← 2014 Paper 2

UPSC 2014 Maths Optional Paper 2 Q2b — Step-by-Step Solution

15 marks · Section A

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

Question

Integrate $\displaystyle\int_0^1 f(x)\,dx$, where

Technique

Antiderivative spotting via reverse product rule; FTC with one-sided limit at the singular endpoint.

Solution

Strategy. Recognise the integrand as the derivative of $x^{2}\sin(1/x)$, then apply FTC with care at the boundary $x=0$.

Step 1 — Spot the antiderivative

For $x>0$,

So the integrand equals $F'(x)$ where $F(x)=x^{2}\sin(1/x)$ on $(0,1]$.

Step 2 — Extend $F$ continuously to $[0,1]$

$\lim_{x\to 0^{+}}x^{2}\sin(1/x)=0$ (since $|x^{2}\sin(1/x)|\le x^{2}\to 0$).

So define $F(0)=0$; $F$ is continuous on $[0,1]$. ($F$ is not differentiable at $0$, but the FTC limit form still works.)

Step 3 — Boundedness and integrability

The integrand $f(x)$ is bounded on $(0,1]$: $|2x\sin(1/x)|\le 2$, $|\cos(1/x)|\le 1$. With $f(0)=0$, $f$ is bounded on $[0,1]$ with a single discontinuity at $x=0$ (measure zero). So $f$ is Riemann-integrable.

Step 4 — Apply FTC on $[\varepsilon,1]$, take limit $\varepsilon\to 0^{+}$

For $\varepsilon>0$:

Letting $\varepsilon\to 0^{+}$: the second term goes to $0$.

Answer