Question

The area enclosed between the curve $ y^2 = 4x, \quad \text{and the line } y = x \text{ is} $

• A. \( \frac{2}{3} \)
• B. \( \frac{4}{3} \)
• C. \( \frac{1}{2} \)
• D. \( \frac{8}{3} \)

Hint:

When finding the area enclosed by curves, first determine the points of intersection, then set up and evaluate the integral of the difference between the functions.

Answer 1

The Correct Option is B

Step 1: Find the Points of Intersection
To find the points of intersection between the curve \( y^2 = 4x \) and the line \( y = x \), substitute \( y = x \) into the equation \( y^2 = 4x \): \[ x^2 = 4x \quad \Rightarrow \quad x(x - 4) = 0 \] Thus, \( x = 0 \) and \( x = 4 \).
Therefore, the points of intersection are \( (0, 0) \) and \( (4, 4) \).
Step 2: Set up the Integral
To find the area enclosed, we integrate the difference between the two curves from \( x = 0 \) to \( x = 4 \): \[ \text{Area} = \int_0^4 \left( x - \sqrt{4x} \right) dx \]
Step 3: Evaluate the Integral
After performing the integration, we find that the area is \( \frac{4}{3} \).
Step 4: Conclusion
Thus, the area enclosed is \( \frac{4}{3} \).