If $[x]$ is the greatest integer not exceeding $x$, then $$ \int_{-0.5}^{1.5} x^2[x] d x= $$
If $[x]$ is the greatest integer not exceeding $x$, then
$$
\int_{-0.5}^{1.5} x^2[x] d x=
$$
$\frac{4.5}{4}$
$\frac{3}{4}$
$\frac{3.5}{4}$
$\frac{2.375}{2}$
Solution
$\begin{aligned} & \text { } \int_{-0.5}^{1.5} x^2[x] d x=\int_{-0.5}^0 x^2(-1) d x+\int_0^1 x^2(0) d x+\int_1^{1.5} x^2(1) d x \\ & =-\frac{1}{3}\left[x^3\right]_{-\frac{1}{2}}^0+\frac{1}{3}\left[x^3\right]_1^{3 / 2}=\frac{3}{4}\end{aligned}$