Question

integration of |sinx| from 0 to 2pi​

Answer 1

We see that,

$\sf{For\quad\!x\in[0,\ \pi],\quad|\sin(x)|=\sin(x)\quad\because\ \sin(x)\geq0}$

$\sf{For\quad\!x\in[\pi,\ 2\pi],\quad|\sin(x)|=-\sin(x)\quad \because\ \sin(x)\leq0}$

Therefore,

$\displaystyle\longrightarrow\sf{\int\limits_0^{2\pi}|\sin(x)|\ dx=\int\limits_0^{\pi}|\sin(x)|\ dx+\int\limits_{\pi}^{2\pi}|\sin(x)|\ dx}$

$\displaystyle\longrightarrow\sf{\int\limits_0^{2\pi}|\sin(x)|\ dx=\int\limits_0^{\pi}\sin(x)\ dx-\int\limits_{\pi}^{2\pi}\sin(x)\ dx}$

Since $\displaystyle\sf{\int\sin(x)\ dx=-\cos(x),}$

$\displaystyle\longrightarrow\sf{\int\limits_0^{2\pi}|\sin(x)|\ dx=-\big[\cos(x)\big]_0^{\pi}+\big[\cos(x)\big]_{\pi}^{2\pi}}$

$\displaystyle\longrightarrow\sf{\int\limits_0^{2\pi}|\sin(x)|\ dx=-\big[\cos(\pi)-\cos(0)\big]+\big[\cos(2\pi)-\cos(\pi)\big]}$

$\displaystyle\longrightarrow\sf{\int\limits_0^{2\pi}|\sin(x)|\ dx=-\big[-1-1\big]+\big[1-(-1)\big]}$

$\displaystyle\longrightarrow\underline{\underline{\sf{\int\limits_0^{2\pi}|\sin(x)|\ dx=4}}}$

Hence 4 is the answer.