Question

The area bounded by \[ 2 - 4x \leq y \leq x^2 + 4 \quad \text{and} \quad x = \frac{1}{2}, \, x \geq 0, \, y \geq 0 \] (in square units) is:

• A. \( \frac{25}{37} \, \text{sq. unit} \)
• B. \( \frac{24}{37} \, \text{sq. unit} \)
• C. \( \frac{37}{25} \, \text{sq. unit} \)
• D. \( \frac{37}{24} \, \text{sq. unit} \)

Hint:

When finding the area between curves, set up the integral by subtracting the lower function from the upper function, and use the limits of integration as the points where the curves intersect or where the region is defined.

Answer 1

The Correct Option is D

Step 1: Understand the problem.
The problem asks us to find the area bounded by the given curves: \[ 2 - 4x \leq y \leq x^2 + 4 \] and the vertical line \( x = \frac{1}{2} \), with the region restricted to \( x \geq 0 \) and \( y \geq 0 \). The area can be calculated by integrating the difference between the upper curve (\( y = x^2 + 4 \)) and the lower curve (\( y = 2 - 4x \)).
Step 2: Set up the integral.
To find the area, we first calculate the intersection points of the curves \( y = x^2 + 4 \) and \( y = 2 - 4x \). Equating the two expressions: \[ x^2 + 4 = 2 - 4x \] \[ x^2 + 4x + 2 = 0 \] Solving this quadratic equation: \[ x = \frac{-4 \pm \sqrt{4^2 - 4(1)(2)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 8}}{2} = \frac{-4 \pm \sqrt{8}}{2} = \frac{-4 \pm 2\sqrt{2}}{2} = -2 \pm \sqrt{2} \] Since \( x \geq 0 \), we take the positive root: \[ x = -2 + \sqrt{2} \approx 0.414 \] Thus, the point of intersection is \( x \approx 0.414 \).
Step 3: Integrate to find the area.
The total area between the curves is given by the integral from \( x = 0 \) to \( x = \frac{1}{2} \): \[ A = \int_0^{1/2} \left( (x^2 + 4) - (2 - 4x) \right) dx \] Simplify the integrand: \[ A = \int_0^{1/2} \left( x^2 + 4 - 2 + 4x \right) dx = \int_0^{1/2} \left( x^2 + 4x + 2 \right) dx \]
Step 4: Compute the integral.
Now, integrate: \[ A = \left[ \frac{x^3}{3} + 2x^2 + 2x \right]_0^{1/2} \] Substitute the limits: \[ A = \left( \frac{\left(\frac{1}{2}\right)^3}{3} + 2\left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) \right) - 0 \] \[ A = \left( \frac{1}{24} + \frac{2}{4} + 1 \right) = \frac{1}{24} + \frac{1}{2} + 1 \] \[ A = \frac{1 + 12 + 24}{24} = \frac{37}{24} \] Thus, the area is \( \frac{37}{24} \) square units.