← 2022 Paper 2

UPSC 2022 Maths Optional Paper 2 Q2a — Step-by-Step Solution

15 marks · Section A

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

Question

Let $f(x)=x^2$ on $[0,k]$, $k>0$. Show that $f$ is Riemann integrable on $[0,k]$ and $\int_0^k f\,dx=k^3/3$.

Technique

Darboux sums on equal partition; $f=x^2$ increasing, so $m_i=f(\text{left endpoint})$, $M_i=f(\text{right endpoint})$; use $\sum_{i=1}^n i^2=n(n+1)(2n+1)/6$; take limits to show both sums $\to k^3/3$.

Solution

Strategy. Use Darboux sums: show upper and lower sums converge to the same limit as partition gets fine.

Step 1 — Set up partition

Let $P_n$ be the equal partition of $[0,k]$ into $n$ subintervals, each of width $h=k/n$. Subintervals: $[ih,(i+1)h]$ for $i=0,1,\dots,n-1$.

Step 2 — Compute lower and upper Darboux sums

Since $f(x)=x^2$ is increasing on $[0,k]$ (as $k>0$):

• $m_i=\inf_{[ih,(i+1)h]}f=f(ih)=(ih)^2=i^2 h^2$.
• $M_i=\sup_{[ih,(i+1)h]}f=f((i+1)h)=(i+1)^2 h^2$.

Lower sum:

Using $\sum_{i=0}^{n-1}i^2=\dfrac{(n-1)n(2n-1)}{6}$:

$L(P_n)=h^3\cdot\dfrac{(n-1)n(2n-1)}{6}=\dfrac{k^3}{n^3}\cdot\dfrac{(n-1)n(2n-1)}{6}=\dfrac{k^3(n-1)(2n-1)}{6n^2}$.

Upper sum:

Step 3 — Take limits

$\lim_{n\to\infty}L(P_n)=\lim_{n\to\infty}\dfrac{k^3(n-1)(2n-1)}{6n^2}=\dfrac{k^3\cdot 2}{6}=\dfrac{k^3}{3}$.

$\lim_{n\to\infty}U(P_n)=\lim_{n\to\infty}\dfrac{k^3(n+1)(2n+1)}{6n^2}=\dfrac{k^3\cdot 2}{6}=\dfrac{k^3}{3}$.

Both sums converge to the same value $k^3/3$.

Step 4 — Conclusion

Since $\lim L(P_n)=\lim U(P_n)=k^3/3$, $f$ is Riemann integrable and

Answer