Question

5. Using integration determine the area of the region bounded by y = sinx
between x=0 and x=2π​

Answer 1

Let's check condition for satisfying $\displaystyle\sf {\sin x\leq0.}$

This implies $\displaystyle\sf {x}$ belongs to 3rd or 4th quadrant.

$\displaystyle\sf{\Longrightarrow x\in [(2n-1)\pi,\ 2n\pi]}$

where $\displaystyle\sf {n\in\mathbb{Z}.}$

For $\displaystyle\sf {x\in[0,\ 2\pi],}$

$\displaystyle\sf{\longrightarrow\sin x\leq0\quad\Longrightarrow\quad x\in [\pi,\ 2\pi]}$

Therefore, area bounded by $\displaystyle\sf {y=\sin x}$ with x axis,

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

$\displaystyle\sf{\longrightarrow A=-\big[\cos x\big]_0^{\pi}+\big[\cos x\big]_{\pi}^{2\pi}}$

$\displaystyle\sf{\longrightarrow A=-(-1-1)+(1+1)}$

$\displaystyle\sf {\longrightarrow\underline {\underline {A=4}}}$

Hence 4 is the answer.