Definite Integration Question 152
Question: The area enclosed between the curves $ y=x^{3} $ and $ y=\sqrt{x} $ is, (in square units)
[Karnataka CET 2004]
Options:
A) $ \frac{5}{3} $
B) $ \frac{5}{4} $
C) $ \frac{5}{12} $
D) $ \frac{12}{5} $
Show Answer
Answer:
Correct Answer: C
Solution:
Given curves are, $ y=x^{3} $ and $ y=\sqrt{x} $
On solving, we get $ x=0,x=1 $
Therefore, required area = $ \int_0^{1}{(x^{3}-\sqrt{x})}dx $
$ =[ \frac{x^{4}}{4}-\frac{2x\sqrt{x}}{3} ]_0^{1}=[ \frac{1}{4}-\frac{2}{3} ]=\frac{5}{12} $ ,(Area can-t be negative).
BETA
