A calculus problem by 승원 진

$\begin{aligned} h(x) & = \int_{g(x)}^{g(x+1)} \sin (\pi x) \space dt \\ & = \left[ -\frac{\cos (\pi x)}{\pi} \right] _{x(x+1)}^{(x+1)(x+2)} \\ & = \frac{1}{\pi}\left(\cos ([x^2+x]\pi) - \cos ([x^2+3x+2]\pi)\right) \\ & = \frac{1}{\pi} \left( \cos ([x^2+x]\pi) - \cos ([x^2+3x]\pi)\right) \\ & = \frac{1}{\pi} \left( \cos ([x^2+x]\pi) - \cos ([x^2+x+2x]\pi) \right) \\ & = \frac{1}{\pi} \left( \cos ([x^2+x]\pi) - \cos ([x^2+x]\pi) \cos (2x\pi) + \sin ([x^2+x]\pi) \sin (2x\pi) \right) \\ & = \frac{1}{\pi} \left( \cos ([x^2+x]\pi) (1 - 2 \cos^2 (2x\pi) + 1) + 2 \sin ([x^2+x]\pi) \sin (x\pi) \cos (x\pi) \right) \\ & = \frac{1}{\pi} \left(2\cos ([x^2+x]\pi)\sin^2 (x\pi) + 2 \sin ([x^2+x]\pi) \sin (x\pi) \cos (x\pi) \right) \\ & = \frac{2}{\pi} \sin (x \pi) \left(\cos ([x^2+x]\pi)\sin (x\pi) + \sin ([x^2+x]\pi) \cos (x\pi) \right) \\ & = \frac{2}{\pi} \sin (x \pi) \left(\cos ([x^2+x]\pi)\sin (x\pi) + \sin ([x^2+x]\pi) \cos (x\pi) \right) \\ & = \frac{2}{\pi} \sin (x \pi) \sin ([x^2+2x]\pi) \end{aligned}$

$\Rightarrow h(x) = 0 \quad \Rightarrow \begin{cases} \sin (x\pi) = 0 & \Rightarrow x = -1, 0, 1 \\ \sin ([x^2+2x] \pi) = 0 & \Rightarrow x^2 + 2x = n \quad ...(*) \quad \text{where } n \text{ is an integer} \end{cases}$

$\begin{aligned} (*): \quad x^2 + 2x - n & = 0 \\ x & = \frac{-2 \pm \sqrt{4+4n}}{2} \\ & = -1 + \sqrt{1+n} \quad \quad \small \color{#3D99F6}{-1 - \sqrt{1+n} \le -1 \text{ out of range}} \\ & = \begin{cases} -1 & \text{for } n = -1 \\ 0 & \text{for } n = 0 \\ \sqrt{2} - 1 = 0.4142 & \text{for } n = 1 \\ \sqrt{3} - 1 = 0.7321 & \text{for } n = 2 \\ 1 & \text{for } n = 3 \end{cases} \end{aligned}$

Therefore, $h(x)$ has $\boxed{5}$ real roots $\in [-1,1]$ .