Riemann sums and the definite integral: left, right and midpoint rectangle estimates, total change from a rate, the definite integral as the limit of Riemann sums, signed area, and the link to the Fundamental Theorem of Calculus (the sum notation, total change and limit definition are made explicit in the 2024 Mathematics Advanced syllabus)

Syllabus dot point

“Interpret Riemann sums as estimates of total change, define the definite integral as the limit of Riemann sums, and recognise that the integral is negative where the function is negative”

HSCMaths AdvancedYear 12: Calculus13 min readUpdated 7 Oct 2026

Quick answer

A Riemann sum adds the areas of rectangles under a graph, $\sum f(x_i^*)\,\Delta x$ , and the definite integral is the limit of these sums as the strips become infinitely thin. The integral measures signed area (negative where $f < 0$ ) and, for a rate, the total change. For an increasing function the left sum underestimates and the right sum overestimates. The Fundamental Theorem then computes the limit quickly. The 2017 course already approximated areas with rectangles and trapezia; the 2024 Mathematics Advanced syllabus makes the sum notation, total change and the limit definition explicit.

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What this dot point is asking
The answer
Exam-style questions
Practice questions

What this dot point is asking

In the Mathematics Advanced 11-12 Syllabus (2024), the Integral calculus focus area opens with an explicit account of where the definite integral comes from. NESA expects you to interpret Riemann sums as estimates of the total change in a quantity, to define the definite integral as the limit of Riemann sums, to recognise that the integral is negative when the function is negative, and to approximate areas using rectangles and trapezoids before you use the Fundamental Theorem of Calculus.

The 2017 course already approximated areas with rectangles and trapezia, explored using more shapes, and treated the definite integral as a sum of signed areas. The 2024 course makes the sum notation, total change and the limit definition explicit, so that the rules have a meaning.

Note

Imagine finding the area of a lake on a map by laying a grid of thin vertical strips over it and adding up the strips. With a few wide strips the answer is rough. With thousands of very thin strips it is almost perfect. The definite integral is what the total approaches as the strips become infinitely thin. And since distance is speed times time, adding up thin strips of speed-times-time gives the total distance, which is why an integral can measure more than an area.

The answer

Key fact

A Riemann sum for $f$ on $[a, b]$ with $n$ strips of width $\Delta x = \frac{b - a}{n}$ is $\sum f(x_i^*)\,\Delta x$ , where $x_i^*$ is a point in strip $i$ (left end, right end or midpoint).
The definite integral is the limit of these sums as $n \to \infty$ :
$\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*)\,\Delta x.$
If $f(x) < 0$ the rectangles count negatively, so the integral is negative there (signed area).
If $f$ is a rate of change, $\int_a^b f(x)\,dx$ is the total change in the quantity from $a$ to $b$ .

Left, right and midpoint sums

With equal strips of width $\Delta x$ :

Estimate	Height used	For an increasing function
Left sum $L_n$	$f$ at the left end of each strip	Underestimate
Right sum $R_n$	$f$ at the right end of each strip	Overestimate
Midpoint sum $M_n$	$f$ at the centre of each strip	Usually closest
Trapezoidal	average of $L_n$ and $R_n$	Overestimate if concave up

For an increasing function, $L_n \leq \int \leq R_n$ . For a decreasing function the order reverses. As $n$ grows both sums squeeze towards the same limit.

From a sum to an integral

For a right sum the width is $\Delta x = \frac{b - a}{n}$ and $x_i = a + i\Delta x$ . To evaluate a limit by hand:

Write $\Delta x$ and $x_i$ in terms of $n$ and $i$ .
Write the sum $\sum f(x_i)\Delta x$ and pull out any factors that do not depend on $i$ .
Use a standard sum, such as $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ or $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$ (a question will give these if it requires them).
Simplify, then let $n \to \infty$ , so every term with $n$ in the denominator tends to $0$ .

In exams the full limit calculation is rare, but understanding the structure makes the notation $\int f(x)\,dx$ meaningful: the integral sign is a stretched "S" for sum, and $dx$ is the strip width as it shrinks to zero.

Signed area and total change

Where the graph is above the axis, the integral is positive. Where it is below, the integral is negative. So $\int_a^b f(x)\,dx$ measures signed area. For the area enclosed by the curve and the axis, split at each $x$ -intercept and add the sizes. For a velocity graph, $\int_a^b v\,dt$ gives displacement (signed), while $\int_a^b |v|\,dt$ gives distance travelled.

Connection to the Fundamental Theorem

The theorem says $\int_a^b f(x)\,dx = F(b) - F(a)$ for any primitive $F$ of $f$ . It is the shortcut that replaces the limit of a sum. The Riemann sum is why the integral measures accumulated change; the theorem is how you compute it quickly. Numerical rules, such as the trapezoidal rule, return to the sum idea when no primitive is available or the data is a table.

Worked examples

Estimate with a table and decide the direction of error

The values of $f$ at $x = 0, 1, 2, 3$ are $5, 4, 2, 1$ . Estimate $\int_0^3 f(x)\,dx$ with left and right sums and say which is larger.

Sums

The width is $1$ . Left: $5 + 4 + 2 = 11$ . Right: $4 + 2 + 1 = 7$ .

Compare

The function values decrease, so the left sum is an overestimate ( $11$ ) and the right sum an underestimate ( $7$ ). The integral lies between $7$ and $11$ , and the average $9$ is the trapezoidal estimate.

Signed area from geometry

Evaluate $\int_{-2}^{2} x\,dx$ without a primitive.

Reason with the graph

The line $y = x$ is below the axis on $-2 \leq x < 0$ (a triangle of area $2$ , counted as $-2$ ) and above it on $0 < x \leq 2$ (a triangle of area $2$ ). The integral is $-2 + 2 = 0$ .

Marker's note: state that the first area is counted negatively; the cancellation is the point.

Common traps

Getting the direction of the error wrong: Left sums underestimate for increasing functions and overestimate for decreasing ones. Sketch the rectangles if unsure.
Treating the integral as area: Where $f < 0$ the integral is negative. For the enclosed area, split at the intercepts and add sizes.
Wrong strip width: $\Delta x = \frac{b - a}{n}$ . A question with $4$ strips on $[0, 4]$ has width $1$ , but on $[1, 5]$ with $2$ strips it is $2$ .
Using the wrong endpoints: A left sum omits the last $x$ -value and a right sum omits the first.
Forgetting units: Total change has the units of rate multiplied by the units of time.

Exam technique

In a table question, write the width, then list which function values go in each sum before you add. Check the direction of the error by asking whether the function is increasing or decreasing on the interval. For a "limit of Riemann sums" question, set out $\Delta x$ , $x_i$ , the sum, the simplified expression and the limit as five separate lines so that each earns its mark.

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

HSC-style3 marks

A car's speed in metres per second is measured every

5

seconds:

0, 8, 14, 18, 20

t = 0, 5, 10, 15, 20

. Use the trapezoidal rule to estimate the distance travelled in

20

seconds.

Show worked answer →

With $h = 5$ : $\frac{5}{2}\left(0 + 2(8 + 14 + 18) + 20\right) = \frac{5}{2}(0 + 80 + 20) = \frac{5}{2} \times 100 = 250$ metres.

Markers give one mark for the trapezoidal structure (ends once, middle values twice), one for substituting correctly and one for the answer with units. Distance is the area under a speed graph.

HSC-style3 marks

Without evaluating, explain why

\displaystyle\int_0^{2\pi} \sin x\,dx = 0

even though the area enclosed between

y = \sin x

and the

x

-axis is not

0

Show worked answer →

On $0 \leq x \leq \pi$ the curve is above the $x$ -axis, so its integral is positive; on $\pi \leq x \leq 2\pi$ it is below, so its integral is negative. By symmetry the two halves have the same size, $2$ , so $\int_0^{\pi} \sin x\,dx = 2$ and $\int_{\pi}^{2\pi} \sin x\,dx = -2$ and the total is $0$ . The enclosed area is $2 + 2 = 4$ .

Markers award one mark for identifying the sign on each half, one for the equal sizes and one for linking signed integral to area.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marks

Use four rectangles of equal width to estimate

\displaystyle\int_0^4 x^2 \, dx

with (a) left endpoints and (b) right endpoints. Compare with the exact value.

Show worked solution →

Set up. The width is $\Delta x = \frac{4 - 0}{4} = 1$ and the strip boundaries are $0, 1, 2, 3, 4$ .

(a) Left endpoints use $x = 0, 1, 2, 3$ : $L_4 = 1 \times (0 + 1 + 4 + 9) = 14$ .

(b) Right endpoints use $x = 1, 2, 3, 4$ : $R_4 = 1 \times (1 + 4 + 9 + 16) = 30$ .

Exact value. $\int_0^4 x^2\,dx = \left[\frac{x^3}{3}\right]_0^4 = \frac{64}{3} \approx 21.33$ . Because $x^2$ is increasing on $[0, 4]$ , the left sum $14$ is an underestimate and the right sum $30$ is an overestimate.

Marker's note: one mark for each sum and one for the comparison.

core3 marks

For

\displaystyle\int_0^4 x^2 \, dx

, use four rectangles with midpoint heights.

Show worked solution →

Midpoints of the four strips are $0.5, 1.5, 2.5, 3.5$ with width $1$ .

M_4 = 1 \times \left(0.25 + 2.25 + 6.25 + 12.25\right) = 21.

The exact value is $\frac{64}{3} \approx 21.33$ , so the midpoint estimate of $21$ is much closer than the left ( $14$ ) or right ( $30$ ) sums.

Marker's note: one mark for the strip width and midpoints, one for the sum of the function values, one for multiplying by the width to give the estimate. Midpoint estimates usually beat endpoint estimates at the same number of strips.

core4 marks

The rate of flow of water into a tank, in litres per minute, is recorded at two-minute intervals: |

t

(min) | 0 | 2 | 4 | 6 | 8 | | --- | --- | --- | --- | --- | --- | |

r(t)

(L/min) | 3 | 5 | 6 | 4 | 2 | Estimate the total volume entering over the first

8

minutes using (a) left endpoints and (b) right endpoints, and (c) the average of the two.

Show worked solution →

The area under a rate graph is total change: Each strip has width $2$ minutes.
(a) Left: $2 \times (3 + 5 + 6 + 4) = 2 \times 18 = 36$ litres.
(b) Right: $2 \times (5 + 6 + 4 + 2) = 2 \times 17 = 34$ litres.
(c) Average: $\frac{36 + 34}{2} = 35$ litres. This is the trapezoidal rule estimate, since $2\left(\frac{3}{2} + 5 + 6 + 4 + \frac{2}{2}\right) = 2 \times 17.5 = 35$ .

Marker's note: one mark for each of (a) and (b), one for (c) and one for units and the interpretation as total volume.

exam4 marks

Using the limit of right-endpoint Riemann sums, show that

\displaystyle\int_0^2 x \, dx = 2

. (You may use

\sum_{i=1}^{n} i = \frac{n(n+1)}{2}

Show worked solution →

Divide $[0, 2]$ into $n$ strips of width $\Delta x = \frac{2}{n}$ . The right endpoint of strip $i$ is $x_i = \frac{2i}{n}$ , so the height is $f(x_i) = \frac{2i}{n}$ .

Form the sum.

R_n = \sum_{i=1}^{n} \frac{2i}{n} \cdot \frac{2}{n} = \frac{4}{n^2} \sum_{i=1}^{n} i = \frac{4}{n^2} \cdot \frac{n(n+1)}{2} = \frac{2(n+1)}{n} = 2 + \frac{2}{n}.

Take the limit. As $n \to \infty$ , $\frac{2}{n} \to 0$ , so $R_n \to 2$ . Hence $\int_0^2 x\,dx = 2$ , which matches the area of the triangle with base $2$ and height $2$ .

Marker's note: one mark for the width and endpoint, one for the sum, one for simplifying and one for the limit.

exam5 marks

Using the limit of right-endpoint Riemann sums, evaluate

\displaystyle\int_0^3 x^2 \, dx

. (You may use

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

Show worked solution →

Strips. $\Delta x = \frac{3}{n}$ and $x_i = \frac{3i}{n}$ , so $f(x_i) = \frac{9i^2}{n^2}$ .

Sum.

R_n = \sum_{i=1}^{n} \frac{9i^2}{n^2} \cdot \frac{3}{n} = \frac{27}{n^3} \sum_{i=1}^{n} i^2 = \frac{27}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{9(n+1)(2n+1)}{2n^2}.

Expand and take the limit.

R_n = \frac{9\left(2n^2 + 3n + 1\right)}{2n^2} = 9 + \frac{27}{2n} + \frac{9}{2n^2} \;\to\; 9 \quad \text{as } n \to \infty.

So $\int_0^3 x^2\,dx = 9$ , in agreement with $\left[\frac{x^3}{3}\right]_0^3 = 9$ .

Marker's note: one mark each for the width and heights, the sum, the use of the formula, the simplified expression and the limit.

exam4 marks

(a) Evaluate

\displaystyle\int_0^5 (x - 2)\, dx

and explain, using a sketch, why the answer is smaller than the total area enclosed between the line and the

x

-axis. (b) Find that total area.

Show worked solution →

(a) Evaluate: $\int_0^5 (x - 2)\,dx = \left[\frac{x^2}{2} - 2x\right]_0^5 = \frac{25}{2} - 10 = \frac{5}{2}$ .
Interpret: The line $y = x - 2$ crosses the $x$ -axis at $x = 2$ . On $0 \leq x \leq 2$ it is below the axis, forming a triangle of area $\frac{1}{2} \times 2 \times 2 = 2$ which counts as negative, $-2$ . On $2 \leq x \leq 5$ it is above the axis, forming a triangle of area $\frac{1}{2} \times 3 \times 3 = \frac{9}{2}$ , which counts as positive. So the integral is $-2 + \frac{9}{2} = \frac{5}{2}$ .
(b) Total area: Add the sizes: $2 + \frac{9}{2} = \frac{13}{2}$ .

Marker's note: one mark for the integral, one for the sign explanation, one for the two triangle areas and one for the total.

exam4 marks

For

f(x) = \sqrt{x}

[0, 4]

, use four strips of equal width to find a left sum

L_4

and a right sum

R_4

and show that

L_4 < \displaystyle\int_0^4 \sqrt{x}\,dx < R_4

. Give each to 3 decimal places.

Show worked solution →

Strips have width $1$ with boundaries $0, 1, 2, 3, 4$ .

Left sum: $L_4 = \sqrt{0} + \sqrt{1} + \sqrt{2} + \sqrt{3} = 1 + \sqrt{2} + \sqrt{3} \approx 4.146$ .
Right sum: $R_4 = \sqrt{1} + \sqrt{2} + \sqrt{3} + \sqrt{4} = 3 + \sqrt{2} + \sqrt{3} \approx 6.146$ .
Exact value: $\int_0^4 \sqrt{x}\,dx = \left[\frac{2}{3}x^{3/2}\right]_0^4 = \frac{2}{3} \times 8 = \frac{16}{3} \approx 5.333$ .

Since $\sqrt{x}$ is increasing, each left rectangle lies under the curve and each right rectangle lies over it, and indeed $4.146 < 5.333 < 6.146$ . (The two sums differ by $f(4) - f(0) = 2$ exactly, which is $\Delta x \times (f(b) - f(a))$ .)

Marker's note: one mark for each sum, one for the exact value and one for the reason the inequality holds.

Practise this

Sources & how we know this

Mathematics Advanced 11-12 Syllabus (2024): Year 12 Integral calculus — NESA (2024)
Mathematics Advanced 11-12 Syllabus (2024): course overview and implementation — NESA (2024)
Mathematics Advanced Stage 6 Syllabus (2017) — NESA
Mathematics Advanced HSC exam papers — NESA