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Question: \(\int_{0}^{1.5}{\lbrack x^{2}\rbrack dx}\), where [.] denotes the greatest integer function, equals...

∫01.5[x2]dx\int_{0}^{1.5}{\lbrack x^{2}\rbrack dx}, where [.] denotes the greatest integer function, equals

A

2+22 + \sqrt{2}

B

2−22 - \sqrt{2}

C

1+21 + \sqrt{2}

D

2−1\sqrt{2} - 1

Answer

2−22 - \sqrt{2}

Explanation

Solution

I=∫01.5[x2]dx=∫01[x2]dx+∫12[x2]dx+∫21.5[x2]dxI = \int_{0}^{1.5}{\lbrack x^{2}\rbrack dx = \int_{0}^{1}{\lbrack x^{2}\rbrack dx +}\int_{1}^{\sqrt{2}}{\lbrack x^{2}\rbrack dx}} + \int_{\sqrt{2}}^{1.5}{\lbrack x^{2}\rbrack dx} ⇒ I=0+∫121dx+∫21.52dx=2−1+3−22I = 0 + \int_{1}^{\sqrt{2}}{1dx + \int_{\sqrt{2}}^{1.5}{2dx = \sqrt{2} - 1 + 3 - 2\sqrt{2}}} ⇒ I=2−2I = 2 - \sqrt{2}