The area of the region bounded by the curve y | vedclass

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(C) The area $A$ is given by the integral $\int_{-1}^{2} |\sin(\pi x)| \, dx$. Since $\sin(\pi x)$ changes sign at $x = 0$ and $x = 1$,we split the in

Vedclass · Accepted Answer

$\frac{6}{\pi}$

Vedclass · Answer

(C) The area $A$ is given by the integral $\int_{-1}^{2} |\sin(\pi x)| \, dx$. Since $\sin(\pi x)$ changes sign at $x = 0$ and $x = 1$,we split the integral: $A = \int_{-1}^{0} |\sin(\pi x)| \, dx + \int_{0}^{1} |\sin(\pi x)| \, dx + \int_{1}^{2} |\sin(\pi x)| \, dx$. For $x \in [-1, 0]$,$\sin(\pi x) \leq 0$,so $|\sin(\pi x)| = -\sin(\pi x)$. For $x \in [0, 1]$,$\sin(\pi x) \geq 0$,so $|\sin(\pi x)| = \sin(\pi x)$. For $x \in [1, 2]$,$\sin(\pi x) \leq 0$,so $|\sin(\pi x)| = -\sin(\pi x)$. $A = \int_{-1}^{0} -\sin(\pi x) \, dx + \int_{0}^{1} \sin(\pi x) \, dx + \int_{1}^{2} -\sin(\pi x) \, dx$. Evaluating these: $\int -\sin(\pi x) \, dx = \frac{\cos(\pi x)}{\pi}$. $A = [\frac{\cos(\pi x)}{\pi}]_{-1}^{0} + [-\frac{\cos(\pi x)}{\pi}]_{0}^{1} + [\frac{\cos(\pi x)}{\pi}]_{1}^{2}$. $A = (\frac{1}{\pi} - \frac{-1}{\pi}) + (-(\frac{-1}{\pi} - \frac{1}{\pi})) + (\frac{1}{\pi} - \frac{-1}{\pi}) = \frac{2}{\pi} + \frac{2}{\pi} + \frac{2}{\pi} = \frac{6}{\pi}$. Thus,the correct option is $C$.

The area of the region bounded by the curve $y = \sin(\pi x)$ and the $X$-axis for $x \in [-1, 2]$ is . . . . . . sq. units.

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