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UPSC 2024 Maths Optional Paper 2 Q4b — Step-by-Step Solution

15 marks · Section A

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

Question

Find the upper and lower Riemann integrals for

on $[0,1]$. Hence show that $f$ is not Riemann integrable on $[0,1]$.

Technique

Use the density of rationals and irrationals to identify the supremum and infimum on every subinterval; compute the resulting Darboux sums.

Solution

Step 1 — Pointwise comparison

For $x\in(0,1)$: $\sqrt{1-x^2}=\sqrt{(1-x)(1+x)}$. Since $1+x>1$:

so $\sqrt{1-x^2}>(1-x)$ on $(0,1)$. At endpoints both values agree.

Step 2 — Sup and inf on any subinterval $[\alpha,\beta]\subseteq[0,1]$

By density of both rationals and irrationals:

Step 3 — Upper and lower integrals

As the mesh of a partition $\to 0$, the upper Darboux sums approach $\displaystyle\int_0^1\sqrt{1-x^2}\,dx=\dfrac{\pi}{4}$ and the lower Darboux sums approach $\displaystyle\int_0^1(1-x)\,dx=\dfrac{1}{2}$.

Since $\pi/4\ne 1/2$, the upper and lower integrals differ and $f$ is not Riemann integrable.

Answer