Question

If the area of the region bounded by the curves $ y = 4 - \frac{x^2}{4} $ and $ y = \frac{x - 4}{2} $ is equal to $ \alpha $, then $ 6\alpha $ equals:

• A. \( 250 \)
• B. \( 210 \)
• C. \( 240 \)
• D. \( 220 \)

Hint:

To find the area between curves, set up an integral for the difference of the curves over the given interval. Make sure to simplify the integrand before solving the integral.

Answer 1

The Correct Option is A

Step 1: Find Points of Intersection

Set the two equations equal to find intersection points:
\[ 4 - \frac{x^2}{4} = \frac{x - 4}{2} \]
Multiply through by 4:
\[ 16 - x^2 = 2(x - 4) \]
\[ 16 - x^2 = 2x - 8 \]
Rearrange:
\[ x^2 + 2x - 24 = 0 \]
Solve the quadratic equation:
\[ x = \frac{-2 \pm \sqrt{4 + 96}}{2} = \frac{-2 \pm 10}{2} \]
\[ x = 4 \quad \text{or} \quad x = -6 \]

Step 2: Determine Upper and Lower Curves

Test at \(x = 0\):
\[ \begin{align} \text{Parabola:} & \quad y = 4 - 0 = 4 \\ \text{Line:} & \quad y = \frac{0 - 4}{2} = -2 \end{align} \]
The parabola is above the line in \([-6, 4]\).

Step 3: Set Up the Integral

The area \(\alpha\) is:
\[ \alpha = \int_{-6}^{4} \left[\left(4 - \frac{x^2}{4}\right) - \left(\frac{x - 4}{2}\right)\right] dx \]
Simplify the integrand:
\[ 6 - \frac{x^2}{4} - \frac{x}{2} \]

Step 4: Compute the Integral

Break into three parts:
\[ \begin{align} \int_{-6}^{4} 6 \, dx &= 6(4 - (-6)) = 60 \\ \int_{-6}^{4} \frac{x^2}{4} \, dx &= \frac{1}{12}\left[x^3\right]_{-6}^{4} = \frac{1}{12}(64 - (-216)) = \frac{280}{12} = \frac{70}{3} \\ \int_{-6}^{4} \frac{x}{2} \, dx &= \frac{1}{4}\left[x^2\right]_{-6}^{4} = \frac{1}{4}(16 - 36) = -5 \end{align} \]
Combine results:
\[ \alpha = 60 - \frac{70}{3} - (-5) = 65 - \frac{70}{3} = \frac{125}{3} \]

Step 5: Calculate \(6\alpha\)

\[ 6\alpha = 6 \times \frac{125}{3} = 250 \]

Step 6: Match with Options

The correct answer is option (1) 250.

Final Answer (1) 250