$

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

$\mathrm{Simplify}$

$=2$