Definite Integration Question 420
Question: The area bounded by the two branches of curve $ {{(y-x)}^{2}}=x^{3} $ and the straight line x = 1 is
Options:
A) 1/5 sq. units
B) 3/5 sq. units
C) 4/5 sq. units
D) 8/4 sq. units
Show Answer
Answer:
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}{[ (x+{x^{3/2}})-(x-{x^{3/2}}) ]}dx $
$ =2\int\limits_0^{1}{{x^{3/2}}dx} $
$ =2{{[ \frac{{x^{5/2}}}{5/2} ]}_0}^{1}=\frac{4}{5} $ sq. units.
BETA
