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Q.
The area bounded by the curve $a^{2}y=x^{2}\left(x + a\right)$ and the $x$ -axis is
NTA AbhyasNTA Abhyas 2022
Solution:
The curve is $y=\frac{x^{2} \left(x + a\right)}{a^{2}}$ which is a cubic polynomial.
Since, $\frac{x^{2} \left(x + a\right)}{a^{2}}=0$ has a repeated root $x=0$ ,
it touches $x$ -axis at $\left(0 , 0\right)$ and intersects at $\left(- a , 0\right)$ .
Hence, required area $=\int\limits _{- a}^{0}y \, dx$
$=\int\limits_{- a}^{0}\left[\frac{x^{2} \left(x + a\right)}{a^{2}}\right]dx$
$=\frac{1}{a^{2}}\left\{\frac{x^{4}}{4} + \frac{a x^{3}}{3}\right\}_{- a}^{0}$
$=\frac{a^{2}}{12} \,$ sq.units
