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Cambridge IGCSE (CIE) · Additional Maths

Integration

The Big Idea: Integration is differentiation in reverse — if differentiating a function tells you its gradient, integrating a gradient tells you back the original function (plus a mystery constant, because flat shifts vanish when you differentiate).

Powers of x

Areas & Definite Integrals

Trig, e^x & 1/x

Reverse Chain Rule

Summary — Everything In This Chapter

Integration undoes differentiation — it's the inverse process, formalised by the Fundamental Theorem of Calculus.
Indefinite integrals always need a "+c" (constant of integration) because constant terms vanish when you differentiate.
To integrate xⁿ: raise the power by 1, then divide by the new power (never works when n = −1).
Definite integrals have limits (a to b) — you evaluate the antiderivative at both ends and subtract. The "+c" cancels out, so you don't need it.
Definite integration gives you the area under a curve — but watch out: area below the x-axis comes out negative, so you take the modulus.
To find area between a curve and a line, or between two curves, integrate "upper − lower" between the intersection points.
Special standard integrals exist for sin x, cos x, sec²x, eˣ, and 1/x.
The Reverse Chain Rule lets you integrate composite functions like (ax+b)ⁿ or e^(ax+b) by "adjusting and compensating."

1. Introduction to Integration

What is integration, really?

Think of differentiation and integration as two doors on the same corridor, facing each other. Walk through the "differentiation" door and you turn a function y = f(x) into its gradient function dy/dx. Walk back through the "integration" door and you should land right back where you started... almost.

Here's the catch. Imagine three parallel lines: y = 5x, y = 5x + 2, and y = 5x − 3. They're all different lines, but they all have the exact same gradient: dy/dx = 5. If someone hands you "dy/dx = 5" and asks "what was y?", you genuinely cannot know which of these (or infinitely many others) it was. That's why every indefinite integral ends with + c — a placeholder acknowledging "there could have been a constant here that we lost."

Fundamental Theorem of Calculus (Indefinite form) Integration is the inverse process of differentiation.

Notation-wise, an integral is written ∫ f(x) dx. Break that symbol down:

The elongated "S" symbol ∫ literally means "integrate" — it's a stretched S for "Sum," historically tied to summing infinitely many thin slices.
f(x) is the function you're integrating (called the integrand).
dx tells you which variable you're integrating with respect to — "do this whole process treating x as the variable."

Why does the constant "disappear" on differentiation?

Graphically, a constant term shifts a curve up or down — it doesn't change the curve's shape or steepness anywhere. Since the gradient measures steepness, a vertical shift is invisible to differentiation. Once you differentiate, that vertical-shift information is gone forever, which is exactly why you can't recover it when integrating back.

Practice

Find (a) dy/dx for y = 6x + 3, and (b) ∫6 dx.

2. Integrating Powers of x

The core rule

This is the rule you'll use more than any other in this entire chapter, so it needs to become pure muscle memory. Two simple actions:

Increase the power by 1.
Divide by that new power.

Power Rule for Integration ∫ xⁿ dx = xⁿ⁺¹ / (n+1) + c Valid for any rational n, EXCEPT n = −1 (you'd be dividing by zero)

If there's a constant multiplying the power of x, it just tags along for the ride and multiplies the whole result:

∫ axⁿ dx = a·xⁿ⁺¹ / (n+1) + c

One especially useful special case: integrating a plain constant just multiplies it by x.

∫ a dx = ax + c e.g. ∫4 dx = 4x + c

Three disguises to watch for

Roots: ⁿ√x must be rewritten as a fractional power first, e.g. ∛x = x^(1/3) before you can integrate it.

Fractions with x in the denominator: 4/x² must become 4x⁻² first — negative powers, not fractions.

Products/quotients: You can NEVER integrate a product like 8x²(2x−3) directly term-by-term — you must expand it into separate power-of-x terms first.

Worked Example — expanding a product first

Find ∫ 8x²(2x − 3) dx

Expand the brackets first (products can't be integrated directly):

8x²(2x − 3) = 16x³ − 24x²

Now integrate term by term: ∫(16x³ − 24x²) dx = 16x⁴/4 − 24x³/3 + c

= 4x⁴ − 8x³ + c

Practice

Given dy/dx = 3x⁴ − 2x² + 3 − 1/√x, find y in terms of x.

Finding the constant of integration

Whenever a question gives you extra information — like a point the curve passes through — you can pin down the exact value of c. This turns your general family of curves into one specific curve.

3-Step Method 1. Rewrite into integrable form (powers of x) → 2. Integrate each term (+c) → 3. Substitute the given point to solve for c

Worked Example

A curve passes through (2, 6) and has gradient function 3x²(2x − 1). Find the equation of the curve.

Step 1 — expand: dy/dx = 3x²(2x−1) = 6x³ − 3x²

Step 2 — integrate: y = 6x⁴/4 − 3x³/3 + c = (3/2)x⁴ − x³ + c

Step 3 — substitute (2,6): 6 = (3/2)(2)⁴ − (2)³ + c → 6 = 24 − 8 + c → c = −10

∴ y = (3/2)x⁴ − x³ − 10

Exam tip

If a question can be answered with a specific numeric value of c, it MUST give you extra information (a coordinate, or another condition). If you find yourself needing "c" but there's no extra info given, that's a strong signal your question wants a definite integral instead — where c isn't needed at all.

3. Definite Integrals

A definite integral has two numbers attached — a lower limit a and an upper limit b — written as ∫ from a to b. Unlike an indefinite integral (which gives you a whole family of functions), a definite integral spits out a single, actual number. That number can be positive, negative, or zero.

Fundamental Theorem of Calculus (Definite form) ∫ₐᵇ f'(x) dx = f(b) − f(a)

Here's the beautiful bit: why don't we need "+c" here? Because when you compute f(b) − f(a), the "+c" appears in BOTH f(b) and f(a) — and then they literally cancel each other out during the subtraction. So it's not that the constant doesn't exist; it's that it's guaranteed to cancel, so including it would just be extra work for nothing.

4-Step Method 1. Name the integral (e.g. "let I = ...") → 2. Rewrite into integrable form → 3. Integrate using square brackets with limits outside → 4. Substitute top limit, then bottom limit, subtract

Worked Example

Find the value of ∫₂⁴ 3x(x² − 2) dx

Expand first: 3x(x²−2) = 3x³ − 6x

Integrate (without applying limits yet): [ (3/4)x⁴ − 3x² ]₂⁴

Substitute x=4: (3/4)(256) − 3(16) = 192 − 48 = 144

Substitute x=2: (3/4)(16) − 3(4) = 12 − 12 = 0

Subtract: 144 − 0 = 144

Special properties worth knowing cold

Property	Why it's useful
∫ₐᵇ k·f(x) dx = k∫ₐᵇ f(x) dx	Pull constants outside — especially handy with fractions/negatives
∫ₐᵇ [f(x)+g(x)] dx = ∫ₐᵇf(x)dx + ∫ₐᵇg(x)dx	Integrate term by term (works for subtraction too)
∫ₐᵃ f(x) dx = 0	Same upper and lower limit → subtracting a value from itself
∫ₐᵇ f(x) dx = −∫ᵦᵃ f(x) dx	Swapping the limits flips the sign
∫ₐᵇ f(x)dx = ∫ₐᶜf(x)dx + ∫ᶜᵇf(x)dx	Split the interval — essential for area problems spanning the x-axis

Watch for this exam trap

Some questions ask for an indefinite integral in part (a) — so "+c" is required — then reuse the same integral as a definite integral in part (b), where "+c" must NOT appear. Read each part carefully; don't carry the habit from one part into the next.

4. Finding Areas with Integration

"Area under a curve" has a precise meaning: it's the region trapped between the curve y = f(x), the x-axis, and two vertical boundary lines x = a and x = b. The value of the definite integral ∫ₐᵇ f(x) dx is that area.

What if I'm not told the limits?

If no limits are given, they're almost always the x-axis intercepts. Set y = 0 and solve to find them.

Worked Example — finding your own limits

Find the shaded area R under y = x(5 − x).

Set y=0: x(5−x)=0 → x=0 or x=5. So limits are a=0, b=5.

Expand and integrate: ∫₀⁵(5x − x²)dx = [5x²/2 − x³/3]₀⁵

= (125/2 − 125/3) − (0−0) = 125/6

Area = 125/6 square units

The negative area trap

Critical rule

If the curve dips below the x-axis over your interval, the definite integral comes out negative — but a real geometric area can never be negative! You must take the modulus (absolute value) of that result. If the shaded region has parts both above AND below the x-axis, you cannot just integrate across the whole thing in one go — split into separate integrals for each part, take the modulus of any negative ones, then add them together.

Worked Example — area with a negative result

Find the area bounded by y = x² − 4x − 5, the x-axis, and the lines x=0 and x=4.

This parabola factorises as (x−5)(x+1), so it's below the x-axis between x=−1 and x=5 — our interval [0,4] sits entirely below the axis. Expect a negative integral.

∫₀⁴(x²−4x−5)dx = [x³/3 − 2x² − 5x]₀⁴ = (64/3 − 32 − 20) − 0 = −92/3

Take the modulus: Area required = 92/3 square units

Practice

Find the area enclosed between y = x² − 6x + 5, the x-axis, and the lines x = 2 and x = 4 (the curve dips below the axis here).

5. Finding Areas Between Lines & Curves

Area between a curve and a straight line

Picture a curve and a straight line crossing at two points, with a sliver of area trapped between them. The trick is simple: find the area under each one separately, then subtract the smaller region from the larger.

4-Step Method 1. Find intersections → 2. Find area under curve R_C (integration) → 3. Find area under line R_L (integration OR shape formula) → 4. Subtract smaller from larger

Since the "area under a line" is often just a rectangle, triangle, or trapezium shape, you can sometimes skip integration entirely and use basic area formulas — much faster:

Shape under the line	Formula
Rectangle (y = c, constant)	R_L = c(b − a)
Triangle (line through origin)	R_L = ½(b − a)·g(b)
Trapezium (general line)	R_L = [(b−a)/2]·[g(b) + g(a)]

Worked Example

Find the area R enclosed between y = x/3 and y = x^(1/2) (curve on top).

Step 1 — Intersections: x/3 = x^(1/2) → x²/9 = x → x² − 9x = 0 → x=0 or x=9

Step 2 — Area under curve: R_C = ∫₀⁹ x^(1/2) dx = [ (2/3)x^(3/2) ]₀⁹ = (2/3)(27) = 18

Step 3 — Area under line (triangle, base 9, height 3): R_L = ½×9×3 = 27/2

Step 4 — Subtract (curve on top): R = 18 − 27/2 = 9/2 square units

Alternative method

You could also treat the line as a "second curve" and integrate (upper function − lower function) directly in one go — see the next section. It works, but it's easier to slip up with sign errors this way, so many students prefer the subtract-two-areas method above.

Area between two curves

Same core idea, but now BOTH boundaries are curves. In general:

Area = ∫ₐᵇ [upper curve − lower curve] dx

The catch: if the two curves cross over each other within your interval (swapping which one is "on top"), you cannot use a single integral across the whole span. You must split the region at the crossover point, integrate each piece separately (being careful which function is upper/lower in each piece), then add the areas.

Worked Example — curves that cross

Find the shaded area between y = x³−5x²+7x and y = 2+2x²−x³.

Step 1 — Intersections: solving 2x³−7x²+7x−2=0 gives x = 1/2, 1, 2.

Step 2 — First region [0.5, 1]: R₁ = ∫₀.₅¹ (2x³−7x²+7x−2) dx = 5/96

Step 3 — Second region [1, 2]: R₂ = ∫₁² (−2x³+7x²−7x+2) dx = 1/3

Step 4 — Add: 5/96 + 1/3 = 37/96 square units

Practice

Explain, in your own words, why you cannot always compute the total shaded area between two curves as a single integral of (upper − lower) between the outermost limits, even when the curves cross over in the middle.

6. Integrating Trig Functions

These come straight from reversing the standard trig derivatives you already know. Notice the sign flip for sine — a classic place students slip up.

Standard Trig Integrals ∫ sin x dx = −cos x + c | ∫ cos x dx = sin x + c | ∫ sec²x dx = tan x + c

For the linear composite (ax + b), every result also gets divided by a (the "compensate" step you'll meet properly in the Reverse Chain Rule section):

∫ sin(ax+b) dx = −(1/a)cos(ax+b) + c | ∫ cos(ax+b) dx = (1/a)sin(ax+b) + c | ∫ sec²(ax+b) dx = (1/a)tan(ax+b) + c

Radians only

For ANY calculus involving trig functions, angles must always be measured in radians, never degrees. This isn't optional — the derivative/integral formulas above are only true in radian mode.

Worked Example

A curve satisfies y = ∫2sin(2x + π/6) dx and passes through (π/3, √3). Find y.

Factor out the 2, then integrate: y = 2 × [−½cos(2x+π/6)] + c = −cos(2x+π/6) + c

Substitute (π/3, √3): √3 = −cos(2π/3+π/6) + c = −cos(5π/6) + c = √3/2 + c

So c = √3/2

y = −cos(2x + π/6) + √3/2

Practice

Find ∫ sec²(3x − π/3) dx

7. Integrating e^x & 1/x

eˣ is famous for being its own derivative — and pleasingly, it's also its own integral. 1/x is the special case that the power rule couldn't handle (since n = −1 breaks the "divide by n+1" step), and it integrates to a natural log instead.

Standard Integrals ∫ eˣ dx = eˣ + c | ∫ (1/x) dx = ln|x| + c

For the linear version (ax+b), divide by a as usual:

∫ e^(ax+b) dx = (1/a)e^(ax+b) + c | ∫ 1/(ax+b) dx = (1/a)ln|ax+b| + c

A useful spinoff: if the numerator exactly matches the coefficient a from the denominator, the 1/a cancels away entirely:

∫ a/(ax+b) dx = ln|ax+b| + c

The "log constant" trick

When your constant of integration will end up combined with a log term, it's often smarter to write c = ln k instead of a plain c. Using log laws, ln|x| + ln k = ln(k|x|) — a single, cleaner term. This is exactly what happens in the worked example below.

Worked Example

f'(x) = 3/(3x+2) + e^(4−x), and f(1) = ln10 − e³. Find f(x).

Integrate: f(x) = 3×(1/3)ln|3x+2| + (1/(−1))e^(4−x) + c = ln|3x+2| − e^(4−x) + c

Substitute x=1: ln10 − e³ = ln|5| − e³ + c → ln10 − ln5 = c

Using log laws: c = ln(10/5) = ln 2

f(x) = ln|3x+2| − e^(4−x) + ln 2, or combined: ln(2|3x+2|) − e^(4−x)

8. Reverse Chain Rule

Differentiation has a huge toolbox of shortcuts (product rule, quotient rule, chain rule) that let you tear apart almost any combination of functions. Integration is not so generous — most of those shortcuts simply don't reverse cleanly. The Reverse Chain Rule is one of the few integration techniques that does work reliably, and it's used constantly.

It applies specifically when your integrand looks like a composite function multiplied by the derivative of its "inner" function. In formal notation:

Reverse Chain Rule (formal form) I = ∫ g'(x)·f'(g(x)) dx

The "adjust and compensate" technique

In practice, exam questions rarely hand you a perfectly matching coefficient. Say you spot that e^(x²) differentiates (via chain rule) to 2x·e^(x²). That means ∫2x·e^(x²) dx = e^(x²) + c is trivial. But what if you're asked to integrate 5x·e^(x²) instead? The coefficient doesn't match (5 ≠ 2). Here's the fix:

Take out the constant: 5∫x·e^(x²) dx
Adjust — force in the "2" you need: 5 × (1/2) ∫2x·e^(x²) dx
Compensate — divide outside by that same 2 to balance it out
Now the integral inside matches the reverse chain rule pattern exactly, and integrates to: (5/2)e^(x²) + c

The most useful special case to recognise instantly

If the numerator is (almost) the exact derivative of the denominator, the integral is a natural log:

I = ∫ f'(x)/f(x) dx = ln|f(x)| + c

Worked Example

f'(x) = 5x²sin(2x³), curve passes through (0,1). Find f(x).

f(x) = 5∫x²sin(2x³) dx

The "sin" part suggests the antiderivative involves −cos. 2x³ differentiates to 6x², so we adjust/compensate with a factor of 1/6:

f(x) = 5 × (1/6) ∫6x²sin(2x³) dx = (5/6)×[−cos(2x³)] + c

f(x) = −(5/6)cos(2x³) + c (then substitute (0,1) to find c if required)

Integrating (ax + b)ⁿ — the composite power rule

∫(ax + b)ⁿ dx = [(ax+b)^(n+1) / (n+1)] × (1/a) + c

In plain words: raise the power by 1, divide by the new power (exactly like the basic power rule) — then ALSO divide the whole thing by the coefficient of x inside the bracket. This only works when the inside of the bracket is linear (ax+b form) — for anything more complex inside, you'd need substitution methods beyond this chapter.

Practice

Find ∫ 3(7 − 2x)^(5/3) dx

What to Memorise

Function	Integral
xⁿ (n ≠ −1)	xⁿ⁺¹/(n+1) + c
a (constant)	ax + c
sin x	−cos x + c
cos x	sin x + c
sec²x	tan x + c
eˣ	eˣ + c
1/x	ln\|x\| + c
sin(ax+b)	−(1/a)cos(ax+b) + c
cos(ax+b)	(1/a)sin(ax+b) + c
sec²(ax+b)	(1/a)tan(ax+b) + c
e^(ax+b)	(1/a)e^(ax+b) + c
1/(ax+b)	(1/a)ln\|ax+b\| + c
a/(ax+b)	ln\|ax+b\| + c
(ax+b)ⁿ	[(ax+b)^(n+1)/(n+1)] × (1/a) + c
f'(x)/f(x)	ln\|f(x)\| + c

Key terms

Constant of integration (c): the unknown constant added to every indefinite integral, because differentiation destroys information about vertical shifts.

Definite integral: an integral with limits a and b; evaluates to an actual number (the area), and needs no "+c" since it cancels.

Integrand: the function being integrated — the f(x) inside ∫f(x)dx.

Reverse Chain Rule: the technique for integrating a composite function multiplied by the derivative of its inner function, often via "adjust and compensate."

Concepts Checklist

I understand why integration needs a "+c" and can explain it using the constant-terms-vanish idea.
I can integrate any power of x, including negative and fractional powers.
I can rewrite roots and fractions into power-of-x form before integrating.
I always expand products/quotients before attempting to integrate them.
I can use a given point on a curve to solve for the constant of integration.
I understand why "+c" is not needed for definite integrals.
I can evaluate a definite integral using the 4-step method (name, rewrite, integrate, substitute limits).
I know the special properties of definite integrals (equal limits, swapping limits, splitting intervals).
I can find the area under a curve, including finding my own limits from x-axis intercepts.
I remember to take the modulus when a region lies below the x-axis.
I can split an area into multiple integrals when a region has parts above and below the x-axis.
I can find the area between a curve and a line by subtracting one region from another.
I can find the area between two curves, including splitting at crossover points.
I know the standard integrals for sin x, cos x, and sec²x (and their linear versions).
I know the standard integrals for eˣ and 1/x (and their linear versions).
I can recognise when to use the Reverse Chain Rule and apply "adjust and compensate."
I can integrate (ax+b)ⁿ using the composite power rule.
I can spot the f'(x)/f(x) pattern and integrate it as ln|f(x)| + c.

Exam Tips — Common Mistakes & Mark-Scheme Traps

Forgetting "+c"

Losing the "+c" on an indefinite integral is one of the most common ways to drop an easy mark. Get in the habit of writing it the moment you finish integrating, before you even simplify.

Integrating products/quotients directly

You cannot integrate (2x+1)(x−3) term by term as if it were already expanded, and you cannot integrate a fraction like (x²+1)/x term by term without first simplifying/expanding it into separate power terms.

Forgetting the modulus on negative areas

If your definite integral comes out negative but you were asked for an "area," that negative sign is a signal — not the final answer. Take the modulus.

Not adjusting for the coefficient of x in (ax+b)

Students often integrate (3x+2)⁴ as if it were x⁴, forgetting the extra ÷a step. Always ask: "is the bracket linear, and if so, what's its coefficient of x?"

Degrees instead of radians

All trig integration formulas assume radians. If you see an angle written in degrees in a calculus question, something is off — convert first.

Ignoring where curves cross when finding area between them

Always check whether "upper" and "lower" swap partway through the interval. If they do, split the integral at the crossover point — otherwise you'll silently undercount the area.

What examiners reward

Sketching a diagram (even a rough one) when none is given, labelling limits and intersection points clearly, and showing the "[ ]" notation with limits attached before substituting — these earn method marks even if your final number is slightly off.

Also in the full note

5. Finding Areas Between Lines & Curves
7. Integrating e^x & 1/x
Exam Tips — Common Mistakes & Mark-Scheme Traps

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