Question

Evaluate the integral
${\int }_0^{1.5}\left[x^2\right]dx$

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

Answer 1

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

${\int }_0^{1.5}\left[x^2\right]dx$

${\int }_0^1\left[x^2\right]dx+{\int }_1^{\sqrt{2}}\left[x^2\right]dx+{\int }_{\sqrt{2}}^{1-5}\left[x^2\right]dx$

$=0+{\int }_1^{\sqrt{2}}dx+{\int }_{\sqrt{2}}^{\left(1.5\right)}2dx$

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

$\Rightarrow 2-\sqrt{2}$