$\int_{0.2}^{3.5}[x] \mathrm{d} x=$ (where $[x]=$ greatest integer not greater than $x$ )

Question

Aicharya · Accepted Answer

$\begin{aligned} \int_{0.2}^{3.5}[x] \mathrm{d} x & =\int_{0.2}^1(0) \mathrm{d} x+\int_1^2(1) \mathrm{d} x+\int_2^3 2 \mathrm{~d} x+\int_3^{3.5} 3 \mathrm{~d} x \ & =0+[x]_1^2+2[x]_2^3+3[x]_3^{3.5} \ & =0+(2-1)+2(3-2)+3(3.5-3) \ & =0+1+2+1.5 \ & =4.5\end{aligned}$

$\int_{0.2}^{3.5}[x] \mathrm{d} x=$ (where $[x]=$ greatest integer not greater than $x$ )

Solution