← 2013 Paper 2

UPSC 2013 Maths Optional Paper 2 Q1c — Step-by-Step Solution

10 marks · Section A

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

Question

Let $f(x)=\begin{cases}\dfrac{x^{2}}{2}+4 & \text{if }x\ge 0\\ -\dfrac{x^{2}}{2}+2 & \text{if }x<0\end{cases}$. Is $f$ Riemann integrable in the interval $[-1,2]$? Why? Does there exist a function $g$ such that $g'(x)=f(x)$? Justify your answer.

Technique

Lebesgue criterion for Riemann integrability; Darboux’s theorem rules out the antiderivative.

Solution

Part 1 — Riemann integrability

Step 1 — Boundedness on $[-1,2]$.

On $[0,2]$: $f(x)=x^{2}/2+4\in[4,\,6]$. On $[-1,0)$: $f(x)=-x^{2}/2+2\in[3/2,\,2)$.

So $f$ is bounded on $[-1,2]$ (between 3/2 and 6).

Step 2 — Set of discontinuities.

On $(0,2]$ and on $[-1,0)$, $f$ is polynomial — hence continuous.

At $x=0$: $\lim_{x\to 0^+}f(x)=4$ but $\lim_{x\to 0^-}f(x)=2$. Jump discontinuity at $x=0$.

So $f$ has exactly one discontinuity on $[-1,2]$, namely $x=0$ — a measure-zero set.

Step 3 — Apply Lebesgue criterion.

A bounded function on a closed interval is Riemann integrable iff its set of discontinuities has Lebesgue measure zero. The single point $\{0\}$ has measure zero.

Part 2 — Existence of an antiderivative

Claim. No function $g$ with $g'(x)=f(x)$ for all $x\in[-1,2]$ exists.

Proof via Darboux’s theorem (intermediate value property of derivatives).

Darboux’s theorem: if $g$ is differentiable on an interval $I$, then $g'$ has the intermediate value property on $I$ — i.e., for any $a,b\in I$ and any value $c$ strictly between $g'(a)$ and $g'(b)$, there exists some $\xi\in(a,b)$ with $g'(\xi)=c$.

In particular, $g'$ cannot have a jump discontinuity.

Our $f$ has a jump discontinuity at $0$: $f(0^-)=2,\,f(0^+)=4$, $f(0)=4$. Any value $c\in(2,4)$ — say $c=3$ — is not attained by $f$ on $[-1,2]$ (check: for $x<0$, $f(x)\in[3/2,2)$, so $\le 2$; for $x\ge 0$, $f(x)\ge 4$; the value $3$ is skipped). So $f$ does not have the intermediate value property.

By Darboux, $f$ cannot be the derivative of any function.

Answer