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Riemann Sums — Question Types

Type 1: Upper Bound Using Rectangles (4 marks)​

Example: s20/21 Q4(a) — Use rectangles to find an upper bound for ∫01x2 dx\int_0^1 x^2\,dx.

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f(x)=x2f(x) = x^2 is increasing on [0,1][0,1]. Divide [0,1][0,1] into nn equal strips of width Δx=1n\Delta x = \frac{1}{n}.

For an upper bound (increasing function), use right endpoints:

xi=inx_i = \frac{i}{n} for i=1,2,…,ni = 1, 2, \ldots, n

Un=∑i=1nf(xi)Δx=∑i=1n(in)2⋅1nU_n = \sum_{i=1}^n f(x_i)\Delta x = \sum_{i=1}^n \left(\frac{i}{n}\right)^2 \cdot \frac{1}{n}

=1n3∑i=1ni2=1n3⋅n(n+1)(2n+1)6= \frac{1}{n^3}\sum_{i=1}^n i^2 = \frac{1}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}

=(n+1)(2n+1)6n2= \frac{(n+1)(2n+1)}{6n^2}

As n→∞n\to\infty, Un→26=13U_n \to \frac{2}{6} = \frac{1}{3}, which is the exact value of ∫01x2 dx\int_0^1 x^2\,dx.

M1 — Δx=1n\Delta x = \frac{1}{n} with correct endpoints M1 — Form sum ∑f(xi)Δx\sum f(x_i)\Delta x with correct xix_i A1 — Correct expression (n+1)(2n+1)6n2\frac{(n+1)(2n+1)}{6n^2} A1 — Simplify and state upper bound

Example: w20/21 Q4 — Find an upper bound for ∫01(1−x3) dx\int_0^1 (1 - x^3)\,dx using nn equal subintervals.

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f(x)=1−x3f(x) = 1 - x^3 is decreasing on [0,1][0,1]. For decreasing ff, the upper bound uses left endpoints.

Δx=1n\Delta x = \frac{1}{n}

Left endpoints: xi=i−1nx_i = \frac{i-1}{n} for i=1,2,…,ni = 1, 2, \ldots, n

Un=∑i=1nf(i−1n)1n=1n∑i=0n−1(1−i3n3)U_n = \sum_{i=1}^n f\left(\frac{i-1}{n}\right)\frac{1}{n} = \frac{1}{n}\sum_{i=0}^{n-1} \left(1 - \frac{i^3}{n^3}\right)

=1n[n−1n3∑i=0n−1i3]=1n[n−1n3⋅(n−1)2n24]= \frac{1}{n}\left[n - \frac{1}{n^3}\sum_{i=0}^{n-1} i^3\right] = \frac{1}{n}\left[n - \frac{1}{n^3}\cdot\frac{(n-1)^2 n^2}{4}\right]

=1−(n−1)24n2=1−n2−2n+14n2= 1 - \frac{(n-1)^2}{4n^2} = 1 - \frac{n^2 - 2n + 1}{4n^2}

=3n2+2n−14n2= \frac{3n^2 + 2n - 1}{4n^2}

M1 — Recognize decreasing so use left endpoints M1 — Correct sum with Δx=1/n\Delta x = 1/n A1 — Correct sigma sum and formula A1 — Simplify to 3n2+2n−14n2\frac{3n^2 + 2n - 1}{4n^2}

Type 2: Lower Bound Using Rectangles (4 marks)​

Example: s20/21 Q4(b) — Find a lower bound for ∫01x2 dx\int_0^1 x^2\,dx using nn equal subintervals.

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f(x)=x2f(x) = x^2 is increasing. For lower bound, use left endpoints.

xi=i−1nx_i = \frac{i-1}{n} for i=1,2,…,ni = 1, 2, \ldots, n

Ln=∑i=1nf(xi)Δx=∑i=1n(i−1n)2⋅1nL_n = \sum_{i=1}^n f(x_i)\Delta x = \sum_{i=1}^n \left(\frac{i-1}{n}\right)^2 \cdot \frac{1}{n}

=1n3∑i=1n(i−1)2=1n3∑j=0n−1j2= \frac{1}{n^3}\sum_{i=1}^n (i-1)^2 = \frac{1}{n^3}\sum_{j=0}^{n-1} j^2

=1n3⋅(n−1)n(2n−1)6= \frac{1}{n^3}\cdot\frac{(n-1)n(2n-1)}{6}

=(n−1)(2n−1)6n2= \frac{(n-1)(2n-1)}{6n^2}

As n→∞n\to\infty, Ln→13L_n \to \frac{1}{3}.

M1 — Use left endpoints for lower bound M1 — Correct sum A1 — (n−1)(2n−1)6n2\frac{(n-1)(2n-1)}{6n^2} A1 — Lower bound stated

Type 3: Stirling-Type Approximations (ln⁡N!\ln N!) (8 marks)​

Example: s20/23 Q4 — Use rectangles to estimate ln⁡N!\ln N!.

Consider f(x)=ln⁡xf(x) = \ln x on [1,N][1, N]. Since ff is increasing:

Left endpoint sum (lower bound):

∑r=1N−1ln⁡r≤∫1Nln⁡x dx\sum_{r=1}^{N-1} \ln r \le \int_1^N \ln x\,dx

Right endpoint sum (upper bound):

∫1Nln⁡x dx≤∑r=2Nln⁡r\int_1^N \ln x\,dx \le \sum_{r=2}^N \ln r

Since ln⁡1=0\ln 1 = 0, we have ∑r=1N−1ln⁡r=ln⁡(N−1)!\sum_{r=1}^{N-1} \ln r = \ln(N-1)! and ∑r=2Nln⁡r=ln⁡N!\sum_{r=2}^N \ln r = \ln N!.

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∫1Nln⁡x dx=[xln⁡x−x]1N=Nln⁡N−N+1\int_1^N \ln x\,dx = [x\ln x - x]_1^N = N\ln N - N + 1

Lower bound: ln⁡(N−1)!≤Nln⁡N−N+1\ln(N-1)! \le N\ln N - N + 1

ln⁡N!=ln⁡(N−1)!+ln⁡N≤Nln⁡N−N+1+ln⁡N\ln N! = \ln(N-1)! + \ln N \le N\ln N - N + 1 + \ln N

Upper bound: Nln⁡N−N+1≤ln⁡N!N\ln N - N + 1 \le \ln N!

Therefore:

Nln⁡N−N+1≤ln⁡N!≤Nln⁡N−N+1+ln⁡NN\ln N - N + 1 \le \ln N! \le N\ln N - N + 1 + \ln N

For large NN, ln⁡N!≈Nln⁡N−N+12ln⁡(2πN)\ln N! \approx N\ln N - N + \frac{1}{2}\ln(2\pi N) (full Stirling).

M1 — Set f(x)=ln⁡xf(x) = \ln x, note increasing M1 — Write ∫1Nln⁡x dx\int_1^N \ln x\,dx and evaluate to Nln⁡N−N+1N\ln N - N + 1 M1 — Lower bound: ln⁡(N−1)!≤∫1Nln⁡x dx\ln(N-1)! \le \int_1^N \ln x\,dx M1 — Upper bound: ∫1Nln⁡x dx≤ln⁡N!\int_1^N \ln x\,dx \le \ln N! A1 — Correct inequality: Nln⁡N−N+1≤ln⁡N!≤Nln⁡N−N+1+ln⁡NN\ln N - N + 1 \le \ln N! \le N\ln N - N + 1 + \ln N

Example: w20/22 Q8 — Use upper and lower Riemann sums for ∫1nln⁡x dx\int_1^n \ln x\,dx to show that:

nne1−n≤n!≤nn+1e1−nn^n e^{1-n} \le n! \le n^{n+1} e^{1-n}

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From Riemann sums on ln⁡x\ln x:

ln⁡(n−1)!≤∫1nln⁡x dx≤ln⁡n!\ln(n-1)! \le \int_1^n \ln x\,dx \le \ln n!

∫1nln⁡x dx=nln⁡n−n+1\int_1^n \ln x\,dx = n\ln n - n + 1

So: ln⁡(n−1)!≤nln⁡n−n+1≤ln⁡n!\ln(n-1)! \le n\ln n - n + 1 \le \ln n!

From RHS: nln⁡n−n+1≤ln⁡n!⇒ln⁡n!≥nln⁡n−n+1n\ln n - n + 1 \le \ln n! \Rightarrow \ln n! \ge n\ln n - n + 1

⇒n!≥enln⁡n−n+1=nne1−n\Rightarrow n! \ge e^{n\ln n - n + 1} = n^n e^{1-n}

From LHS: ln⁡(n−1)!≤nln⁡n−n+1\ln(n-1)! \le n\ln n - n + 1

ln⁡n!=ln⁡(n−1)!+ln⁡n≤nln⁡n−n+1+ln⁡n\ln n! = \ln(n-1)! + \ln n \le n\ln n - n + 1 + \ln n

⇒n!≤nn+1e1−n\Rightarrow n! \le n^{n+1}e^{1-n}

Therefore: nne1−n≤n!≤nn+1e1−nn^n e^{1-n} \le n! \le n^{n+1} e^{1-n}

M1 — Riemann sum inequalities A1 — Correct integral evaluation M1 — Lower bound manipulation M1 — Upper bound manipulation A1 — Correct final inequality A1 — Exponential form