Definite Integration Question 288
Question: The area enclosed between the parabolas $ y^{2}=4x $ and $ x^{2}=4y $ is
[Karnataka CET 1999, 2003]
Options:
A) $ \frac{14}{3} $ sq. unit
B) $ \frac{3}{4} $ sq. unit
C) $ \frac{3}{16} $ sq. unit
D) $ \frac{16}{3} $ sq. unit
Show Answer
Answer:
Correct Answer: D
Solution:
Equations of curves $ y^{2}=4x $ and $ x^{2}=4y. $ The given equations may be written as $ y=2\sqrt{x} $ and $ y=\frac{x^{2}}{4}. $
We know that area enclosed by the parabolas $ =\int _{0}^{4}{2\sqrt{x}dx-}\int _{0}^{4}{\frac{x^{2}}{4}dx=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}} $ sq. unit.
BETA
