GATE 2018 EE Q.18: solution and answer

Official answer: 0.49 to 0.51Source: Official answer key, GATE 2018 EE (IIT Guwahati)

1. $f(x)=x-[x]$ is the fractional part of $x$. The greatest integer function $[x]$ stays constant between consecutive integers and jumps at each integer. The only integer inside $[0.25,1.25]$ is $x=1$.

2. Split the range at $x=1$. On $0.25\le x<1$, $[x]=0$, so $f(x)=x$. On $1\le x\le 1.25$, $[x]=1$, so $f(x)=x-1$. The value of $f$ at the single point $x=1$ does not affect the integral.

3. First piece: $${\int }_{0.25}^{1}xdx={\left[\frac{x^2}{2}\right]}_{0.25}^1=\frac{1-0.0625}{2}=0.46875$$

4. Second piece: $${\int }_1^{1.25}(x-1)dx={\left[\frac{(x-1{)}^2}{2}\right]}_1^{1.25}=\frac{0.0625-0}{2}=0.03125$$

5. Add the pieces: $${\int }_{0.25}^{1.25}f(x)dx=0.46875+0.03125=0.5$$

6. Check with periodicity: $[x+1]=[x]+1$, so $f(x+1)=f(x)$ and $f$ has period 1. A periodic function has the same integral over every interval that is one period long, and $[0.25,1.25]$ is exactly one period long. So the integral equals ${\int }_0^{1}xdx=\frac{1}{2}=0.50$.