Integral of floor function

Question

Integral of floor function, $$\int_{0}^{1} \lfloor4x\rfloor dx.$$

I tried to solve this question like this:

$4x = 1\Rightarrow x = \frac{1}{4} = 0.25$

and total area is equal to $6$? Is this right or wrong?

Answer 1

Integral $$\int_{0}^{1} \lfloor4x\rfloor dx=\int_{1/4}^{2/4} 1dx+\int_{2/4}^{3/4} 2dx+\int_{3/4}^1 3dx=\frac{1}{4}(1+2+3)=\frac{3}{2}$$

Answer 2

$$x\in[0,1]\Longrightarrow4x\in[0,4]$$ We pose, $y=4x$ $$\int_0^1\lfloor4x\rfloor dx=\frac{1}{4}\int_0^4\lfloor y\rfloor dx=\frac{1}{4}(0+1+2+3)=\frac{3}{2}$$