Question

The area enclosed between the parabola \( y^2 = 4x \) and the line \( y = 2x - 4 \) is:

• A. \( \frac{17}{3} \) sq. units
• B. 15 sq. units
• C. \( \frac{19}{3} \) sq. units
• D. 9 sq. units

Hint:

To find the area between curves, first identify the points of intersection, then set up and evaluate the integral of the difference between the functions.

Answer 1

The Correct Option is A

We are asked to find the area enclosed between the parabola \( y^2 = 4x \) and the line \( y = 2x - 4 \). ### Step 1: Find the points of intersection To find the points of intersection, substitute the equation of the line into the equation of the parabola. \[ y = 2x - 4 \] Substitute \( y \) into \( y^2 = 4x \): \[ (2x - 4)^2 = 4x \] Expanding the equation: \[ 4x^2 - 16x + 16 = 4x \] Simplifying the equation: \[ 4x^2 - 20x + 16 = 0 \] Dividing the entire equation by 4: \[ x^2 - 5x + 4 = 0 \] Factoring the quadratic: \[ (x - 4)(x - 1) = 0 \] Thus, \( x = 4 \) and \( x = 1 \). These are the points of intersection. ### Step 2: Set up the integral for the area The area between the curves can be calculated by integrating the difference between the functions for \( x \) from 1 to 4. The area between the curves is given by: \[ A = \int_{1}^{4} \left( (2x - 4) - \sqrt{4x} \right) dx \] ### Step 3: Calculate the integral First, we calculate the integral of \( 2x - 4 \): \[ \int (2x - 4) dx = x^2 - 4x \] Next, calculate the integral of \( \sqrt{4x} = 2\sqrt{x} \): \[ \int 2\sqrt{x} dx = \frac{4}{3} x^{3/2} \] Now, substitute the limits of integration: \[ A = \left[ (x^2 - 4x) - \frac{4}{3} x^{3/2} \right]_{1}^{4} \] Substituting \( x = 4 \): \[ A = \left( (16 - 16) - \frac{4}{3} (8) \right) = 0 - \frac{32}{3} \] Substituting \( x = 1 \): \[ A = \left( (1 - 4) - \frac{4}{3} (1) \right) = -3 - \frac{4}{3} = -\frac{13}{3} \] Thus, the total area is: \[ A = \frac{17}{3} \, \text{sq. units} \]