Definite Integration Question 463
Question: $ \int_0^{1.5}{[x^{2}]dx} $ , where $ [.] $ denotes the greatest integer function, equals
[IIT 1988; DCE 2000, 01]
Options:
A) $ 2+\sqrt{2} $
B) $ 2-\sqrt{2} $
C) $ -2+\sqrt{2} $
D) $ -2-\sqrt{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_0^{1.5}{[x^{2}]dx=\int_0^{1}{[x^{2}]dx+\int_1^{\sqrt{2}}{[x^{2}]dx+\int _{\sqrt{2}}^{1.5}{[x^{2}]dx}}}} $
$ =0+\int_1^{\sqrt{2}}{1dx+\int _{\sqrt{2}}^{1.5}{2dx=\sqrt{2}-1+3-2\sqrt{2}=2-\sqrt{2}}} $ .
BETA
