Area of the region bounded by the curve $y=\sin \left(\dfrac{x}{2}\right)$ between $-4 \pi$ and 0 is

The area $A$ of the region bounded by the curve $y = \sin \left( \dfrac{x}{2} \right)$ between $x = -4\pi$ and $x = 0$ is given by the integral:

$A = \left| \int_{-4\pi}^{0} \sin \left( \dfrac{x}{2} \right) dx \right|$

Evaluating the integral:

$\int \sin \left( \dfrac{x}{2} \right) dx = -2 \cos \left( \dfrac{x}{2} \right)$

Applying the limits from $-4\pi$ to $0$:

$\left[ -2 \cos \left( \dfrac{x}{2} \right) \right]_{-4\pi}^{0} = -2 \cos(0) - (-2 \cos(-2\pi))$

Since $\cos(0) = 1$ and $\cos(-2\pi) = 1$:

$-2(1) - (-2(1)) = -2 + 2 = 0$

The integral evaluates to $0$ because the function $\sin \left( \dfrac{x}{2} \right)$ is negative on the interval $(-2\pi, 0)$ and positive on $(-4\pi, -2\pi)$. The area is the sum of the absolute values of the integrals over these sub-intervals:

$A = \left| \int_{-4\pi}^{-2\pi} \sin \left( \dfrac{x}{2} \right) dx \right| + \left| \int_{-2\pi}^{0} \sin \left( \dfrac{x}{2} \right) dx \right|$

$\int_{-4\pi}^{-2\pi} \sin \left( \dfrac{x}{2} \right) dx = \left[ -2 \cos \left( \dfrac{x}{2} \right) \right]_{-4\pi}^{-2\pi} = -2 \cos(-\pi) - (-2 \cos(-2\pi)) = -2(-1) - (-2(1)) = 2 + 2 = 4$

$\int_{-2\pi}^{0} \sin \left( \dfrac{x}{2} \right) dx = \left[ -2 \cos \left( \dfrac{x}{2} \right) \right]_{-2\pi}^{0} = -2 \cos(0) - (-2 \cos(-\pi)) = -2(1) - (-2(-1)) = -2 - 2 = -4$

The area is $|4| + |-4| = 4 + 4 = 8$ sq units.

Answer: 8 sq units