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UPSC Maths 2013 Paper 1 Q1c — Solution

10 marks · Section A

Question

Evaluate $\displaystyle\int_0^1\!\left(2x\sin\tfrac{1}{x}-\cos\tfrac{1}{x}\right)dx$.

Technique

Antiderivative spotting (product-rule reverse) + FTC with a one-sided limit at the singular endpoint.

Solution

Strategy. Recognise the integrand as the derivative of $x^{2}\sin(1/x)$, then apply (an improper-integral form of) the Fundamental Theorem of Calculus.

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 — Handle the endpoint $x=0$

Extend $F$ to $[0,1]$ by setting $F(0)=0$. Then $F$ is continuous at $0$ because

(However $F$ is not differentiable at $0$: $F'(0)$ doesn’t exist because $\cos(1/x)$ has no limit at $0$.)

The integrand is bounded on $(0,1]$: $|2x\sin(1/x)|\le 2x\le 2$ and $|\cos(1/x)|\le 1$. So it’s a bounded function with a single essential discontinuity at $x=0$ — Riemann integrable on $[0,1]$.

Step 3 — Evaluate as a limit

Apply FTC on $[\varepsilon,1]$ for $\varepsilon>0$ (where $F$ is differentiable):

Let $\varepsilon\to 0^{+}$: the right-hand side tends to $\sin 1-0=\sin 1$ (since $|\varepsilon^{2}\sin(1/\varepsilon)|\le\varepsilon^{2}\to 0$). Therefore

Answer