Day 17: Riemann Sums & Numerical Approximation

The Definite Integral as a Limit of Riemann Sums

The definite integral $\int_a^b f(x)\,dx$ is defined as the limit of sums of thin rectangles. That limit is the exact signed area between the graph and the $x$ -axis — everything else in the unit is a shortcut for computing it.

Core Theorem

\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*)\,\Delta x, \qquad \Delta x = \frac{b - a}{n}

where

x_i^*

is any sample point in the

i

-th subinterval. Area above the

x

-axis counts as positive, area below as negative.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Write the limit of a right Riemann sum that equals

\displaystyle\int_1^5 \sqrt{x}\,dx

, using

n

equal subintervals.

How to think about it

Key idea: $\Delta x = \frac{5 - 1}{n} = \frac{4}{n}$ , and the right endpoint of the $i$ -th subinterval is $x_i = 1 + i\,\Delta x$ .

Full solution

1. Step 1: Width and Sample Points

\Delta x = \dfrac{4}{n}

and

x_i = 1 + \dfrac{4i}{n}

.

2. Step 2: Assemble the Sum

\displaystyle\int_1^5 \sqrt{x}\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} \sqrt{\,1 + \dfrac{4i}{n}\,}\cdot \dfrac{4}{n}

.

Step-by-Step SOP

1
Partition
Divide $[a, b]$ into $n$ subintervals of width $\Delta x = \frac{b-a}{n}$ .
2
Sample and Sum
Pick a sample point $x_i^*$ in each subinterval, then form $\sum f(x_i^*)\,\Delta x$ .
3
Take the Limit
$\int_a^b f\,dx = \lim_{n \to \infty}$ of that sum, when it exists.
4
Convert Both Directions
Be able to turn a given limit-of-a-sum into an integral (read off $a$ , $b$ , $f$ ) and vice versa.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Hard

Express

\displaystyle\lim_{n \to \infty} \sum_{i=1}^{n} \left(2 + \dfrac{3i}{n}\right)^2 \cdot \dfrac{3}{n}

as a definite integral.

Need a hint?

Read off

\Delta x = \frac{3}{n}

, so

b - a = 3

. The sample point

2 + \frac{3i}{n}

starts at

2

, so

a = 2

and

b = 5

; the function is

(\ )^2

.

Show solution

1. Step 1: Identify the Interval

\Delta x = \dfrac{3}{n}

gives

b - a = 3

. The right endpoints run from just above

2

to

2 + 3 = 5

, so

[a, b] = [2, 5]

.

2. Step 2: Identify the Function

The factor being squared,

2 + \dfrac{3i}{n}

, is the sample point

x

. So

f(x) = x^2

.

3. Step 3: Write the Integral

\displaystyle\int_2^5 x^2\,dx

.

Practice 02Easy

Evaluate

\displaystyle\int_{-2}^{2} x\,dx

using the signed-area interpretation.

Need a hint?

Sketch

y = x

from

-2

to

2

. The area below the axis on

[-2, 0]

cancels the area above it on

[0, 2]

.

Show solution

1. Step 1: Split into Signed Pieces

On

[-2, 0]

the region is a triangle below the axis: signed area

-2

. On

[0, 2]

it is a triangle above the axis: signed area

+2

.

2. Step 2: Add

\displaystyle\int_{-2}^{2} x\,dx = -2 + 2 = 0

.

Common Pitfalls

⚠
Reading a Sum's Interval Off WrongFrom $\sum f(a + i\Delta x)\Delta x$ with $\Delta x = \frac{k}{n}$ , the interval has length $k$ and starts at $a$ — so $b = a + k$ , not $b = k$ .
⚠
Forgetting Signed Area $\int_a^b f\,dx$ is not total area unless $f \ge 0$ on $[a, b]$ . Where $f < 0$ , the integral subtracts.

Left, Right, Midpoint, and Trapezoidal Sums

With finitely many subintervals you get an approximation, and the four standard choices — left endpoint, right endpoint, midpoint, and trapezoidal — each have a predictable bias. For a monotonic function you can say for certain whether an estimate is too big or too small.

Core Theorem

For

n

subintervals of width

\Delta x

on

[a, b]

:
Left sum

L_n = \Delta x\big[f(x_0) + f(x_1) + \cdots + f(x_{n-1})\big]

Right sum

R_n = \Delta x\big[f(x_1) + f(x_2) + \cdots + f(x_n)\big]

Midpoint sum

M_n = \Delta x\big[f(\bar{x}_1) + \cdots + f(\bar{x}_n)\big]

, where

\bar{x}_i

is the midpoint of subinterval

i

Trapezoidal sum

T_n = \dfrac{\Delta x}{2}\big[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)\big] = \dfrac{L_n + R_n}{2}

Bias (for equal widths): if

f

is increasing,

L_n

underestimates and

R_n

overestimates (reversed if decreasing). If

f

is concave up,

T_n

overestimates and

M_n

underestimates (reversed if concave down).

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Estimate

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

with

n = 4

using (a) a left sum and (b) a right sum. State whether each is an over- or underestimate.

How to think about it

Key idea: $\Delta x = 0.5$ . Left uses $f(0), f(0.5), f(1), f(1.5)$ ; right uses $f(0.5), f(1), f(1.5), f(2)$ . $x^2$ is increasing on $[0, 2]$ .

Full solution

1. Step 1: Left Sum

L_4 = 0.5\,[0 + 0.25 + 1 + 2.25] = 0.5(3.5) = 1.75

.

2. Step 2: Right Sum

R_4 = 0.5\,[0.25 + 1 + 2.25 + 4] = 0.5(7.5) = 3.75

.

3. Step 3: Bias

x^2

is increasing on

[0, 2]

, so

L_4 = 1.75

underestimates and

R_4 = 3.75

overestimates. (Exact value:

\frac{8}{3} \approx 2.667

.)

Step-by-Step SOP

1
Find Δx and the Grid Points
$\Delta x = \frac{b - a}{n}$ ; the endpoints are $a, a + \Delta x, a + 2\Delta x, \dots, b$ .
2
Pick the Heights
Left = left endpoints; right = right endpoints; midpoint = subinterval centers; trapezoid = ends counted once, interior points doubled, all over 2.
3
Judge the Bias
Monotonic ⇒ left/right bias from the sign of $f'$ . Concavity ⇒ trapezoid/midpoint bias from the sign of $f''$ .

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

For

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

with

n = 4

, compute the trapezoidal estimate

T_4

and state whether it is an over- or underestimate.

Need a hint?

T_4 = \frac{L_4 + R_4}{2}

, or use the

\frac{\Delta x}{2}[f_0 + 2f_1 + 2f_2 + 2f_3 + f_4]

formula.

x^2

is concave up.

Show solution

1. Step 1: Trapezoidal Formula

T_4 = \dfrac{0.5}{2}\big[f(0) + 2f(0.5) + 2f(1) + 2f(1.5) + f(2)\big] = 0.25\,[0 + 0.5 + 2 + 4.5 + 4] = 0.25(11) = 2.75

.

2. Step 2: Bias

x^2

is concave up on

[0, 2]

, so the trapezoids lie above the curve and

T_4 = 2.75

overestimates. (It also equals

\frac{L_4 + R_4}{2} = \frac{1.75 + 3.75}{2}

.)

Practice 02Hard

g

is decreasing and concave down on

[1, 5]

. A student computes a midpoint sum

M_2

and a trapezoidal sum

T_2

for

\int_1^5 g(x)\,dx

. Which is larger, and which (if either) overestimates the integral?

Need a hint?

Concavity, not monotonicity, controls midpoint vs. trapezoid. Concave down flips the concave-up rules.

Show solution

1. Step 1: Apply the Concave-Down Rule

For a concave-down function, the trapezoid rule underestimates and the midpoint rule overestimates.

2. Step 2: Conclude

M_2 > \int_1^5 g\,dx > T_2

. So

M_2

is larger,

M_2

overestimates, and

T_2

underestimates. (The decreasing behavior is irrelevant to this comparison.)

Common Pitfalls

⚠
Using Monotonicity to Judge the Trapezoid RuleLeft/right bias comes from whether $f$ is increasing or decreasing. Trapezoid/midpoint bias comes from concavity — a different property.
⚠
Doubling the Endpoints in the Trapezoid FormulaIn $T_n$ , the two outer values $f(x_0)$ and $f(x_n)$ are counted once; only the interior values are doubled.

Approximating an Integral from a Table of Values

AP free-response questions routinely give you a function only as a table — a few $(t, f(t))$ pairs, often at unequal spacings — and ask for a Riemann or trapezoidal estimate of an integral, plus its meaning with units. The subinterval widths come straight from the table.

Core Theorem

With table points

t_0 < t_1 < \cdots < t_n

(widths

\Delta t_i = t_i - t_{i-1}

, possibly unequal):
Left sum

= \sum \Delta t_i \, f(t_{i-1})

; Right sum

= \sum \Delta t_i \, f(t_i)

.
Trapezoid on each subinterval:

\dfrac{\Delta t_i}{2}\big[f(t_{i-1}) + f(t_i)\big]

, then add.

\dfrac{1}{b - a}\displaystyle\int_a^b f(t)\,dt

is the average value of

f

over

[a, b]

, in the units of

f

.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Water flows into a tank at rate

r(t)

liters/hour. The table gives

r

:

t = 0, 2, 5, 6

hours with

r(t) = 3, 7, 10, 8

liters/hour. Use a left Riemann sum with the three subintervals in the table to estimate

\displaystyle\int_0^6 r(t)\,dt

, and say what it represents with units.

How to think about it

Key idea: The widths are $2 - 0 = 2$ , $5 - 2 = 3$ , and $6 - 5 = 1$ . A left sum uses the value at the left end of each subinterval.

Full solution

1. Step 1: Widths and Left Values

Subintervals

[0,2], [2,5], [5,6]

with widths

2, 3, 1

. Left values:

r(0) = 3

,

r(2) = 7

,

r(5) = 10

.

2. Step 2: Left Sum

2(3) + 3(7) + 1(10) = 6 + 21 + 10 = 37

.

3. Step 3: Interpret

\displaystyle\int_0^6 r(t)\,dt \approx 37

liters — the approximate total amount of water that flowed into the tank during the first 6 hours.

Step-by-Step SOP

1
Read the Widths from the Table
$\Delta t_i = t_i - t_{i-1}$ for each consecutive pair — do not assume they are equal.
2
Choose the Method's Heights
Left = left value; right = right value; trapezoid = average of the two end values.
3
Multiply by Width and Add
One term per subinterval, then sum.
4
Interpret with Units
$\int r(t)\,dt$ is a total amount (units of $r$ × units of $t$ ); $\frac{1}{b-a}\int$ is an average value (units of $f$ ).

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Using the same table (

t = 0, 2, 5, 6

;

r = 3, 7, 10, 8

), estimate

\displaystyle\int_0^6 r(t)\,dt

with a trapezoidal sum.

Need a hint?

On each subinterval, average the two endpoint values and multiply by that subinterval's width.

Show solution

1. Step 1: Trapezoid on Each Subinterval

[0,2]

:

\dfrac{2}{2}(3 + 7) = 10

.

[2,5]

:

\dfrac{3}{2}(7 + 10) = 25.5

.

[5,6]

:

\dfrac{1}{2}(10 + 8) = 9

.

2. Step 2: Add

10 + 25.5 + 9 = 44.5

liters.

Practice 02Hard

The table gives the temperature

H(t)

in °C of a pool:

t = 0, 3, 6, 9

hours with

H = 18, 21, 24, 23

. Using a trapezoidal sum over

[0, 9]

, estimate the average temperature of the pool over those 9 hours.

Need a hint?

Average value

= \frac{1}{9}\int_0^9 H(t)\,dt

. Estimate the integral with trapezoids first (equal width 3), then divide by 9.

Show solution

1. Step 1: Trapezoidal Estimate of the Integral

Equal width

3

:

\dfrac{3}{2}\big[H(0) + 2H(3) + 2H(6) + H(9)\big] = \dfrac{3}{2}\big[18 + 42 + 48 + 23\big] = \dfrac{3}{2}(131) = 196.5

.

2. Step 2: Divide by the Interval Length

Average temperature

\approx \dfrac{196.5}{9} \approx 21.83\ ^\circ\mathrm{C}

.

Common Pitfalls

⚠
Assuming Equal Subinterval WidthsTable $t$ -values are often unevenly spaced. Multiply each term by its own $\Delta t$ , not by a single shared width.
⚠
Answering with No Units or Context" $\approx 37$ " loses points. Write " $\approx 37$ liters, the total water in over the first 6 hours."

Worked Examples

1. Step 1: Width and Sample Points

2. Step 2: Assemble the Sum

Partition

Sample and Sum

Take the Limit

Convert Both Directions

Practice — Your Turn

1. Step 1: Identify the Interval

2. Step 2: Identify the Function

3. Step 3: Write the Integral

1. Step 1: Split into Signed Pieces

2. Step 2: Add

Worked Examples

1. Step 1: Left Sum

2. Step 2: Right Sum

3. Step 3: Bias

Find Δx and the Grid Points

Pick the Heights

Judge the Bias

Practice — Your Turn

1. Step 1: Trapezoidal Formula

2. Step 2: Bias

1. Step 1: Apply the Concave-Down Rule

2. Step 2: Conclude

Worked Examples

1. Step 1: Widths and Left Values

2. Step 2: Left Sum

3. Step 3: Interpret

Read the Widths from the Table

Choose the Method's Heights

Multiply by Width and Add

Interpret with Units

Practice — Your Turn

1. Step 1: Trapezoid on Each Subinterval

2. Step 2: Add

1. Step 1: Trapezoidal Estimate of the Integral

2. Step 2: Divide by the Interval Length

Gary Chang