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Q.
The area bounded by the curve $a^{2}y=x^{2}\left(\right.x+a\left.\right)$ and the $x$ -axis is
NTA AbhyasNTA Abhyas 2020Application of Integrals
Solution:
The curve is $y=\frac{x^{2} \left(\right. x + a \left.\right)}{a^{2}}$ which is a cubic polynomial.
Since $\frac{x^{2} \left(\right. x + a \left.\right)}{a^{2}}=0$ has a repeated root $x=0$ ,
it touches $x$ -axis at $\left(\right.0,0\left.\right)$ and intersects at $\left(\right.-a, \, 0\left.\right)$ .
Hence, required area $=\displaystyle \int _{- a}^{0} y \, d x=\displaystyle \int _{- a}^{0} \left[\frac{x^{2} \left(x + a\right)}{a^{2}}\right]dx=\frac{1}{a^{2}}\left(\left\{\frac{x^{4}}{4} + \frac{a x^{3}}{3}\right\}\right)_{- a}^{0}$
$=\frac{a^{2}}{12} \, sq. \, units$
