$\int_0^2[2 x] \mathrm{d}=$ (where [.] denotes the greatest integer function.)

Question

Aicharya · Accepted Answer

$\begin{aligned} & \int_0^2[2 x] \mathrm{d} x=\int_0^{1 / 2} 0 \mathrm{~d} x+\int_{\frac{1}{2}}^1 1 \cdot \mathrm{d} x+\int_1^{3 / 2} 2 \mathrm{~d} x+\int_{\frac{3}{2}}^2 3 \mathrm{~d} x \ & =0+\left(1-\frac{1}{2}ight)+2\left(\frac{3}{2}-1ight)+3\left(2-\frac{3}{2}ight) \ & =0+\frac{1}{2}+2 	imes \frac{1}{2}+3 	imes \frac{1}{2}=3\end{aligned}$

$\int_0^2[2 x] \mathrm{d}=$ (where [.] denotes the greatest integer function.)

Solution