Question

Let $f:R\to R$ be a function defined by $f\left(x\right)=\{\begin{array}{c}\left[x\right],x\le 2\\ 0,x>2\end{array},$
where $\left[x\right]$ is the greatest integer less than or equal to $x.$ If $I=\underset{-1}{\overset{2}{\int }}\frac{xf\left(x^2\right)}{2+f(x+1)}dx,$ then the value of $(4I–1)$ is

Answer 1

$I=\underset{-1}{\overset{2}{\int }}\frac{xf\left(x^2\right)}{2+f(x+1)}dx$

Given $f\left(x\right)=\{\begin{array}{c}\left[x\right],x\le 2\\ 0,x>2\end{array}$
$f(x+1)=\{\begin{array}{c}[x+1],x+1\le 2\\ 0,x+1>2\end{array}\Rightarrow f(x+1)=\{\begin{array}{c}\left[x\right]+1,x\le 1\\ 0,x>1\end{array}$
Now,
$f\left(x^2\right)=\{\begin{array}{c}\left[x^2\right],x^2\le 2\\ 0,x^2>2\end{array}\Rightarrow f\left(x^2\right)=\{\begin{array}{c}\left[x^2\right],x\in [-\sqrt{2},\sqrt{2}]\\ 0,x\in (-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty )\end{array}$

Now,
$I=\underset{-1}{\overset{2}{\int }}\frac{xf\left(x^2\right)}{2+f(x+1)}dx\Rightarrow I=\underset{-1}{\overset{\sqrt{2}}{\int }}\frac{xf\left(x^2\right)}{2+f(x+1)}dx\Rightarrow I=\underset{-1}{\overset{0}{\int }}\frac{xf\left(x^2\right)}{2+f(x+1)}dx+\underset{0}{\overset{1}{\int }}\frac{xf\left(x^2\right)}{2+f(x+1)}dx+\underset{1}{\overset{\sqrt{2}}{\int }}\frac{xf\left(x^2\right)}{2+f(x+1)}dx\Rightarrow I=\underset{-1}{\overset{0}{\int }}\frac{x\cdot 0}{2+0}dx+\underset{0}{\overset{1}{\int }}\frac{x\cdot 0}{2+1}dx+\underset{1}{\overset{\sqrt{2}}{\int }}\frac{x\cdot 1}{2+0}dx\Rightarrow I=\underset{1}{\overset{\sqrt{2}}{\int }}\frac{x}{2}dx\Rightarrow I={\left[\frac{x^2}{4}\right]}_1^{\sqrt{2}}\Rightarrow I=\frac{1}{4}\therefore 4I-1=0$