Evaluate $\int_{[0,6]} \sin \frac{\lfloor x \rfloor \pi}{4} d\lambda$ / Notation check

Question

Given a measure space $(S, \Sigma, \mu) = ([0,6], \mathscr B([0,6]), \lambda)$,

evaluate $$\int_{[0,6]} \sin \frac{\lfloor x \rfloor \pi}{4} \lambda(dx)$$

---

$$\int_{[0,6]} \sin \frac{\lfloor x \rfloor \pi}{4} \lambda(dx)$$

$$ = \int_{[0,6]} \sin \frac{\lfloor x \rfloor \pi}{4} dx$$

$$= \int_{[0,1)} \sin \frac{\lfloor x \rfloor \pi}{4} dx + \int_{[1,2)} \sin \frac{\lfloor x \rfloor \pi}{4} dx + \int_{[2,3)} \sin \frac{\lfloor x \rfloor \pi}{4} dx$$

$$+ \int_{[3,4)} \sin \frac{\lfloor x \rfloor \pi}{4} dx + \int_{[4,5)} \sin \frac{\lfloor x \rfloor \pi}{4} dx + \int_{[5,6)} \sin \frac{\lfloor x \rfloor \pi}{4} dx$$

$$+ \underbrace{\int_{\{6\}} \sin \frac{\lfloor x \rfloor \pi}{4} dx}_{0}$$

$$= \int_{[0,1)} 0 dx + \int_{[1,2)} \frac{\sqrt{2}}{2} dx + \int_{[2,3)} 1 dx$$

$$+ \int_{[3,4)} \frac{\sqrt{2}}{2} dx + \int_{[4,5)} 0 dx + \int_{[5,6)} -\frac{\sqrt{2}}{2} dx$$

$$= \frac{\sqrt{2}}{2} + 1 $$

$$+ \frac{\sqrt{2}}{2} + -\frac{\sqrt{2}}{2} $$

$$= \frac{\sqrt{2}}{2} + 1 $$

Is that right? Value, notation, steps and all?

Answer 1

Looks correct to me, assuming $x$ and $\lambda$ are the same thing. WolframAlpha also agrees with you.