Question

The area enclosed between the curves \( y^2 = 4x \) and \( y = |x| \) is

• A. \( \frac{8}{3} \) sq. units
• B. \( \frac{5}{3} \) sq. units
• C. \( \frac{4}{3} \) sq. units
• D. \( \frac{2}{3} \) sq. units

Hint:

The area between \( y^2 = 4ax \) and \( y = mx \) is \( \frac{8a^2}{3m^3} \).

Answer 1

The Correct Option is A

Step 1: Concept Area between curves is calculated using integration: \(\int_{x_1}^{x_2} (y_{upper} - y_{lower}) dx\).

Step 2: Meaning The curve \( y^2 = 4x \) is a parabola opening right, and \( y = |x| \) consists of lines \( y = x \) and \( y = -x \).

Step 3: Analysis The curves intersect where \( x^2 = 4x \) (for \( x \ge 0 \)), which gives \( x = 0 \) and \( x = 4 \).
In the first quadrant, \( \sqrt{4x} \ge x \).
Area = \(\int_{0}^{4} (2\sqrt{x} - x) dx\).
Calculation: \([ \frac{4}{3}x^{3/2} - \frac{x^2}{2} ]_0^4 = \frac{4}{3}(8) - \frac{16}{2} = \frac{32}{3} - 8 = \frac{8}{3}\).


Step 4: Conclusion The area enclosed by the two curves is \( \frac{8}{3} \) square units. Final Answer: (A)