question_answer

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

A) 1/5 sq. units      

B) 3/5 sq. units

C) 4/5 sq. units      

D) 8/4 sq. units

Correct Answer: C

Solution :

[c] \[{{(y-x)}^{2}}={{x}^{3}}\], where \[x\ge 0\] \[\Rightarrow y-x=\pm {{x}^{3/2}}\]                  ...(1) Or \[y=x+{{x}^{3/2}}\]                       ...(2) \[y=x-{{x}^{3/2}}\] Function (1) is an increasing function. Function (2) meets x-axis, when \[x-{{x}^{3/2}}=0\]or \[x=0,1\]. Also, for \[0<x<1,\]\[x-{{x}^{3/2}}>0\] and for \[x>1,x-{{x}^{3/2}}<0.\] When \[x\to \infty ,x-{{x}^{3/2}}\to -\infty .\] From these information, we can plot the graph as shown. Required area\[=\int\limits_{0}^{1}{\left[ (x+{{x}^{3/2}})-(x-{{x}^{3/2}}) \right]}dx\] \[=2\int\limits_{0}^{1}{{{x}^{3/2}}dx}\]\[=2{{\left[ \frac{{{x}^{5/2}}}{5/2} \right]}_{0}}^{1}=\frac{4}{5}\]sq. units.