GATE 2026 IN Q.57: solution and answer

Official answer: 2.4 to 2.6Source: Official answer key, GATE 2026 IN (IIT Guwahati)

1. $f$ is $0$ outside $[0,3]$, so the infinite integral reduces to a finite one: ${\int }_{-\infty }^{\infty }f(\tau )d\tau ={\int }_0^{3}f(\tau )d\tau$.

2. Split at the breakpoint $\tau =2$ using additivity of the definite integral: ${\int }_0^{3}fd\tau ={\int }_0^{2}1d\tau +{\int }_2^3(3-\tau )d\tau$. Both formulas give $f(2)=1$, so the two pieces agree at the joint. A single point has zero width and could not change the integral anyway.

3. First piece, a rectangle of height $1$ and width $2$: ${\int }_0^{2}1d\tau =2$.

4. Second piece, a straight line falling from $1$ at $\tau =2$ to $0$ at $\tau =3$: $${\int }_2^3(3-\tau )d\tau ={\left[3\tau -\frac{{\tau }^2}{2}\right]}_2^3=\left(9-\frac{9}{2}\right)-(6-2)=4.5-4=0.5$$ This matches the area of a triangle with base $1$ and height $1$: $\frac{1}{2}\times 1\times 1=0.5$.

5. Add the pieces: ${\int }_{-\infty }^{\infty }f(\tau )d\tau =2+0.5=2.5$.