Explanation
The graph of $$y=\sin x$$ is shown in image.
$$A=\int_{0} ^{2\pi}\bigg(\sin x\bigg) $$
$$A=\int_{0} ^{\pi}\bigg(\sin x\bigg)+\int_{\pi} ^{2\pi}\bigg(\sin x\bigg) $$
Since, $$\Bigg|\int_{0} ^{\pi}\bigg(\sin x\bigg)\Bigg| =\Bigg|\int_{0} ^{2\pi}\bigg(\sin x\bigg)\Bigg| $$
Hence, $$A=2\Bigg(\int_{0} ^{\pi}\bigg(\sin x\bigg) \Bigg) $$
$$A=-2\bigg[\cos x\bigg]\Bigg|^0_{\pi}$$
$$A=2\times 2$$
$$A=4$$ sq. unit
$$\textbf{Step-1: Finding the bounded Area}$$
$$\text{On plotting these curves we can easily see area bounded by them}$$
$$\textbf{Step-2: And Area bounded by them is given as}$$
$$\Rightarrow 2\int\limits_{0}^{1}{[(2-{{x}^{2}})-{{x}^{2}}]}dx$$
$$\Rightarrow 2\int\limits_{0}^{1}{[2-2{{x}^{2}}]}dx$$
$$\Rightarrow 2\times \left| 2x-\dfrac{2{{x}^{3}}}{3} \right|_{0}^{1}$$
$$\Rightarrow 2\left[ \left( 2-\dfrac{2}{3} \right)-0 \right]$$
$$\Rightarrow \dfrac{8}{3}$$
$$\textbf{Therefore, area bounded by these curves is (A) }$$ $$\mathbf {\dfrac{8}{3}}$$
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