What is the basic concept of integration in Pure Mathematics 1?

In Pure Mathematics 1, integration is introduced as the reverse process of differentiation. It is used to find the original function given its derivative. This process is essential for calculating areas under curves and solving problems involving accumulation.

How do you perform indefinite integration in Edexcel International A Levels?

Indefinite integration involves finding the antiderivative of a function. For Edexcel International A Levels, this typically involves applying basic integration rules, such as the power rule, which states that the integral of x^n is (x^(n+1))/(n+1) + C, where C is the constant of integration.

What are some common techniques for solving integration problems in Integration 1?

Common techniques include substitution, where you replace a part of the integrand with a single variable to simplify the integral, and integration by parts, which is useful for products of functions. These methods are crucial for tackling more complex integrals in the Edexcel curriculum.

Why is the constant of integration important in indefinite integrals?

The constant of integration, denoted as C, is important because indefinite integration represents a family of functions. Since differentiation of a constant is zero, the original function could have any constant value added to it, making C essential for expressing all possible antiderivatives.

How is integration used to find the area under a curve in Pure Mathematics 1?

Integration is used to calculate the area under a curve by evaluating the definite integral of a function over a specified interval. This process involves finding the antiderivative and then applying the limits of integration to determine the exact area, which is a key application in Pure Mathematics 1.

Why is "Omitting $+C$ on an indefinite integral" a common mistake in Integration 1?

Why it happens: Rushing. How to avoid it: Indefinite integrals ALWAYS need +C — non-negotiable. Definite integrals don't (it cancels). Burn this in: 'indef = + C, def = no C'. Source: WMA11/01 examiner reports — flagged on every paper

Why is "Trying to integrate $1/x$ using the power rule" a common mistake in Integration 1?

Why it happens: Applying x^n dx when n = -1. How to avoid it: Power rule excludes n = -1. x^-1 dx = |x| + C — but this is P2 content, NOT P1. If 1/x appears in a P1 problem, the question is wrong (or you've made an error).

Why is "Giving a negative value as 'area'" a common mistake in Integration 1?

Why it happens: Just integrating, ignoring sign. How to avoid it: AREA is always positive. If the curve is below the x-axis, the definite integral is negative — take its absolute value. Split the integral at each axis crossing. Source: WMA11/01 examiner reports — recurring annual issue

Why is "Evaluating $F(a) - F(b)$ instead of $F(b) - F(a)$" a common mistake in Integration 1?

Why it happens: Swapping upper and lower limits. How to avoid it: Definite integral \int_a^b: TOP LIMIT FIRST. F(b) - F(a) — always. Writing the limits explicitly [F(x)]_a^b helps.

Why is "Using $n + 1$ as the strip count when given $n$ ordinates (or vice versa)" a common mistake in Integration 1?

Why it happens: Confusion about strips vs ordinates. How to avoid it: If the question gives n STRIPS, there are n + 1 ordinates (function values), from y_0 to y_n. If it gives n ordinates, you have n - 1 strips. Strip width h = (b - a)/(number of strips).

Why is "Leaving $+C$ in the answer when a specific point is given" a common mistake in Integration 1?

Why it happens: Forgetting to solve for C. How to avoid it: If the question gives 'curve passes through (x_0, y_0)', substitute that point AFTER integrating to solve for C. The final answer is a SPECIFIC equation, not a family.

Edexcel International A Levels Mathematics (XMA01-YMA01)

Integration 1

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Short Notes - Integration 1

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Detailed Study Notes

Detailed notes on Pure Mathematics 1 for Edexcel International A Levels Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Integration 1 Study Notes — Edexcel IAL Mathematics P1 (WMA11/01, 2018 spec — 2026 onwards)

Indefinite and definite integrals of powers, finding a curve from its gradient, area under a curve, and the trapezium rule. Differentiation in reverse.

At a glance

Power rule: $\int x^{n} dx = \dfrac{x^{n+1}}{n+1} + C$ ( $n \ne -1$ ).
Sum rule: integrate term by term.
Indefinite integral: $+C$ required.
Definite integral: $[F(x)]_{a}^{b} = F(b) - F(a)$ .
Area: $\int|f(x)| dx$ — split at axis crossings.
Constant of integration: solve using a boundary point.
Trapezium rule (in booklet): $\dfrac{h}{2}[y_{0} + 2(y_{1}+\ldots+y_{n-1}) + y_{n}]$ .

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XMA01-YMA01 syllabus (2026 onwards).

9.1 — Understand and use integration as the reverse of differentiation; integrate $x^{n}$ for $n \ne -1$ , sums, and constant multiples.
9.2 — Evaluate definite integrals using $[F(x)]_{a}^{b} = F(b) - F(a)$ .
9.3 — Use integration to find the area under a curve (above the $x$ -axis or signed area).
9.4 — Use the trapezium rule to estimate the area under a curve numerically.

Integration as the reverse of differentiation

Raise power by 1, divide by new power, add $+C$ .

Definition. $\int f(x) \, dx$ is the family of functions whose derivative is $f(x)$ .

Power rule (the inverse of differentiation).

$\int x^{n} \, dx = \dfrac{x^{n + 1}}{n + 1} + C, \quad n \ne -1.$

Mnemonic. Raise the power by 1. Divide by the new power. Add $+C$ .

$f(x)$	$\int f(x) \, dx$
$x^{3}$	$\tfrac{x^{4}}{4} + C$
$x^{-2}$	$-x^{-1} + C = -1/x + C$
$x^{1/2}$	$\tfrac{2}{3}x^{3/2} + C$
$1$	$x + C$
$5$	$5x + C$

Why $+C$ ? Differentiating a constant gives zero. So $F(x) + 1$ , $F(x) + 100$ , and $F(x) - 42$ are all antiderivatives of $f(x)$ . We capture the whole family with one constant $C$ .

Why $n \ne -1$ ? $\int x^{-1} dx$ would give $\dfrac{x^{0}}{0}$ , which is undefined. The case $n = -1$ has its own antiderivative: $\int 1/x \, dx = \ln|x| + C$ — but that's P2 content.

Sum and constant-multiple rules. Just like differentiation:

$\int (af + bg) dx = a \int f \, dx + b \int g \, dx.$

Worked example. $\int (5\sqrt{x} - \dfrac{6}{x^{2}} + 1) dx$ .

Rewrite: $5x^{1/2} - 6x^{-2} + 1$ .
Integrate each term:
- $5 \cdot \dfrac{x^{3/2}}{3/2} = \dfrac{10}{3}x^{3/2}$
- $-6 \cdot \dfrac{x^{-1}}{-1} = 6x^{-1} = 6/x$ (the two minuses cancel!)
- $\int 1 \, dx = x$
Combine: $\dfrac{10}{3}x^{3/2} + \dfrac{6}{x} + x + C$ .

The definite integral ∫ₐᵇ f(x) dx equals the signed area between the curve y = f(x) and the x-axis, between x = a and x = b.

Raise power by 1, divide, add $+C$ .
Works for any real $n$ except $-1$ .
Rewrite roots/reciprocals as powers first.
Sum rule: integrate term by term.

Definite integrals — Fundamental Theorem of Calculus

Substitute upper and lower limits, subtract.

Definite integral.

$\int_{a}^{b} f(x) \, dx = F(b) - F(a),$

where $F$ is any antiderivative of $f$ .

Geometrically: the SIGNED area between $y = f(x)$ and the $x$ -axis from $x = a$ to $x = b$ . 'Signed' means areas BELOW the axis count NEGATIVE.

Method.

Integrate to get $F(x)$ (no $+C$ — it cancels).
Compute $F(b)$ .
Compute $F(a)$ .
Subtract: $F(b) - F(a)$ .

Worked example. $\int_{1}^{3} (4x - 5) dx$ .

Integrate: $2x^{2} - 5x$ .
Substitute: $[2x^{2} - 5x]_{1}^{3} = (18 - 15) - (2 - 5) = 3 - (-3) = 6$ .

Notation. Square brackets indicate the antiderivative with limits. Then subtract value-at-upper-limit minus value-at-lower-limit.

Why $+C$ cancels.

$(F(b) + C) - (F(a) + C) = F(b) - F(a)$ .

So you don't need $C$ for definite integrals — saves time.

Useful properties.

Property	Statement
Same limits	$\int_{a}^{a} f \, dx = 0$
Swap limits	$\int_{b}^{a} f \, dx = -\int_{a}^{b} f \, dx$
Splitting	$\int_{a}^{b} f \, dx = \int_{a}^{c} f \, dx + \int_{c}^{b} f \, dx$
Linearity	$\int_{a}^{b}(\alpha f + \beta g) dx = \alpha\int_{a}^{b} f dx + \beta\int_{a}^{b} g dx$

$\int_{a}^{b} = F(b) - F(a)$ .
TOP limit first.
$+C$ cancels — don't include.
Splitting: $\int_{a}^{b} = \int_{a}^{c} + \int_{c}^{b}$ .

Area under a curve — absolute value matters

Above axis = positive integral. Below = negative integral. Split, take absolute values, sum.

Signed vs absolute area.

Definite integral $\int_{a}^{b} f \, dx$ = SIGNED area. Negative for regions below $x$ -axis.
AREA (geometric, positive) = $\int_{a}^{b} |f(x)| \, dx$ .

Method.

Find where the curve meets the $x$ -axis between $a$ and $b$ (set $f(x) = 0$ ).
Split the interval at these crossings.
Integrate each piece.
Take absolute values of any negative pieces.
Sum.

Worked example. Area between $y = x^{2} - 2x$ and the $x$ -axis, from $x = 0$ to $x = 3$ .

$x$ -intercepts: $x(x - 2) = 0 \Rightarrow x = 0, 2$ .
Split: $[0, 2]$ and $[2, 3]$ .
Antiderivative: $F(x) = x^{3}/3 - x^{2}$ .

Piece 1: $[0, 2]$ .

$F(2) - F(0) = (8/3 - 4) - 0 = -4/3$ . Curve is BELOW axis here.

Take absolute value: $4/3$ .

Piece 2: $[2, 3]$ .

$F(3) - F(2) = (9 - 9) - (8/3 - 4) = 0 - (-4/3) = 4/3$ . Already positive.

Total area: $4/3 + 4/3 = 8/3$ .

Wording trap. 'Find the area' vs 'find $\int_{a}^{b} f \, dx$ ':

Phrasing	Use
"Area enclosed by ..."	Absolute value, split at crossings
"Evaluate $\int_{a}^{b}$ ..."	Signed value, no absolute
"Total area between curve and axis"	Absolute, split

Bounded by two curves. Area = $\int_{a}^{b} (\text{top} - \text{bottom}) dx$ between intersection points $a, b$ . (Beyond P1 in detail; first appears in P2/P3.)

AREA is always positive.
Split at axis crossings.
Take absolute values of pieces.
Sum at the end.

Finding a curve from its gradient

Integrate to get the family. Use a point to pin down $C$ .

Setup. Given $\dfrac{dy}{dx} = \ldots$ and a point on the curve, find the equation of the curve.

Method.

Integrate the gradient function to get $y = F(x) + C$ .
Substitute the given point to find $C$ .
Write the specific equation.

Worked example. Curve passes through $(2, 5)$ with $\dfrac{dy}{dx} = 3x^{2} - 4x + 1$ . Find $y$ .

Integrate: $y = x^{3} - 2x^{2} + x + C$ .
Substitute $(2, 5)$ : $5 = 8 - 8 + 2 + C \Rightarrow C = 3$ .
Equation: $y = x^{3} - 2x^{2} + x + 3$ .

Why a point is essential. Without a point, you have a FAMILY of curves all with the same gradient (just shifted vertically). The point picks one specific curve.

Two-point version. Sometimes the question gives two points; you solve for $C$ as in standard substitution.

Modelling context. If $\dfrac{ds}{dt}$ = velocity and you know position at $t = 0$ :

Integrate to get position $s(t) + C$ .
Use $s(0)$ to find $C$ .
Now you have $s(t)$ .

Integrate to family $F(x) + C$ .
Substitute given point to find $C$ .
Specific equation = $F(x) + C_{0}$ .

Trapezium rule — numerical integration

Split into strips, treat each as a trapezium, sum areas.

Trapezium rule (in the IAL formula booklet).

$\int_{a}^{b} y \, dx \approx \dfrac{h}{2}\left[y_{0} + 2(y_{1} + y_{2} + \ldots + y_{n - 1}) + y_{n}\right],$

where:

$n$ = number of strips.
$h = (b - a)/n$ = strip width.
$y_{i} = f(a + ih)$ = function value at the $i$ -th ordinate.

$n$ strips → $n + 1$ ordinates. This is the key relationship.

Picture. Divide $[a, b]$ into $n$ equal-width strips. Each strip is approximated by a trapezium (top edge a straight line). The trapezium rule formula sums the areas of all trapeziums.

Worked example. $\int_{1}^{3} 1/(1 + x^{2}) dx$ with 4 strips.

$h = (3 - 1)/4 = 0.5$ .
$x$ -values: $1, 1.5, 2, 2.5, 3$ .
$y$ -values: $0.5, 0.3077, 0.2, 0.1379, 0.1$ .
Apply formula: $\dfrac{0.5}{2}[0.5 + 2(0.3077 + 0.2 + 0.1379) + 0.1] = 0.25 \times 1.8912 = 0.4728$ .

Accuracy. Trapezium rule is exact for linear $f$ , approximate for everything else. Error scales as $h^{2}$ — halving the strip width quarters the error.

Over-estimate vs under-estimate. Depends on curvature:

Curve is	Trapezium rule
Concave up ( $f'' > 0$ )	OVER-estimate
Concave down ( $f'' < 0$ )	UNDER-estimate
Linear	Exact

Edexcel tip. Tabulate the $y$ -values BEFORE applying the formula. A 5-row table for 4 strips: $x_{i}$ , $y_{i}$ . This earns method marks even if you make an arithmetic error in the final sum.

$n$ strips → $n + 1$ ordinates.
$h = (b - a)/n$ .
Endpoints get weight 1; middle ordinates get weight 2.
Exact for linear functions.
Over-/under-estimates depend on concavity.

Quick recap

Power rule: $\int x^{n} dx = x^{n+1}/(n+1) + C$ .
Indefinite: $+C$ . Definite: no $C$ (cancels).
Definite integral: $F(b) - F(a)$ .
Area: absolute value, split at axis crossings.
Constant of integration: solved using a given point.
Trapezium rule: $n$ strips, $n + 1$ ordinates, endpoints weight 1, middles weight 2.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Power rule: $\int x^{n} dx = \dfrac{x^{n+1}}{n+1} + C$ , $n \ne -1$ .
Definite: $\int_{a}^{b} = F(b) - F(a)$ .
$+C$ : indefinite YES; definite NO.
Area: absolute value, split at crossings.
Trapezium rule (booklet): $\dfrac{h}{2}[y_{0} + 2(\Sigma y_{\text{mid}}) + y_{n}]$ .
$n$ strips → $n + 1$ ordinates.

How it’s examined

Integration is the second-highest-weighted topic in WMA11/01 P1 (after differentiation), typically 10-15 marks. Standard questions: indefinite integral with mixed powers (4-5 marks); definite integral (3-4 marks); area under curve, sometimes crossing the axis (5-7 marks); finding a curve from its gradient and a point (4-5 marks); trapezium rule (4-6 marks, appears almost every paper). Most-dropped marks: forgetting $+C$ , computing signed-not-absolute area, mis-counting strips vs ordinates in the trapezium rule. A* candidates show all algebraic steps and tabulate trapezium-rule data clearly.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA11/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

Take this whole topic with you

Step-by-step worked examples — Integration 1

Step-by-step solutions to past-paper-style questions on integration 1, written exactly the way a tutor would explain them at the board.

1Indefinite integral of a polynomial (4 marks, P1)
Core• Style: WMA11/01 January 2024 Q4• indefinite integral
Question
Find $\displaystyle\int (6x^{2} - 4x + 7) \, dx$ .
Step-by-step solution
1. Step 1
  Apply $\int x^{n} dx = \dfrac{x^{n+1}}{n+1} + C$ to each term:
2. Step 2
  $\int 6x^{2} dx = 6 \cdot \dfrac{x^{3}}{3} = 2x^{3}$ .
3. Step 3
  $\int (-4x) dx = -4 \cdot \dfrac{x^{2}}{2} = -2x^{2}$ .
4. Step 4
  $\int 7 \, dx = 7x$ .
5. Step 5
  Combine and add $+C$ :
  $\int (6x^{2} - 4x + 7) \, dx = 2x^{3} - 2x^{2} + 7x + C$
Answer
$2x^{3} - 2x^{2} + 7x + C$
Examiner tip
M1 for at least one correctly integrated term; A1 A1 for two correct terms; A1 for the constant of integration. Forgetting the $+C$ on an INDEFINITE integral costs the final A1 — non-negotiable.
2Integrate mixed powers (5 marks, P1)
Core• Style: WMA11/01 June 2024 Q5• indefinite integral, fractional powers
Question
Find $\displaystyle\int \left( 5\sqrt{x} - \dfrac{6}{x^{2}} + 1 \right) dx$ .
Step-by-step solution
1. Step 1
  Rewrite as powers: $5x^{1/2} - 6x^{-2} + 1$ .
2. Step 2
  Integrate term by term:
  $= 5 \cdot \dfrac{x^{3/2}}{3/2} - 6 \cdot \dfrac{x^{-1}}{-1} + x + C$
3. Step 3
  Simplify: $\dfrac{5x^{3/2}}{3/2} = \dfrac{10}{3}x^{3/2}$ ; $\dfrac{-6x^{-1}}{-1} = 6x^{-1} = \dfrac{6}{x}$ .
4. Step 4
  Final:
  $\dfrac{10}{3}x^{3/2} + \dfrac{6}{x} + x + C$
Answer
$\dfrac{10}{3}x^{3/2} + \dfrac{6}{x} + x + C$ (or $\dfrac{10\sqrt{x^{3}}}{3} + \dfrac{6}{x} + x + C$ ).
Examiner tip
M1 for rewriting; M1 for power-up by 1; A1 A1 A1 for each term; A1 for $+C$ . Common dropped marks: forgetting that dividing by $n + 1 = -1$ flips the sign on the $-6x^{-2}$ term.
3Definite integral (4 marks, P1)
Core• Style: WMA11/01 January 2023 Q5• definite integral
Question
Evaluate $\displaystyle\int_{1}^{3} (4x - 5) \, dx$ .
Step-by-step solution
1. Step 1
  Integrate: $\int (4x - 5) dx = 2x^{2} - 5x$ . (No $+C$ needed for a definite integral — it cancels.)
2. Step 2
  Evaluate at limits:
  $\left[2x^{2} - 5x\right]_{1}^{3} = (18 - 15) - (2 - 5) = 3 - (-3) = 6$
Answer
$6$
Examiner tip
M1 for integrating; A1 for the antiderivative; M1 for substituting both limits; A1 for the answer. Definite integrals don't need $+C$ . Watch the order: TOP limit FIRST.
4Area under a curve (6 marks, P1)
Extended• Style: WMA11/01 June 2023 Q11• area
Question
Find the area of the region bounded by the curve $y = x^{2} - 2x$ , the $x$ -axis, and the lines $x = 0$ and $x = 3$ .
Step-by-step solution
1. Step 1
  Find where the curve meets the $x$ -axis between 0 and 3: $x^{2} - 2x = 0 \Rightarrow x(x - 2) = 0$ . Roots at $0$ and $2$ .
2. Step 2
  Between $x = 0$ and $x = 2$ : the curve is BELOW the $x$ -axis (test $x = 1$ : $1 - 2 = -1 < 0$ ). Integral is NEGATIVE → take absolute value.
3. Step 3
  Between $x = 2$ and $x = 3$ : curve is ABOVE the $x$ -axis. Integral is positive.
4. Step 4
  Integrate: $\int (x^{2} - 2x) dx = \tfrac{x^{3}}{3} - x^{2}$ .
5. Step 5
  First piece (with absolute value): $\left|\left[\tfrac{x^{3}}{3} - x^{2}\right]_{0}^{2}\right| = |\tfrac{8}{3} - 4 - 0| = |-\tfrac{4}{3}| = \tfrac{4}{3}$ .
6. Step 6
  Second piece: $\left[\tfrac{x^{3}}{3} - x^{2}\right]_{2}^{3} = (9 - 9) - (\tfrac{8}{3} - 4) = 0 - (-\tfrac{4}{3}) = \tfrac{4}{3}$ .
7. Step 7
  Total area: $\tfrac{4}{3} + \tfrac{4}{3} = \tfrac{8}{3}$ .
Answer
Area $= \dfrac{8}{3}$ (square units).
Examiner tip
M1 for finding the $x$ -intercept inside the range; M1 for splitting the integral; A1 for the antiderivative; M1 for evaluating each piece; A1 A1 for both areas; A1 for the total. Critical: 'AREA' means absolute value, never negative. Areas below the $x$ -axis contribute POSITIVELY to total area.
5Find a curve given its gradient function (5 marks, P1)
Core• Style: WMA11/01 January 2024 Q7• constant of integration
Question
A curve passes through $(2, 5)$ and has gradient function $\dfrac{dy}{dx} = 3x^{2} - 4x + 1$ . Find the equation of the curve.
Step-by-step solution
1. Step 1
  Integrate: $y = \int (3x^{2} - 4x + 1) dx = x^{3} - 2x^{2} + x + C$ .
2. Step 2
  Use the point $(2, 5)$ to find $C$ : $5 = 8 - 8 + 2 + C \Rightarrow C = 3$ .
3. Step 3
  Equation:
  $y = x^{3} - 2x^{2} + x + 3$
Answer
$y = x^{3} - 2x^{2} + x + 3$ .
Examiner tip
M1 for integrating; A1 for the antiderivative (with $+C$ ); M1 for substituting the point; A1 for $C = 3$ ; A1 for the final equation. The reason we need a SPECIFIC point: infinitely many curves have the same gradient function; the point pins down which one.
6Trapezium rule (6 marks, P1)
Extended• Style: WMA11/01 June 2024 Q10• trapezium rule
Question
Use the trapezium rule with 4 strips to estimate $\displaystyle\int_{1}^{3} \dfrac{1}{1 + x^{2}} dx$ . Give your answer to 4 decimal places.
Step-by-step solution
1. Step 1
  Strip width $h = (3 - 1)/4 = 0.5$ . So $x$ -values are $1, 1.5, 2, 2.5, 3$ .
2. Step 2
  Compute $y_{i} = 1/(1 + x_{i}^{2})$ at each:
  $y_{0} = 0.5, \; y_{1} = 0.3077, \; y_{2} = 0.2, \; y_{3} = 0.1379, \; y_{4} = 0.1$
3. Step 3
  Trapezium rule:
  $\int \approx \dfrac{h}{2}\left[y_{0} + 2(y_{1} + y_{2} + y_{3}) + y_{4}\right]$
4. Step 4
  Sum: $0.5 + 0.1 = 0.6$ ; $2(0.3077 + 0.2 + 0.1379) = 2(0.6456) = 1.2912$ . Total: $0.6 + 1.2912 = 1.8912$ .
5. Step 5
  Multiply by $h/2 = 0.25$ :
  $0.25 \times 1.8912 = 0.4728$
Answer
$\approx 0.4728$ (to 4 d.p.).
Examiner tip
B1 for strip width; M1 for all five $y$ -values; A1 for at least four correct $y$ -values; M1 for trapezium-rule formula; M1 for substitution; A1 for the answer. Comparison with the exact: $\int_{1}^{3} 1/(1 + x^{2}) dx = \arctan 3 - \arctan 1 \approx 0.4636$ (but $\arctan$ is beyond P1; the trapezium rule is the IAL P1 tool).

Key Formulae — Integration 1

The formulae you need to memorise for integration 1 on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Power rule for integration
$\int x^{n} \, dx = \dfrac{x^{n + 1}}{n + 1} + C, \quad n \ne -1$
$n$
any real number except $-1$
$C$
constant of integration
When to use
Integrating any power of $x$ . The $n = -1$ case ( $\int 1/x \, dx = \ln|x| + C$ ) belongs to P2.
Example
$\int x^{3} dx = \tfrac{x^{4}}{4} + C$ ; $\int x^{-2} dx = -x^{-1} + C$ .
Sum and constant-multiple rules
$\int (af + bg) \, dx = a \int f \, dx + b \int g \, dx$
$a, b$
constants
$f, g$
functions of $x$
When to use
Integrate polynomial sums term by term, exactly like differentiation.
Example
$\int (3x^{2} - 5x + 7) dx = x^{3} - \tfrac{5x^{2}}{2} + 7x + C$ .
Definite integral (Fundamental Theorem of Calculus)
$\int_{a}^{b} f(x) \, dx = F(b) - F(a)$
$F$
any antiderivative of $f$
$a, b$
lower and upper limits
When to use
Evaluate a definite integral. The $+C$ cancels.
Example
$\int_{1}^{3}(4x - 5) dx = [2x^{2} - 5x]_{1}^{3} = 3 - (-3) = 6$ .
Trapezium rule
$\int_{a}^{b} y \, dx \approx \dfrac{h}{2}\left[y_{0} + 2(y_{1} + y_{2} + \ldots + y_{n-1}) + y_{n}\right]$
$h$
strip width = $(b - a)/n$
$n$
number of strips
$y_i$
function value at $x = a + ih$
When to use
Numerical estimate of a definite integral, especially when antiderivative isn't easy. IS in the IAL formula booklet.
Example
$\int_{1}^{3} 1/(1 + x^{2}) dx \approx 0.4728$ with 4 strips.
Area under a curve
$\text{Area} = \int_{a}^{b} |f(x)| \, dx$
$[a, b]$
interval of integration
When to use
When asked for the AREA (positive). If the curve crosses the $x$ -axis in $[a, b]$ , split the integral at each crossing and take absolute values.
Example
Area of $y = x^{2} - 2x$ between $x = 0$ and $x = 3$ : split at $x = 2$ where the curve meets the axis, compute each piece separately, sum absolute values.

Key Definitions and Keywords — Integration 1

Definitions to memorise and the exact keywords mark schemes credit for integration 1 answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Indefinite integral
Examiner keyword
The family of antiderivatives of $f$ : $\int f(x) \, dx = F(x) + C$ , where $F'(x) = f(x)$ .
Definite integral
Examiner keyword
A specific number: $\int_{a}^{b} f(x) \, dx = F(b) - F(a)$ . Geometrically, the signed area between the curve and the $x$ -axis from $x = a$ to $x = b$ .
Constant of integration ( $C$ )
Examiner keyword
An unspecified constant added to every indefinite integral, since differentiating any constant gives zero. Determined by a boundary condition.
Antiderivative
A function $F$ whose derivative is $f$ . The 'reverse' of differentiation.
Strip (trapezium rule)
One of the equal-width subintervals used in the trapezium rule. $n$ strips means $n + 1$ ordinates (function values).

Common Mistakes and Misconceptions — Integration 1

The traps other students keep falling into on integration 1 questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Omitting $+C$ on an indefinite integral
WMA11/01 examiner reports — flagged on every paper
Why it happens
Rushing.
How to avoid it
Indefinite integrals ALWAYS need $+C$ — non-negotiable. Definite integrals don't (it cancels). Burn this in: 'indef = + C, def = no C'.
✕Trying to integrate $1/x$ using the power rule
Why it happens
Applying $\int x^{n} dx$ when $n = -1$ .
How to avoid it
Power rule excludes $n = -1$ . $\int x^{-1} dx = \ln|x| + C$ — but this is P2 content, NOT P1. If $1/x$ appears in a P1 problem, the question is wrong (or you've made an error).
✕Giving a negative value as 'area'
WMA11/01 examiner reports — recurring annual issue
Why it happens
Just integrating, ignoring sign.
How to avoid it
AREA is always positive. If the curve is below the $x$ -axis, the definite integral is negative — take its absolute value. Split the integral at each axis crossing.
✕Evaluating $F(a) - F(b)$ instead of $F(b) - F(a)$
Why it happens
Swapping upper and lower limits.
How to avoid it
Definite integral $\int_{a}^{b}$ : TOP LIMIT FIRST. $F(b) - F(a)$ — always. Writing the limits explicitly $[F(x)]_{a}^{b}$ helps.
✕Using $n + 1$ as the strip count when given $n$ ordinates (or vice versa)
Why it happens
Confusion about strips vs ordinates.
How to avoid it
If the question gives $n$ STRIPS, there are $n + 1$ ordinates (function values), from $y_{0}$ to $y_{n}$ . If it gives $n$ ordinates, you have $n - 1$ strips. Strip width $h = (b - a)/(\text{number of strips})$ .
✕Leaving $+C$ in the answer when a specific point is given
Why it happens
Forgetting to solve for $C$ .
How to avoid it
If the question gives 'curve passes through $(x_{0}, y_{0})$ ', substitute that point AFTER integrating to solve for $C$ . The final answer is a SPECIFIC equation, not a family.

Integration 1 — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Integration 1

Short Notes - Integration 1

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Indefinite integral of a polynomial (4 marks, P1)

2Integrate mixed powers (5 marks, P1)

3Definite integral (4 marks, P1)

4Area under a curve (6 marks, P1)

5Find a curve given its gradient function (5 marks, P1)

6Trapezium rule (6 marks, P1)

Power rule for integration

Sum and constant-multiple rules

Definite integral (Fundamental Theorem of Calculus)

Trapezium rule

Area under a curve

Indefinite integral

Definite integral

Constant of integration (CCC)

Antiderivative

Strip (trapezium rule)

✕Omitting +C+C+C on an indefinite integral

✕Trying to integrate 1/x1/x1/x using the power rule

✕Giving a negative value as 'area'

✕Evaluating F(a)−F(b)F(a) - F(b)F(a)−F(b) instead of F(b)−F(a)F(b) - F(a)F(b)−F(a)

✕Using n+1n + 1n+1 as the strip count when given nnn ordinates (or vice versa)

✕Leaving +C+C+C in the answer when a specific point is given

Integration 1 — frequently asked questions

Constant of integration ( $C$ )

✕Omitting $+C$ on an indefinite integral

✕Trying to integrate $1/x$ using the power rule

✕Evaluating $F(a) - F(b)$ instead of $F(b) - F(a)$

✕Using $n + 1$ as the strip count when given $n$ ordinates (or vice versa)

✕Leaving $+C$ in the answer when a specific point is given