so 
.![$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}}](/tpl/images/0264/3877/9b1e4.png)
![\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}](/tpl/images/0264/3877/f1458.png)
Anybody know the integration for the answer is k=3
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Answers: 1
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