Question:medium

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}

Show Hint

If $x=\alpha$ divides the area from $0$ to $a$ into half, then $\alpha^3 = a^3/2$. Here $4^3/2 = 32$.
Updated On: Jun 19, 2026
  • $2\sqrt[3]{2}$
  • $2\sqrt[3]{4}$
  • $\sqrt[3]{32}$
  • $32$
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Question:
The problem involves dividing an area under a parabola into two equal parts using a vertical line $x = \alpha$.

Step 2: Key Formula or Approach:

Area under $y = f(x)$ from $a$ to $b$ is $\int_a^b f(x) dx$.

Step 3: Detailed Explanation:

The curve is $y = \frac{x^2}{4}$.
Total area $A$ from $x=0$ to $x=4$: \[ A = \int_0^4 \frac{x^2}{4} dx = [\frac{x^3}{12}]_0^4 = \frac{64}{12} = \frac{16}{3} \] The line $x = \alpha$ divides it into equal areas, so the area from $0$ to $\alpha$ is $A/2$: \[ \int_0^\alpha \frac{x^2}{4} dx = \frac{1}{2} \times \frac{16}{3} = \frac{8}{3} \] \[ [\frac{x^3}{12}]_0^\alpha = \frac{8}{3} \] \[ \frac{\alpha^3}{12} = \frac{8}{3} \] \[ \alpha^3 = \frac{8 \times 12}{3} = 32 \] \[ \alpha = (32)^{\frac{1}{3}} \]

Step 4: Final Answer:

The value of $\alpha$ is $(32)^{\frac{1}{3}}$.
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