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

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(D) The area $A$ is given by the integral of the absolute value of the function over the interval $[1, 3]$.$A = \int_{1}^{3} |\sin(\pi x)| \, dx$.Sinc

Vedclass · Accepted Answer

$\frac{4}{\pi}$

Vedclass · Answer

(D) The area $A$ is given by the integral of the absolute value of the function over the interval $[1, 3]$.$A = \int_{1}^{3} |\sin(\pi x)| \, dx$.Since $\sin(\pi x)$ changes sign at $x = 2$,we split the integral:$A = \int_{1}^{2} |\sin(\pi x)| \, dx + \int_{2}^{3} |\sin(\pi x)| \, dx$.In the interval $[1, 2]$,$\sin(\pi x) \leq 0$,so $|\sin(\pi x)| = -\sin(\pi x)$.In the interval $[2, 3]$,$\sin(\pi x) \geq 0$,so $|\sin(\pi x)| = \sin(\pi x)$.$A = \int_{1}^{2} -\sin(\pi x) \, dx + \int_{2}^{3} \sin(\pi x) \, dx$.Evaluating the integrals:$A = \left[ \frac{\cos(\pi x)}{\pi} \right]_{1}^{2} + \left[ -\frac{\cos(\pi x)}{\pi} \right]_{2}^{3}$.$A = \frac{1}{\pi} (\cos(2\pi) - \cos(\pi)) - \frac{1}{\pi} (\cos(3\pi) - \cos(2\pi))$.$A = \frac{1}{\pi} (1 - (-1)) - \frac{1}{\pi} (-1 - 1) = \frac{2}{\pi} + \frac{2}{\pi} = \frac{4}{\pi}$.Thus,the correct option is $D$.

The area of the region bounded by the curve $f(x) = \sin(\pi x)$ and the $X$-axis for $x \in [1, 3]$ is . . . . . . sq. units.

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