Question

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

• A. 4 sq units
• B. 8 sq units
• C. 6 sq units
• D. 1 sq units

Hint:

For area involving sine or cosine over a full period, consider symmetry and take absolute values where needed.

Answer 1

The Correct Option is B

Step 1: Write the required area integral.
Area between curve and x-axis is:
\[ A = \int_{-4\pi}^{0} \left|\sin\left(\frac{x}{2}\right)\right| dx. \]

Step 2: Understand the behavior of the function.
The function \( \sin\left(\frac{x}{2}\right) \) completes one full cycle when \( x \) changes by \( 4\pi \).
So, interval \( [-4\pi,0] \) represents one full period.

Step 3: Split the interval.
\[ \sin\left(\frac{x}{2}\right) \leq 0 \text{ on } [-4\pi,-2\pi] \] \[ \sin\left(\frac{x}{2}\right) \geq 0 \text{ on } [-2\pi,0]. \]

Step 4: Write area as sum of two integrals.
\[ A = -\int_{-4\pi}^{-2\pi} \sin\left(\frac{x}{2}\right) dx + \int_{-2\pi}^{0} \sin\left(\frac{x}{2}\right) dx. \]

Step 5: Substitute \( u = \frac{x}{2} \).
\[ dx = 2du. \]
So,
\[ \int \sin\left(\frac{x}{2}\right)dx = -2\cos\left(\frac{x}{2}\right). \]

Step 6: Evaluate the integrals.
First part:
\[ -\left[-2\cos\left(\frac{x}{2}\right)\right]_{-4\pi}^{-2\pi} = 2[\cos(-2\pi)-\cos(-\pi)]. \]
\[ = 2(1 - (-1)) = 4. \]
Second part:
\[ [-2\cos\left(\frac{x}{2}\right)]_{-2\pi}^{0} = -2[1 - (-1)] = -4. \]
Taking absolute value contribution:
\[ A = 4 + 4 = 8. \]

Step 7: Final conclusion.
Thus, the total bounded area is 8 sq units.
Final Answer:
\[ \boxed{8}. \]