Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function defined as

$f(x)=a \sin \left(\frac{\pi[x]}{2}\right)+[2-x], a \in \mathbb{R}$ where $[t]$ is the greatest integer less than or equal to $t$. If $\mathop {\lim }\limits_{x \to -1 } f(x)$ exists, then the value of $\int\limits_{0}^{4} f(x) d x$ is equal to

<p>$f(x) = a\sin \left( {{{\pi [x]} \over 2}} \right) + [2 - x]\,a \in R$</p> <p>Now,</p> <p>$\because$ $\mathop {\lim }\limits_{x \to - 1} f(x)$ exist</p> <p>$\therefore$ $$\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to - {1^ + }} f(x)$$</p> <p>$$ \Rightarrow a\sin \left( {{{ - 2\pi } \over 2}} \right) + 3 = a\sin \left( {{{ - \pi } \over 2}} \right) + 2$$</p> <p>$\Rightarrow - a = 1 \Rightarrow a = - 1$</p> <p>Now, $$\int_0^4 {f(x)dx = \int_0^4 {\left( { - \sin \left( {{{\pi [x]} \over 2}} \right) + [2 - x]} \right)dx} } $$</p> <p>$$ = \int_0^1 {1dx + \int_1^2 { - 1dx + \int_2^3 { - 1dx + \int_3^4 {(1 - 2)dx} } } } $$</p> <p>$= 1 - 1 - 1 - 1 = - 2$</p>