Question

Let $R\to R$ be function defined as
$f\left(x\right)=a\sin \left(\frac{\pi \left[x\right]}{2}\right)+[2-x],a\in R$, where $\left[t\right]$ is the greatest integer less than or equal to $t$. If $\underset{x\to -1}{lim}f\left(x\right)$ exists, then the value of $\underset{0}{\overset{4}{\int }}f\left(x\right)dx$ is equal to

Answer 1

$f\left(x\right)=a\sin \left(\frac{\pi \left[x\right]}{2}\right)+[2-x],a\in R$
Now,
$\because \underset{x\to -1}{lim}f\left(x\right)$ exists
$\therefore \underset{x\to -1^{-}}{lim}f\left(x\right)=\underset{x\to -1^{+}}{lim}f\left(x\right)$
$\Rightarrow a\sin \left(\frac{-2\pi }{2}\right)+3=a\sin \left(\frac{-\pi }{2}\right)+2$
$\Rightarrow -a=1\Rightarrow a=-1$
Now,
$\underset{0}{\overset{4}{\int }}f\left(x\right)dx=\underset{0}{\overset{4}{\int }}(-\sin \left(\frac{\pi \left[x\right]}{2}\right)+[2-x\left]\right)dx$
$=\underset{0}{\overset{1}{\int }}1dx+\underset{1}{\overset{2}{\int }}-1dx+\underset{2}{\overset{3}{\int }}-1dx+\underset{3}{\overset{4}{\int }}(1-2)dx$
$=1-1-1-1=-2$