Question

The area bounded by the curves \( y = 4 - \frac{x^2}{4} \) and \( y = \frac{x - 4}{2} \) (in square units) is:

• A. \( \frac{20}{3} \)
• B. \( \frac{120}{3} \)
• C. \( \frac{80}{3} \)
• D. \( \frac{125}{3} \)

Hint:

To find the area bounded by curves, first find the points of intersection and then integrate the difference between the functions over the given range.

Answer 1

The Correct Option is A

To find the area bounded by the curves, we first need to determine the points of intersection of the curves \( y = 4 - \frac{x^2}{4} \) and \( y = \frac{x - 4}{2} \). Set the equations equal to each other: \[ 4 - \frac{x^2}{4} = \frac{x - 4}{2} \] Multiply through by 4 to eliminate the fractions: \[ 16 - x^2 = 2(x - 4) \] \[ 16 - x^2 = 2x - 8 \] \[ x^2 + 2x - 24 = 0 \] Solve for \( x \) using the quadratic formula: \[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-24)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 96}}{2} = \frac{-2 \pm \sqrt{100}}{2} = \frac{-2 \pm 10}{2} \] Thus, the points of intersection are \( x = 4 \) and \( x = -6 \). Now, we calculate the area between the curves by integrating the difference of the functions from \( x = -6 \) to \( x = 4 \): \[ A = \int_{-6}^{4} \left[ \left( 4 - \frac{x^2}{4} \right) - \frac{x - 4}{2} \right] \, dx \] Simplify the integrand: \[ A = \int_{-6}^{4} \left[ 4 - \frac{x^2}{4} - \frac{x}{2} + 2 \right] \, dx \] \[ A = \int_{-6}^{4} \left[ 6 - \frac{x^2}{4} - \frac{x}{2} \right] \, dx \] Now, integrate: \[ A = \left[ 6x - \frac{x^3}{12} - \frac{x^2}{4} \right]_{-6}^{4} \] Substitute the limits: \[ A = \left( 6(4) - \frac{(4)^3}{12} - \frac{(4)^2}{4} \right) - \left( 6(-6) - \frac{(-6)^3}{12} - \frac{(-6)^2}{4} \right) \] \[ A = \left( 24 - \frac{64}{12} - \frac{16}{4} \right) - \left( -36 - \frac{-216}{12} - \frac{36}{4} \right) \] \[ A = \left( 24 - \frac{16}{3} - 4 \right) - \left( -36 + 18 - 9 \right) \] \[ A = \left( 20 - \frac{16}{3} \right) - \left( -27 \right) \] \[ A = 20 - \frac{16}{3} + 27 = 47 - \frac{16}{3} \] \[ A = \frac{141}{3} - \frac{16}{3} = \frac{125}{3} \] Therefore, the area bounded by the curves is \( \frac{125}{3} \). Thus, the correct answer is (D) \( \frac{125}{3} \).