For 0 le x le pi , the area bounded by y = x | vedclass

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(A) The curves $y = x$ and $y = x + \sin x$ intersect at $x = 0$ and $x = \pi$.In the interval $[0, \pi]$,$\sin x \ge 0$,so $x + \sin x \ge x$.The are

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$2$

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(A) The curves $y = x$ and $y = x + \sin x$ intersect at $x = 0$ and $x = \pi$.In the interval $[0, \pi]$,$\sin x \ge 0$,so $x + \sin x \ge x$.The area $A$ bounded by the two curves is given by the integral:$A = \int_{0}^{\pi} ((x + \sin x) - x) \, dx$$A = \int_{0}^{\pi} \sin x \, dx$$A = [-\cos x]_{0}^{\pi}$$A = -(\cos \pi - \cos 0)$$A = -(-1 - 1) = -(-2) = 2$.Thus,the area is $2$ square units.

For $0 \le x \le \pi ,$ the area bounded by $y = x$ and $y = x + \sin x$ is

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