Definite Integration Question 328
Question: Area bounded by curves $ y=x^{2} $ and $ y=2-x^{2} $ is
[Orissa JEE 2005]
Options:
A) 8/3
B) 3/8
C) 3/2
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ y=x^{2} $ ……(i) $ y=2-x^{2} $ ……(ii)
$ \therefore $ By equation (i) and (ii) , we get,
$ x=\pm 1 $
$ \therefore $ $ y=\pm 1 $
$ \therefore $ Required area $ =2[ \int_0^{1}{(2-x^{2})dx-\int_0^{1}{x^{2}dx}} ] $
$ =2[ 2x-\frac{2x^{3}}{3} ]_0^{1}=4[ x-\frac{x^{3}}{3} ]_0^{1}=4( \frac{2}{3} )=\frac{8}{3} $ .
BETA
