Definite Integration Question 67
Question: The value of $ \int _{0}^{\sqrt{2}}{[x^{2}]dx}, $ where [.] is the greatest integer function
[AIEEE 2002]
Options:
A) $ 2-\sqrt{2} $
B) $ 2+\sqrt{2} $
C) $ \sqrt{2}-1 $
D) $ \sqrt{2}-2 $
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Answer:
Correct Answer: C
Solution:
$ I=\int_0^{\sqrt{2}}{[x^{2}]dx} $
$ =\int _{0}^{1}{[x^{2}]dx+}\int _{1}^{\sqrt{2}}{[x^{2}]dx} $
$ =\int _{0}^{1}{0dx+}\int _{1}^{\sqrt{2}}{dx} $
$ =[x]_1^{\sqrt{2}}=\sqrt{2}-1 $ .
BETA
