Question

Area bounded by the curve $f\left(x\right)=\frac{1}{x^2+[x{]}^2+2\{x\}+1-2x[x]}$ and x-axis between$x=\frac{-3}{2}$ and $x=\frac{5}{2}$ is equal to
(where [ ] denotes greatest integer function and { } denotes fractional part function)

Answer 1

$f\left(x\right)=\frac{1}{x^2+[x{]}^2+2\{x\}+1-2x[x]}\Rightarrow f\left(x\right)=\frac{1}{{(x-[x\left]\right)}^2+2\left\{x\right\}+1}\Rightarrow f\left(x\right)=\frac{1}{\{x{\}}^2+2\{x\}+1}\Rightarrow f\left(x\right)=\frac{1}{\left(\right\{x\}+1{)}^2}$
So,area bounded =$\underset{\frac{-3}{2}}{\overset{\frac{5}{2}}{\int }}\frac{1}{\left(\right\{x\}+1{)}^2}dx=4\underset{0}{\overset{1}{\int }}\frac{1}{(x+1{)}^2}dx=4{\left(\frac{-1}{1+x}\right)}_0^1=2$