Home / A-Level Further Maths 9231 / Further integration / Rectangles, bounds and sums
Cambridge International AS & A Level · Topic 2.4
A-Level Further Maths 9231 / Topic 2.4 / AS & A Level

Trap the area between two staircases.

A sum of rectangle areas sits just above or just below the area under a curve. Use that to estimate an integral, to squeeze a sum such as \(\sum 1/r\) between two logarithms, and to turn the limit of a sum with width \(\frac{1}{n}\) into a definite integral.

Topic 2.4All further integration lessons →
The key idea

For a positive decreasing function, a rectangle whose height is the value at its LEFT end lies above the curve and one using the RIGHT end lies below it (reverse this for an increasing function). With unit width each rectangle's area equals its height, so for \(f\) positive and decreasing \(\displaystyle\int_{1}^{n+1} f(x)\,\mathrm{d}x < \sum_{r=1}^{n} f(r) < f(1) + \int_{1}^{n} f(x)\,\mathrm{d}x\). With width \(\frac{1}{n}\), \(\displaystyle\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n} f\!\left(\frac{r}{n}\right) = \int_{0}^{1} f(x)\,\mathrm{d}x\).

x y O 1 2 3 4 5 6 1 upper: heights 1, 1/2, ..., 1/5 lower: heights 1/2, ..., 1/6 y = 1/x, area ln 6 1.45 < ln 6 ≈ 1.79 < 2.28
Fig. 1: y = 1/x from x = 1 to x = 6 with unit-width rectangles. Each orange upper rectangle takes its height from the left end of its strip and pokes above the curve; each teal lower rectangle takes its height from the right end and sits under it. So 1/2 + 1/3 + ... + 1/6 is less than ln 6, which is less than 1 + 1/2 + ... + 1/5
Section 01

Rectangles above and below the curve.

Split the interval \(a \le x \le b\) into \(n\) strips of width \(h = \dfrac{b - a}{n}\) and stand a rectangle on each strip. If the curve is decreasing, the highest point of a strip is at its left end, so a rectangle using the left-end height covers the curve (an upper sum); the right-end height gives a rectangle that fits under it (a lower sum). For an increasing curve the two swap. The true area is trapped between the two sums.

Function on the intervalUpper sum usesLower sum uses
decreasingleft-end heightsright-end heights
increasingright-end heightsleft-end heights

For a decreasing \(f\) with \(n\) strips of width \(h\):

\[ h\sum_{r=1}^{n} f(a + rh) < \int_{a}^{b} f(x)\,\mathrm{d}x \] \[ \int_{a}^{b} f(x)\,\mathrm{d}x < h\sum_{r=0}^{n-1} f(a + rh) \]

The two sums share every height except the end ones, so they differ by only \(h\,|f(b) - f(a)|\). Narrower strips squeeze the gap.

Worked example Bounds for ln 5
Use four rectangles of unit width to find a lower and an upper bound for \(\displaystyle\int_{1}^{5} \frac{1}{x}\,\mathrm{d}x\).
  1. \(y = \dfrac{1}{x}\) is decreasing for \(x > 0\), so left-end heights give the upper sum and right-end heights the lower sum (Fig. 1).
  2. Upper sum: \(1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} = \dfrac{25}{12} \approx 2.083\).
  3. Lower sum: \(\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} = \dfrac{77}{60} \approx 1.283\).
  4. The integral is \(\ln 5 \approx 1.609\), which does lie between them.
Answer: \(\dfrac{77}{60} < \ln 5 < \dfrac{25}{12}\)
Worked example An increasing function
Use four strips of width \(0.5\) to find a lower and an upper bound for \(\displaystyle\int_{0}^{2} x^{2}\,\mathrm{d}x\).
  1. \(y = x^{2}\) is increasing on \(0 \le x \le 2\), so left ends give the LOWER sum.
  2. Heights at \(x = 0, 0.5, 1, 1.5, 2\): \(0,\ 0.25,\ 1,\ 2.25,\ 4\).
  3. Lower: \(0.5(0 + 0.25 + 1 + 2.25) = 1.75\).
  4. Upper: \(0.5(0.25 + 1 + 2.25 + 4) = 3.75\).
  5. Check: the gap is \(h\,(f(2) - f(0)) = 0.5 \times 4 = 2\), and the exact value \(\dfrac{8}{3} \approx 2.667\) lies between.
Answer: \(1.75 < \displaystyle\int_{0}^{2} x^{2}\,\mathrm{d}x < 3.75\)
Your turn
Use three rectangles of unit width under and over \(y = \dfrac{1}{x}\) to show that \(\dfrac{13}{12} < \ln 4 < \dfrac{11}{6}\).
Examiner trap

Left-end heights give the UPPER sum only for a decreasing function. For an increasing one they give the lower sum. Sketch the curve and one rectangle, and see which corner of the rectangle touches the curve, before you label anything.

Section 02

Turning a sum into an inequality.

With strips of width 1, a rectangle's area equals its height, so a sum such as \(f(1) + f(2) + \dots + f(n)\) IS the total area of \(n\) rectangles. Compare that area with an integral you can evaluate and you get bounds for a sum you cannot.

For \(f\) positive and decreasing for \(x \ge 1\):

For an increasing \(f\) the inequalities point the other way, for example \(\displaystyle\int_{0}^{n} f(x)\,\mathrm{d}x < \sum_{r=1}^{n} f(r) < \int_{1}^{n+1} f(x)\,\mathrm{d}x\).

Worked example The harmonic sum
Show that \(\displaystyle\ln(n + 1) < \sum_{r=1}^{n} \frac{1}{r} < 1 + \ln n\) for \(n \ge 2\).
  1. \(f(x) = \dfrac{1}{x}\) is positive and decreasing for \(x > 0\).
  2. Rectangles on \([r, r + 1]\) of height \(\dfrac{1}{r}\), \(r = 1\) to \(n\), lie above the curve: \[ \sum_{r=1}^{n} \frac{1}{r} > \int_{1}^{n+1} \frac{1}{x}\,\mathrm{d}x = \ln(n + 1) \]
  3. Rectangles on \([r - 1, r]\) of height \(\dfrac{1}{r}\), \(r = 2\) to \(n\), lie below the curve: \[ \sum_{r=2}^{n} \frac{1}{r} < \int_{1}^{n} \frac{1}{x}\,\mathrm{d}x = \ln n \]
  4. Add the first term, \(1\), to both sides of the second inequality and combine.
Answer: \(\displaystyle\ln(n + 1) < \sum_{r=1}^{n} \frac{1}{r} < 1 + \ln n\)
Worked example Bounding a numerical sum
Find bounds for \(\displaystyle S = \sum_{r=1}^{100} \frac{1}{\sqrt{r}}\).
  1. \(\dfrac{1}{\sqrt{x}}\) is positive and decreasing, and \(\displaystyle\int \frac{1}{\sqrt{x}}\,\mathrm{d}x = 2\sqrt{x}\).
  2. Lower bound: \[ S > \int_{1}^{101} \frac{\mathrm{d}x}{\sqrt{x}} = 2\sqrt{101} - 2 \] which is \(18.0997\dots\)
  3. Upper bound: keep the first term separate. \[ S < 1 + \int_{1}^{100} \frac{\mathrm{d}x}{\sqrt{x}} \] \[ = 1 + (20 - 2) = 19 \]
  4. (For the record, \(S = 18.5896\dots\), comfortably inside.)
Answer: \(18.09 < S < 19\)
Your turn
Show that \(\displaystyle\sum_{r=1}^{n} \frac{1}{r^{2}} < 2\) for every positive integer \(n\).
Examiner trap

Never start the upper-bound integral at 0 for \(\dfrac{1}{x}\) or \(\dfrac{1}{x^{2}}\): \(\int_{0}^{n}\) does not exist. Keep the first term out of the sum and start the integral at 1, exactly as in the examples.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

\(f\) is positive and decreasing. A rectangle on \(r \le x \le r + 1\) has height \(f(r)\). Where does it lie?

Quick check+10 XP

Using unit-width rectangles with left-end heights, what is the upper sum for \(\int_{1}^{4} \frac{1}{x}\,\mathrm{d}x\)?

Quick check+10 XP

Find \(\displaystyle\lim_{n\to\infty} \frac{1}{n}\sum_{r=1}^{n} \left(\frac{r}{n}\right)^{3}\).

Quick check+10 XP

For \(y = x^{2}\) on \(0 \le x \le 2\) with \(n\) strips of equal width, what is the difference between the upper and lower sums?

Section 03

Let the width tend to zero.

On \(0 \le x \le 1\) take \(n\) strips of width \(\frac{1}{n}\), with heights \(f\!\left(\frac{r}{n}\right)\). The sum \(\frac{1}{n}\sum f\!\left(\frac{r}{n}\right)\) is an upper or lower sum, and for a monotonic \(f\) the two sums differ by \(\frac{1}{n}|f(1) - f(0)|\), which tends to 0. Both are squeezed onto the area.

\[ \lim_{n\to\infty} \frac{1}{n}\sum_{r=1}^{n} f\!\left(\frac{r}{n}\right) = \int_{0}^{1} f(x)\,\mathrm{d}x \]

To use it: take out a factor \(\frac{1}{n}\), write what is left using \(\frac{r}{n}\) only, call \(\frac{r}{n} = x\), and read the limits from the first and last values of \(\frac{r}{n}\).

Worked example A limit equal to ln 2
Find \(\displaystyle\lim_{n\to\infty} \sum_{r=1}^{n} \frac{1}{n + r}\).
  1. Take out \(\frac{1}{n}\): \(\dfrac{1}{n + r} = \dfrac{1}{n}\cdot\dfrac{1}{1 + \frac{r}{n}}\).
  2. So the sum is \(\dfrac{1}{n}\displaystyle\sum_{r=1}^{n} f\!\left(\frac{r}{n}\right)\) with \(f(x) = \dfrac{1}{1 + x}\), and \(\frac{r}{n}\) runs from \(\frac{1}{n}\) up to 1.
  3. The limit is \[ \int_{0}^{1} \frac{\mathrm{d}x}{1 + x} = \Big[\ln(1 + x)\Big]_{0}^{1} \] which is \(\ln 2\).
  4. Check with \(n = 1000\): the sum is \(0.69290\dots\), close to \(\ln 2 = 0.69315\dots\)
Answer: \(\ln 2\)
Worked example A limit equal to a quarter of pi
Find \(\displaystyle\lim_{n\to\infty} \sum_{r=1}^{n} \frac{n}{n^{2} + r^{2}}\).
  1. Divide top and bottom by \(n^{2}\): \[ \frac{n}{n^{2} + r^{2}} = \frac{1}{n}\cdot\frac{1}{1 + \left(\frac{r}{n}\right)^{2}} \]
  2. So \(f(x) = \dfrac{1}{1 + x^{2}}\) on \(0 \le x \le 1\).
  3. The limit is \[ \int_{0}^{1} \frac{\mathrm{d}x}{1 + x^{2}} = \Big[\tan^{-1} x\Big]_{0}^{1} \] which is \(\dfrac{\pi}{4}\).
Answer: \(\dfrac{\pi}{4}\)
Your turn
Find \(\displaystyle\lim_{n\to\infty} \frac{1}{n^{3}}\sum_{r=1}^{n} r^{2}\).
Examiner trap

After taking out \(\frac{1}{n}\), every remaining \(n\) must sit inside \(\frac{r}{n}\). A stray \(n\) means the factorising is wrong. Also read the limits from \(\frac{r}{n}\): if \(r\) runs from 1 to \(2n\), then \(x\) runs from 0 to 2, not 0 to 1.

Stage 2 · Exam

Exam-style questions

Unlocks once the four checks above are done. Worth more XP, written to AS & A Level Further Mathematics Paper 1 to 4 standard.

Finish the four checks above to unlock the exam questions
Exam style+20 XP

Find \(\displaystyle\lim_{n\to\infty} \frac{1}{n}\sum_{r=1}^{n} e^{r/n}\).

Exam style+20 XP

\(\displaystyle S = \sum_{r=1}^{50} \frac{1}{\sqrt{r}}\). Which bounds follow correctly from unit-width rectangles?

Exam style+20 XP

Find \(\displaystyle\lim_{n\to\infty} \sum_{r=1}^{n} \frac{n}{(n + r)^{2}}\).

Exam style+20 XP

\(f\) is positive and decreasing for \(x \ge 0\). Which inequality is always true for \(n \ge 1\)?

Skill unlocked

Sums trapped between integrals.

This skill is now recorded on your profile. Keep the chain going.

-Rank -Level -Score -Topics
Go deeper · practice
Six original Cambridge-style questions
Upper and lower sums, the sum of \(1/r^{2}\), bounds for a sum of square roots, two limits of sums with width \(1/n\), and how many harmonic terms pass 5.