Question

The area of the figure bounded by two branches of the curve ${(y-x)}^2=x^3$ and the straight line $x=1$ is:

• A. $\frac{1}{3}$ sq.unit
• B. $\frac{4}{5}$ sq.unit
• C. $\frac{5}{4}$ sq.unit
• D. $3$ sq.unit

Answer 1

The correct option is B $\frac{4}{5}$ sq.unit

Given curve is ${(y-x)}^2=x^3$
$\Rightarrow y-x=\pm x\sqrt{x}$
or $y=x\pm x\sqrt{x}$
$\therefore$ Required Area$=\left|{\int }_0^1\left[\right(x+x\sqrt{x})-(x-x\sqrt{x})]dx\right|$
$=\left|2{\int }_0^1{x}^{\frac{3}{2}}dx\right|$
$=2{\left[\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right]}_0^1$
$=\left|2.\frac{2}{5}\right|=\frac{4}{5}$.sq.unit.