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(C) To find the area bounded by the curves $y = x + \sin x$ and $y = x$,we first find their points of intersection by setting $x + \sin x = x$.This si

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$4$

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(C) To find the area bounded by the curves $y = x + \sin x$ and $y = x$,we first find their points of intersection by setting $x + \sin x = x$.This simplifies to $\sin x = 0$,which gives $x = 0, \pi, 2\pi$ in the interval $[0, 2\pi]$.The area $A$ is given by the integral of the absolute difference between the curves:$A = \int_{0}^{2\pi} |(x + \sin x) - x| \, dx = \int_{0}^{2\pi} |\sin x| \, dx$.We split the integral at the point where $\sin x$ changes sign:$A = \int_{0}^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} (-\sin x) \, dx$.Evaluating the integrals:$\int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = -(-1 - 1) = 2$.$\int_{\pi}^{2\pi} -\sin x \, dx = [\cos x]_{\pi}^{2\pi} = (1 - (-1)) = 2$.Total Area $A = 2 + 2 = 4$.

For $x \in [0, 2\pi]$,the area of the region bounded by the curves $y = x + \sin x$ and $y = x$ is:

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