Question

Anybody know the integration for this... The answer is k=3

Answer 1

You know that $y^2=4x$ so $x=\dfrac{y^2}{4}$.
Area of a shaded region is:

$A=\int\limits_0^k\dfrac{y^2}{4}\,dy=\frac{1}{4}\int\limits_0^ky^2\,dy=\frac{1}{4}\left[\frac{y^3}{3}\right]_0^k=\frac{1}{4}\left[\frac{k^3}{3}-\frac{0^3}{3}\right]=\frac{1}{4}\cdot\dfrac{k^3}{3}=\boxed{\frac{k^3}{12}}$

so k:

$\dfrac{k^3}{12}=\dfrac{9}{4}\\\\\\k^3=\dfrac{9\cdot12}{4}\\\\\\k^3=9\cdot3\\\\k^3=3^3\quad|\sqrt[3]{(\ldots)}\\\\\boxed{k=3}$