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UPSC 2013 Maths Optional Paper 2 Q3d — Step-by-Step Solution

10 marks · Section A

Riemann integral · Real Analysis · asked 10× in 13 yrs · Read the full method →

Question

Let $[x]$ denote the integer part of the real number $x$, i.e., if $n\le x<n+1$ where $n$ is an integer, then $[x]=n$. Is the function $f(x)=[x]^{2}+3$ Riemann integrable in $[-1,2]$? If not, explain why. If it is integrable, compute $\displaystyle\int_{-1}^{2}([x]^{2}+3)\,dx$.

Technique

Step-function on partitioned subintervals; finitely-many-discontinuity criterion for Riemann integrability.

Solution

Step 1 — Determine $f$ on each subinterval

Using the convention $[x]=n$ when $n\le x<n+1$:

So $f$ is a step function on $[-1,2]$ taking finitely many values (namely $3,4,7$).

Step 2 — Riemann integrability

$f$ is bounded ($3\le f(x)\le 7$ on $[-1,2]$) and has only finitely many discontinuities — at $x=0,\,1,\,2$ (jump discontinuities where $[x]$ jumps). A bounded function with finitely many discontinuities is Riemann integrable (the discontinuity set $\{0,1,2\}$ has Lebesgue measure zero).

So $f$ is Riemann integrable on $[-1,2]$.

Step 3 — Compute the integral

The single point $x=2$ contributes nothing to a Riemann integral (measure-zero set). So

Answer