Question

The area inside the parabola \(y^2=4 \mathrm{a} x\), between the lines \(x=\mathrm{a}\) and \(x=4 \mathrm{a}\) is equal to

• A. \(4a^{2}\) sq. units
• B. \(8\mathrm{a}^{2}\) sq. units
• C. \(\frac{56\mathrm{a}^{2}}{3}\) sq. units
• D. \(\frac{35\mathrm{a}^{2}}{3}\) sq. units

Hint:

Integrate the upper branch of the parabola between the two limits and double it for symmetry.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The parabola \(y^2=4ax\) is symmetric about the x-axis, so the area is twice the area of the upper half.

Step 2: Set Up the Integral
\(A=2\int_a^{4a}2\sqrt{a}\,\sqrt{x}\,dx=4\sqrt a\left[\dfrac{2}{3}x^{3/2}\right]_a^{4a}\)

Step 3: Calculation
\(=\dfrac{8\sqrt a}{3}\left[(4a)^{3/2}-a^{3/2}\right]=\dfrac{8\sqrt a}{3}\left(8a^{3/2}-a^{3/2}\right)\)

Step 4: Simplify
\(=\dfrac{8\sqrt a}{3}\cdot7a^{3/2}=\dfrac{56a^2}{3}\text{ sq. units}\)

Final Answer:
The area is 56 a squared over 3 square units. \[ \boxed{\text{(C) }\dfrac{56a^{2}}{3}\text{ sq. units}} \]