Question

The area bounded by the curve $a^{2}y=x^2\left(x+a\right)$ and the $x$ -axis is

• A. $\frac{a^2}{3}$ sq.units
• B. $\frac{a^2}{4}$ sq.units
• C. $\frac{3a^2}{4}$ sq.units
• D. $\frac{a^2}{12}$ sq.units

Answer 1

Answer: (D)

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 $=\underset{-a}{\overset{0}{\int }}ydx$
$=\underset{-a}{\overset{0}{\int }}\left[\frac{x^2\left(x+a\right)}{a^2}\right]dx$
$=\frac{1}{a^2}{\left\{\frac{x^4}{4}+\frac{a{x}^3}{3}\right\}}_{-a}^0$
$=\frac{a^2}{12}$ sq.units