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Q.
If the area bounded by the parabola $x^2 = 4y$, the x-axis and the line $x = 4$ is divided into two equal area by the line $x = \alpha$, then the value of a is:
Application of Integrals
Solution:
Given : $x^{2} = 4y \Rightarrow y = \frac{1}{4}x^{2} $
$ \text{Area} = \int^{4}_{0} ydx $
$\Rightarrow \text{Area} = \frac{1}{4} \int^{4}_{0} x^{2}dx = \frac{1}{4} \left[\frac{x^{3}}{3}\right]^{4}_{0} = $
$\frac{1}{12} \left[4.4.4\right] = \frac{16}{3} $
But the line $x = \alpha$ divide above area into two equal parts.
$\therefore \frac{1}{4} \int^{\alpha}_{0} x^{2}dx = \frac{16}{3} \times\frac{1}{2} = \frac{8}{3}$
$ \Rightarrow \frac{1}{4} \times\frac{1}{3}\left[x^{3}\right]^{\alpha}_{0} = \frac{8}{3} $
$\Rightarrow \alpha^{3} = 32 \Rightarrow \alpha^{3} = 2^{5}\Rightarrow \alpha = 2^{\frac{5}{3}} $