Question

If the area bounded by $x^{2}=4y$, X-axis and $x=4$ is divided into equal areas by $x=\alpha$, then the value of $\alpha$ is}

• A. $2\sqrt[3]{2}$
• B. $2\sqrt[3]{4}$
• C. $\sqrt[3]{32}$
• D. $32$

Hint:

If $x=\alpha$ divides the area from $0$ to $a$ into half, then $\alpha^3 = a^3/2$. Here $4^3/2 = 32$.

Answer 1

The Correct Option is C

Step 1: Concept
Area under curve $y = x^2/4$ from $0$ to $k$ is $\int_0^k \frac{x^2}{4} dx$.

Step 2: Analysis
Total Area $A = \int_0^4 \frac{x^2}{4} dx = [\frac{x^3}{12}]_0^4 = \frac{64}{12} = \frac{16}{3}$.
Half Area $= 8/3$.

Step 3: Calculation
$\int_0^\alpha \frac{x^2}{4} dx = \frac{8}{3} \implies [\frac{x^3}{12}]_0^\alpha = \frac{8}{3}$
$\frac{\alpha^3}{12} = \frac{8}{3} \implies \alpha^3 = \frac{8 \times 12}{3} = 32$
$\alpha = \sqrt[3]{32} = (32)^{1/3} = 2\sqrt[3]{4}$.

Step 4: Conclusion
Hence, $\alpha = \sqrt[3]{32}$. Final Answer: (C)