Question

The area bounded by the curve $a^{2}y=x^2(x+a)$ 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 $y=\frac{x^2(x+a)}{a^2}$ which is a cubic polynomial.
Since $\frac{x^2(x+a)}{a^2}=0$ has a repeated root $x=0$ ,
it touches $x$ -axis at $(0,0)$ and intersects at $(-a,0)$ .

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