Let f: ℝ → ℝ be a function defined as f(x) = asin((π[x])/(2)) + [2...

Given,

$f\left(x\right) = asin\left(\dfrac{\pi\left[x\right]}{2}\right) + \left[2 - x\right], a \in \mathbb{R}$

Now given $\lim_{x \to - 1}f\left(x\right)$ exists,

So $\lim_{x \to - {1}^{+}}asin\left(\pi\dfrac{\left[x\right]}{2}\right) + \left[2 - x\right] = - a + 2$

And $\lim_{x \to - {1}^{-}} asin \left(\pi\dfrac{\left[x\right]}{2}\right) + \left[2 - x\right] = 0 + 3 = 3$

So, $\lim_{x \to - 1}f\left(x\right)$ exist when $- a + 2 = 3 ⇒ a = - 1$

Now,

${∫}_{0}^{4}f\left(x\right)dx = {∫}_{0}^{1}f\left(x\right)dx + {∫}_{1}^{2}f\left(x\right)dx + {∫}_{2}^{3}f\left(x\right)dx + {∫}_{3}^{4}f\left(x\right)dx$

$⇒ {∫}_{0}^{4}f\left(x\right)dx = {∫}_{0}^{1} - \sin \left(\dfrac{\pi\left[x\right]}{2}\right) + \left[2 - x\right]dx + {∫}_{1}^{2} - \sin \left(\dfrac{\pi\left[x\right]}{2}\right) + \left[2 - x\right]dx + {∫}_{2}^{3} - \sin \left(\dfrac{\pi\left[x\right]}{2}\right) + \left[2 - x\right]dx + {∫}_{3}^{4} - \sin \left(\dfrac{\pi\left[x\right]}{2}\right) + \left[2 - x\right]dx$

$= {∫}_{0}^{1}\left(0 + 1\right)dx + {∫}_{1}^{2}\left(- 1 + 0\right)dx + {∫}_{2}^{3}\left(0 - 1\right)dx + {∫}_{3}^{4}\left(1 - 2\right)dx$

$= 1 - 1 - 1 - 1 = - 2$