Definite Integration Question 432
Question: Consider two curves $ C_1:y^{2}=4[\sqrt{y}]x $ and $ C_2:x^{2}=4[\sqrt{x}]y $ , where [.] denotes the greatest integer function. Then the area of region enclosed by these two curves within the square formed by the lines x =1, y =1, x = 4, y = 4 is
Options:
A) 8/3 sq. units
B) 10/3 sq. units
C) 11/3 sq. units
D) 11/4 sq. units
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ y^{2}=4[\sqrt{y}]x $
For $ y\in [1,4][\sqrt{y}]=1 $ or $ y^{2}=4x. $
Similarly, for $ x\in [1,4),[ \sqrt{x} ]=1 $ and $ x^{2}=4\lfloor \sqrt{x} \rfloor y $ would Transform into $ x^{2}=4y. $ The required area is .
$ A=\int\limits_1^{2}{(2\sqrt{x}-1)dx+\int\limits_2^{4}{( 2\sqrt{x}-\frac{x^{2}}{4} )dx}} $
$ ={{( \frac{4}{3}{x^{3/2}}-x )}_1}^{2}+{{( \frac{4}{3}{x^{3/2}}-\frac{x^{3}}{12} )}_2}^{4}=\frac{11}{3} $ Sq. units.
BETA
