Use definite integrals and numerical methods to find accumulated change and area

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Calculus

IB Mathematics: Analysis and Approaches SL — Study topic SL 5.9

A definite integral can describe how much a quantity changes over an interval, or the area between a graph and the horizontal axis. Its meaning depends on the context: a negative rate reduces net change, while geometric area is counted as positive. This lesson connects formulas, graphs, measured data, and contexts, then uses exact and numerical methods to find accumulation.

What you will learn

Interpret a definite integral as signed accumulated change over an interval.
Evaluate definite integrals using an antiderivative when one is available.
Distinguish signed area from total geometric area.
Estimate accumulation from tabulated data using the trapezoidal rule.
Use graphing technology to check an integral or numerical estimate while showing the mathematical reasoning.

1. From rate to accumulated change

A rate tells us how quickly a quantity changes. For example, if

v(t)

is velocity in metres per second, adding its contributions over a time interval gives a change in position in metres. More generally, if

r(x)

is a rate of change with respect to

x

, its accumulated change from

x=a

to

x=b

is represented by the definite integral

\int_a^b r(x)\,dx

. Here

a

and

b

are the interval endpoints.

A definite integral can be evaluated using an antiderivative. If

F'(x)=f(x)

, then

\int_a^b f(x)\,dx=F(b)-F(a)

. In words, find an antiderivative, evaluate it at the upper endpoint, and subtract its value at the lower endpoint. The result is signed: parts of the graph below the horizontal axis contribute negatively.

Units help interpret the answer. If a rate is measured in litres per minute and the input is measured in minutes, the integral is measured in litres. In general, multiply the units of the rate by the units of the input interval.

\int_a^b f(x)\,dx=F(b)-F(a),\qquad F'(x)=f(x)

A definite integral represents net accumulated change when its integrand is a rate.
Subtract the antiderivative value at the lower endpoint from its value at the upper endpoint.
The integral’s units combine the units of the integrand and the input.

2. Reading area from a graph

When

f(x)

is above the horizontal axis, its definite integral over an interval is the area between the graph and the axis. When it is below the axis, that part of the integral is negative. Thus, the integral gives signed area, not necessarily the total geometric area.

To find total area, split the interval wherever the graph crosses the horizontal axis. Find the area of each part as a positive quantity, then add the results. For area between two graphs, identify which graph is higher on each part; the vertical distance is the upper function minus the lower function.

A sketch or graph helps identify crossings and which curve is higher. Algebra can locate exact crossing points when the functions are simple. State the interval and give area in square units.

\text{total area}=\int_a^b |f(x)|\,dx

Area above the axis contributes positively to an integral; area below contributes negatively.
Total geometric area is found by adding positive pieces.
Check where graphs cross before calculating area.

3. Estimating accumulation from data

Sometimes a rate is known only at measured points. A graphing calculator can estimate an integral of a given function, but the trapezoidal rule gives a clear method for data. Join consecutive data points with straight line segments. Each neighbouring pair forms a trapezium: its area is the interval width multiplied by the average of the two endpoint heights.

For equally spaced values, let

h

be the common width. The first and last measurements each count half in the weighted sum, while interior measurements count fully. For unequal widths, calculate each trapezium separately using its own width. The result is an estimate because the actual graph between measurements may curve.

Keep the measurements in order and include units. More points can represent the graph more closely, but a numerical estimate is not automatically exact. If technology is used for arithmetic, show the widths and endpoint averages so the setup is clear.

\int_a^b f(x)\,dx\approx \frac{h}{2}\left[y_0+2y_1+\cdots+2y_{n-1}+y_n\right]

The trapezoidal rule approximates the graph by straight segments between measured points.
Keep intermediate values unrounded until the final answer.
Call a result an estimate when it comes from sampled data, and give sensible accuracy.

4. Technology and a reliable checking routine

For a formula, enter the function and interval endpoints into a graphing calculator’s definite-integral tool. Inspect the graph first to check for crossings or negative sections. When an antiderivative is available, compare the calculator output with the exact calculation. A mismatch may indicate an incorrect endpoint, a missed sign, or a wrong interval.

For tabulated rates, organize the input values and rates in a table, then calculate each trapezium. A calculator can help with arithmetic, but the mathematical setup should still show the interval widths and endpoint averages.

Before calculating, decide whether the question asks for net change, total area, or an estimate. In an exam-style response, include a short interpretation, units, and any requested rounding.

Use a graph to identify sign changes and confirm the interval.
Technology checks arithmetic; it does not replace explaining the method.
Match the result to the question’s meaning and requested accuracy.

Interpreting an integral

Situation	Meaning of the integral	Result
A rate is given by a formula	Net accumulated change on the interval	Signed; may be positive, negative, or zero
A graph is compared with the horizontal axis	Signed area between graph and axis	Negative portions subtract
Total geometric area is requested	Add the magnitudes of separate regions	Non-negative area
Only measured rates are given	Approximate accumulation from data	An estimate, often using trapezia

Worked example

Exact accumulated change from a rate

A quantity changes at rate

r(t)=3t^2-2t

units per second for

0\leq t\leq 2

. Find its net change over this interval.

Set up the accumulation
Net change is the definite integral of the rate over the stated time interval. The result is in units of the quantity because the rate is in units per second and time is in seconds.
$\int_0^2 (3t^2-2t)\,dt$
Find an antiderivative
An antiderivative of $3t^2-2t$ is $t^3-t^2$ : differentiating this expression gives the original rate.
$\int (3t^2-2t)\,dt=t^3-t^2$
Evaluate at the endpoints
Subtract the antiderivative value at $t=0$ from its value at $t=2$ .
$[t^3-t^2]_0^2=(8-4)-(0-0)=4$

Answer: The net change is

4

units.

Check: The rate changes sign: it is negative for

0<t<2/3

and positive for

2/3<t\leq2

. The negative contribution is outweighed by the positive contribution, consistent with the net change of

4

.

Worked example

Total area when the graph crosses the axis

Find the total area between

y=x^2-1

and the horizontal axis for

-2\leq x\leq 2

.

Locate the crossings
Set the function equal to zero. The roots divide the interval into sections where the graph is above or below the axis.
$x^2-1=0\quad\Rightarrow\quad x=-1,\ 1$
Identify the signs
The graph is above the axis on $[-2,-1]$ and $[1,2]$ , and below it on $[-1,1]$ . By symmetry, the two outer areas are equal. Calculate the middle area using $1-x^2$ so it is positive.
$A=2\int_1^2(x^2-1)\,dx+\int_{-1}^1(1-x^2)\,dx$
Evaluate and add positive areas
An antiderivative of $x^2-1$ is $x^3/3-x$ , and one of $1-x^2$ is $x-x^3/3$ . The outer area on one side is $4/3$ , so the two outer areas total $8/3$ . The middle area is $4/3$ . Add these positive areas.
$A=2\left[\frac{x^3}{3}-x\right]_1^2+\left[x-\frac{x^3}{3}\right]_{-1}^1=4$

Answer: The total area is

4

square units.

Check: The signed integral is

8/3-4/3=4/3

: the outer regions contribute positively and the middle region negatively. It is not the total area; replacing the negative middle contribution with its positive area gives

8/3+4/3=4

.

Worked example

Estimate accumulation from measured rates

A rate is measured in litres per minute at times

0

,

2

,

4

, and

6

minutes. The respective rates are

5

,

7

,

6

, and

2

litres per minute. Estimate the accumulated amount from

0

to

6

minutes using the trapezoidal rule.

Determine the interval width
The measurements are equally spaced, so each interval has width $2$ minutes. Each trapezium’s area is its width multiplied by the average of its endpoint rates.
$h=2$
Apply the trapezoidal rule
The first and last rates receive half weight, and the two interior rates receive full weight. Multiplication by the time width gives an amount in litres.
$\frac{2}{2}[5+2(7)+2(6)+2]=33$
Interpret the estimate
The estimate assumes that the rate changes linearly between consecutive measurements. The units are litres per minute multiplied by minutes.
$33\text{ litres}$

Answer: The estimated accumulated amount is

33

litres.

Check: The three trapezium areas are

12

,

13

, and

8

litres. Their sum is

33

litres.

Common mistakes and how to avoid them

Treating every definite integral as total geometric area.

Correction: A definite integral is signed. Split at axis crossings and add positive areas if total geometric area is required.

Subtracting endpoint values in the wrong order.

Correction: Use the antiderivative’s upper-end value minus its lower-end value.

Using the trapezoidal rule without multiplying by interval width.

Correction: Each trapezium’s area includes its width; include that factor to account for the input interval and units.

Rounding each trapezium too early or giving a numerical estimate as exact.

Correction: Retain precision during the calculation and label a result from sampled data as an estimate.

Lesson summary

A definite integral of a rate gives net accumulated change over an interval.
Evaluate an integral by subtracting the antiderivative’s lower-end value from its upper-end value.
Signed area can involve cancellation; total area requires adding positive pieces.
The trapezoidal rule estimates accumulation by joining neighbouring data points with straight segments.
Use graphs and technology to check signs and arithmetic, while showing the setup, units, and interpretation.

Check your understanding

Question 1

If

\int_1^4 r(t)\,dt=-3

, what does this tell you about net change?

The net change is $-3$ units.
The total geometric area must be $-3$ square units.
The rate is negative at every point.
The final quantity must be $3$ units.

Show answer and explanation

The net change is

-3

units.

The definite integral of a rate gives signed net change. It does not show whether the rate is negative at every point, or give the final quantity without an initial value.

Question 2

A rate has values

4

and

8

at the endpoints of a

3

-minute interval. What is the trapezoidal estimate of accumulation on that interval?

$18$ units
$12$ units
$36$ units
$6$ units

Show answer and explanation

18

units

Multiply the interval width by the average endpoint rate:

3(4+8)/2=18

units.

Question 3

The graph of a function lies below the horizontal axis throughout

[a,b]

. What is true about its definite integral on that interval?

It is negative.
It is positive because area is always positive.
It is zero.
Its sign cannot be determined from the graph.

Show answer and explanation

It is negative.

The graph lies below the axis throughout, so its signed integral is negative. The geometric area would instead be reported as positive.

Key terms

Accumulated change: The net change in a quantity over an interval, found by adding the contributions of its rate.
Definite integral: A signed total over a specified interval, written with lower and upper endpoints.
Antiderivative: A function whose derivative is the given function.
Signed area: Area above the horizontal axis counted positively and area below it counted negatively.
Trapezoidal rule: A numerical method that estimates an integral by approximating sections of a graph with trapezia.

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