The area bounded by the curve y = x |sin(frac | vedclass

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(D) The area $A$ is given by $\int_{0}^{4\pi} x |\sin(\frac{x}{2})| dx$.Since $\sin(\frac{x}{2}) \ge 0$ for $x \in [0, 2\pi]$ and $\sin(\frac{x}{2}) \

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$16$

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(D) The area $A$ is given by $\int_{0}^{4\pi} x |\sin(\frac{x}{2})| dx$.Since $\sin(\frac{x}{2}) \ge 0$ for $x \in [0, 2\pi]$ and $\sin(\frac{x}{2}) \le 0$ for $x \in [2\pi, 4\pi]$, we have:$A = \int_{0}^{2\pi} x \sin(\frac{x}{2}) dx + \int_{2\pi}^{4\pi} x (-\sin(\frac{x}{2})) dx$.Using integration by parts $\int u dv = uv - \int v du$ with $u=x, dv=\sin(\frac{x}{2})dx$:$\int x \sin(\frac{x}{2}) dx = -2x \cos(\frac{x}{2}) + 4 \sin(\frac{x}{2})$.Evaluating the first integral: $[-2x \cos(\frac{x}{2}) + 4 \sin(\frac{x}{2})]_{0}^{2\pi} = (-2(2\pi)(-1) + 0) - (0) = 4\pi$.Evaluating the second integral: $-[-2x \cos(\frac{x}{2}) + 4 \sin(\frac{x}{2})]_{2\pi}^{4\pi} = -[(-2(4\pi)(1) + 0) - (-2(2\pi)(-1) + 0)] = -[-8\pi - 4\pi] = 12\pi$.Total area $A = 4\pi + 12\pi = 16\pi$ square units.

The area bounded by the curve $y = x |\sin(\frac{x}{2})|$ and the $X$-axis between the lines $x = 0$ and $x = 4\pi$ (in square units) is: (in $\pi$)

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