Question

The value of the integral $\int \left[x\right]dx$ is

• A. $\frac{[x{]}^2}{2}\forall x>0$
• B. $\frac{\left[x^2\right]}{2}\forall x>0$
• C. $5x-15$ for $x\in [0,t],t\in (5,6)$
• D. Cannot be evaluated

Answer 1

Answer: (C)

$5x-15$ for $x\in [0,t],t\in (5,6)$

$\int \left[x\right]dx$

${\int }_0^6\left[x\right]dx={\int }_0^{1}0.dx+{\int }_1^{2}1.dx+{\int }_2^{3}2.dx+{\int }_3^{4}3.dx+{\int }_4^{5}4.dx+{\int }_5^{6}5.dx=0+1+2+3+4+5=15$

${\int }_0^5\left[x\right]dx={\int }_0^{1}0.dx+{\int }_1^{2}1.dx+{\int }_2^{3}2.dx+{\int }_3^{4}3.dx+{\int }_4^{5}4.dx=0+1+2+3+4=10$

${\int }_0^t\left[x\right]dx$ Where $t\in (5,10)$

At $t=5,{\int }_0^t\left[x\right]dx=10$

At $t=6,{\int }_0^t\left[x\right]dx=15$

Hence, $\int \left[x\right]dx=5x-15$ for $x\in [0,t]x\in [5,6]$