Question

Find the area bounded by the curves $y=\sqrt{1-x^2}$ and $y=x^3-x$. Also find the ratio in which the y-axis divide this area

• A. $\frac{\pi }{2}$ , $\frac{\pi -1}{\pi +1}$
• B. $\frac{\pi }{4}$ , $\frac{\pi -1}{\pi +1}$
• C. $\frac{\pi }{2}$ , $\frac{\pi +1}{\pi -1}$
• D. None of these
  

Answer 1

The correct option is A $\frac{\pi }{2}$ , $\frac{\pi -1}{\pi +1}$

The two curves are $y=\sqrt{1-x^2}$ ....(1)

and $y=x^3-x$ ...(2)

The point of intersection are $P(-1,0);Q(1,0)$

Consider $y=\sqrt{1-x^2}$

On squaring both sides, we get

$x^2+y^2=1$

But $y=\sqrt{1-x^2}\ge 0$ by the definition of square root which is a semi-circle with center $(0,0)$ and radius $1$ and above X-axis.

Consider $y=x^3-x=x(x-1)(x+1)$

Now for $x\le -1,0\le x\le 1;y\le 0$

and for $-1\le x\le 0,x\ge 1;y\ge 0$

Taking into account the oddness of the function and the intervals of constant sign.

(We can construct its graph by finding the maxima and minima at $x=\pm \frac{1}{\sqrt{3}}$

Thus the required Area $=A_1+A_2$

where $A_1={\int }_{-1}^0[\sqrt{1-x^2}-x^3+x]dx=\frac{\pi }{4}-\frac{1}{4}$

and $A_2={\int }_0^1[\sqrt{1-x^2}-x^3+x]dx=\frac{\pi }{4}+\frac{1}{4}$

$\Rightarrow$ Required area $=\frac{\pi }{2}$ and required ratio $=\frac{A_1}{A_2}=\frac{\pi -1}{\pi +1}$