Question

The integral $\stackrel{1.5}{\underset{0}{\int }}\left[x^2\right]dx$, where $\left[\right]$ denotes the greatest integer function, equals ______

• A. $2-2\sqrt{2}$
• B. $3-\sqrt{2}$
• C. $2-\sqrt{2}$
• D. $3-2\sqrt{2}$

Answer 1

The correct option is C $2-\sqrt{2}$

${\int }_n^{1.5}\left[x^2\right]dx$

Splitting the integral ${\int }_0^1\left[x^2\right]dx+{\int }_1^{\sqrt{2}}\left[x^2\right]dx+{\int }_{\sqrt{2}}^{1.5}\left[x^2\right]dx$

Between $0$ & $1$ function is $0$, integral $0$.

Between $1$ & $\sqrt{2}$ function is $1$, integral $\left(1\right)(\sqrt{2}-1)$

Between $\sqrt{2}$ & $1.5$ function is $2$, integral $\left(2\right)(1.5-\sqrt{2})$

${\int }_0^1\left[x^2\right]dx+{\int }_1^{\sqrt{2}}\left[x^2\right]dx+{\int }_{\sqrt{2}}^{1.5}\left[x^2\right]dx$

$=0+\left(1\right)(\sqrt{2}-1)+\left(2\right)(1.5-\sqrt{2})$

$=\sqrt{2}-1+3-2\sqrt{2}$

$=2-\sqrt{2}$.