Question

The ratio in which the area bounded by the curve $y^2=xand{x}^2=y$ is divided by the line $x=\frac{1}{2}$ is

• A. $\frac{4\sqrt{2}-1}{9-4\sqrt{2}}$
• B. $\frac{3\sqrt{2}+3}{9-4\sqrt{2}}$
• C. $\frac{\sqrt{2}-1}{\sqrt{3-1}}$
• D. $\frac{2\sqrt{2}-1}{3\sqrt{3-1}}$

Answer 1

The correct option is A $\frac{4\sqrt{2}-1}{9-4\sqrt{2}}$

Area under a curve $f\left(x\right)$ is given by $\int f\left(x\right)dx$ if it is bounded by $x=a,x=b$ lines then area will be ${\int }_a^{b}f\left(x\right)dx$

area bounded by curve $y^2=x$ from $x=a,x=b$ is :

${\int }_a^b\sqrt{x}dx$

area bounded by curve $y=x^2$ from $x=a,x=b$ is :

${\int }_a^b{x}^{2}dx$

area bounded by them becomes their difference from the limits given.Here limits will be from $x=0,x=\frac{1}{2}$ :

${\int }_0^{\frac{1}{2}}(\sqrt{x}-x^2)dx=\frac{4\sqrt{2}-1}{24}$

and $x=\frac{1}{2},x=1$ :

${\int }_{\frac{1}{2}}^1(\sqrt{x}-x^2)dx=\frac{9-4\sqrt{2}}{24}$

ratio means just divide them both which gives you;

$\frac{4\sqrt{2}-1}{9-4\sqrt{2}}$