MCQ
$\int_0^{1.5} {[{x^2}]\,dx} $, where $[\,\,.\,\,]$ denotes the greatest integer function, equals
- A$2 + \sqrt 2 $
- ✓$2 - \sqrt 2 $
- C$ - 2 + \sqrt 2 $
- D$ - 2 - \sqrt 2 $
✓
Answer
Correct option: B.
$2 - \sqrt 2 $
b
(b) $\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} } } } $
(b) $\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 } } $.
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