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Higher Level · Need to Know

Integration.

The condensed rules — stripped of derivations. Every method, every formula, every watch-out you need on exam day.

How to use

Read the heading. Recall the rule in your head first — that's the bit that builds the memory. Then tap to reveal and check. No scoring, no pressure. Peek is allowed.

① Foundations

4 cards
Card 1

Anti-derivative — undoing differentiation

Differentiation goes one way. Anti-differentiation goes the other. If you can differentiate y = x2 to get dydx = 2x, you should be able to start with dydx = 2x and get back to y.

Two opposite moves:

  • Differentiate = Multiply by the power, then reduce the power by 1
  • Anti-differentiate = Increase the power by 1, then divide by the new power
∫ f′(x) dx = f(x) + c
Card 2

The constant + c — never forget it

If y = x2, then dydx = 2x. But also if y = x2 + 1000, then dydx = 2x — the derivative of any constant is 0.

When we anti-differentiate, we can't tell which constant was there. So we add an unknown constant c to cover all possibilities.

dydx = 2x  ⇒  y = x2 + c
c is a constant and must NOT be forgotten. Lose it and you lose 3 marks for one missing constant.
Card 3

Find f(x) when you know f′(x)

Same idea, but with function notation. Given f′(x), find f(x) by anti-differentiating each term.

Increase each power by 1, divide by the new power. The constant term (e.g. −7) becomes a term in x (e.g. −7x) — the anti-derivative of a constant.

f′(x) = x2 + 5x − 7  ⇒  f(x) = x33 + 5x22 − 7x + c
Card 4

Finding c — when you know one point

If the question gives you a point on the curve — like f(1) = 11 — you can pin down c exactly.

Three-step recipe
  1. Anti-differentiate to get f(x) with + c
  2. Sub in the given point to find c
  3. Write f(x) with the actual c

② The 8 Methods

11 cards
Method 1

Multiply out

If you see two brackets multiplied together — (x + a)(x + b) or (x + a)2 — expand them first. Never try to integrate the brackets directly.

The trap. ∫(x + 1)2 dx ≠ (x + 1)33 + c.   The power rule only works on bare powers of x, not on brackets. Always expand first.
Method 2

Factors

If you see a fraction with a polynomial top and a polynomial bottom, look for a common factor that cancels. Once you cancel, the rest is standard.

a2 − b2 = (a − b)(a + b)
a3 − b3 = (a − b)(a2 + ab + b2)
a3 + b3 = (a + b)(a2 − ab + b2)

Difference of squares · difference of cubes · sum of cubes.

Method 3

Indices

Anything that isn't already in xn form — rewrite it. Then the power rule does the work.

1xn = x−n  ·  flip → negative power
√x = x1/2  ·  root → fractional power
n√x = x1/n  ·  nth root → 1/n power

Once it's in xn form, increase the power by 1, divide by the new power.

Method 4

Split the fraction

One thing on the bottom, many on the top. Split the fraction across the top — every term gets the same denominator.

a + b + cd = ad + bd + cd
Only when the single thing is on the bottom. Never the other way around — splitting across the bottom is a famous mistake.

After splitting, any 1x piece integrates to ln x.

Method 5

From the Tables

Some integrals aren't done by rule — they're looked up in the formula booklet. The relevant page is the integration table. Worth knowing the most common ones by heart anyway.

∫ xn dx = xn+1n+1   (n ≠ −1)
∫ 1x dx = ln |x|
∫ ex dx = ex  ·  ∫ cos x dx = sin x
∫ sin x dx = −cos x  ·  ∫ tan x dx = ln |sec x|

The trig and inverse-trig ones get their own cards. First nail 1x and the exponentials.

Method 5a

The 1x rule  →  ln x

This one's special. The power rule fails on 1x because:

∫ x−1 dx would give x00   — dividing by 0 is undefined.

So the power rule has a hole exactly when n = −1. The Tables fill that hole:

∫ 1x dx = ln x + c

The exam tables list ln |x|. For positive x — the only kind that comes up at HL — it's just ln x.

Method 5b

Tables — inverse trig forms

Two more from the tables, and these have to come from the tables — you'd never spot them otherwise.

∫ 1x2 + a2 dx = 1a tan−1 xa + c

∫ 1√(a2 − x2) dx = sin−1 xa + c

a is any positive constant. Match the form, identify a, plug in.

Two traps to avoid. Don't try indices — the power rule doesn't apply to brackets. Don't try to factorise — x2 + a2 is irreducible over the real numbers.
Method 6

ex, eax and ax

From the tables:

∫ ex dx = ex + c
∫ eax dx = 1a eax + c
∫ ax dx = axln a + c

If you see 1e3x, flip first: 1e3x = e−3x. Then use the rule with a = −3 (same idea as indices).

Method 7a

Trig with a coefficient

From the tables, ∫ sin x dx = −cos x + c  and  ∫ cos x dx = sin x + c. But what if the angle is 3x or 5x instead of just x?

∫ cos ax dx = 1a sin ax + c
∫ sin ax dx = −1a cos ax + c

Integrate as normal, then divide by the coefficient a.

Method 7b

Trig: Products → Sums

You cannot integrate a product of trig functions directly. ∫ cos 7x cos x dx is not ∫ cos 7x dx × ∫ cos x dx. There's no product rule for integration.

The trick: convert the product into a sum first, using the formulas from the tables. Then integrate term by term.

2 cos A cos B = cos(A − B) + cos(A + B)
2 sin A cos B = sin(A + B) + sin(A − B)
2 sin A sin B = cos(A − B) − cos(A + B)
2 cos A sin B = sin(A + B) − sin(A − B)
Each formula starts with 2. If your integrand isn't 2 × something, multiply by 22 — keep the 12 out front, the 2 on the inside.
Method 8

Special case (reverse product rule)

Two integrals you'll meet that none of the other methods touch: ∫ xex dx  and  ∫ ln x dx.

The exam gives you a helper part — differentiate something first, then rearrange that result to make the integral fall out.

The recipe
  1. Differentiate the function the question hands you (usually y = xex or f(x) = x ln x) using the product rule.
  2. Spot the awkward integrand somewhere inside that derivative.
  3. Rearrange to isolate it, then integrate both sides.

Memorise the results:

∫ xex dx = xex − ex + c
∫ ln x dx = x ln x − x + c

③ Definite Integrals

1 card
Card 1

Definite integral — notation and method

Area under a curve = anti-derivative at the right end − anti-derivative at the left end.

∫ba f(x) dx = [ F(x) ]ba = F(b) − F(a)

F(x) is any anti-derivative of f(x). a is the lower limit, b the upper.

Method
  1. Integrate
  2. Sub in TOP value
  3. Sub in BOTTOM value
  4. Subtract: top − bottom

No + c needed. The constants would cancel: (F(b) + c) − (F(a) + c) = F(b) − F(a).

Care with signs. Always use brackets when subtracting — especially when F(a) is itself negative.

④ Area

5 cards
Card 1

Area under a curve

From here on, the question usually says: "Find the area of the region bounded by...". Recipe doesn't change — integrate, evaluate, subtract. Just keep track of which curve and which limits.

A = ∫ba f(x) dx = ∫ba y dx

Always draw a diagram first. If the question doesn't give limits, find where the curve cuts the x-axis (let y = 0) — those are your limits.

Card 2

When the curve dips below — split and mod

Area is always positive. The integral can be negative. If the curve goes below the x-axis, the integral over that bit comes out negative — because we're "subtracting" heights below.

The trick: split the integral at the crossing point, and take the modulus (absolute value) of each piece. Then add them.

The rule
  1. Find where the curve crosses the x-axis. That's your split point.
  2. Integrate each piece separately.
  3. Take the modulus of any negative result.
  4. Add them.
Famous mistake. Integrating in one go from a to b would let the two halves cancel — giving zero or a wrong figure. The integral is right, the area is wrong.
Card 3

Area between two curves — top minus bottom

When the region is bounded above by one curve and below by another, the area is the integral of the difference. Top curve minus bottom curve.

A = ∫ba ( ftop(x) − fbottom(x) ) dx

a and b are the x-values where the curves meet — solve ftop = fbottom to find them.

Same rule for line + curve, curve + tangent, two parabolas. Always top minus bottom.

Card 4

Integration with respect to y

Sometimes the natural variable is y, not x — for example when the region is bounded by the y-axis and horizontal lines. The formula flips: integrate x as a function of y, with respect to y.

A = ∫ba x dy

Rearrange the equation so x is the subject. The limits are y-values, not x-values.

Use whichever is easier — typically y-integration when the curve is most naturally written as x in terms of y.

Card 5

The y = xn ratio  —  B : A = n : 1

A lovely result. Take the curve y = xn for n > 0, between x = 0 and x = 1. It goes from (0, 0) to (1, 1) and divides the unit square into two regions:

  • Region A: between the curve and the x-axis (below the curve)
  • Region B: between the curve and the y-axis (above the curve, inside the unit square)
B : A = n : 1

The region above the curve (inside the unit square) is n times the region below it.

Check. For y = x (n = 1): 1 : 1 — diagonal of a square, each triangle is half. ✓ For y = x2 (n = 2): 2 : 1 — below the parabola is 13, above is 23. ✓

⑤ Applications

3 cards
Card 1

The Trapezoidal Rule

Useful when integration doesn't work neatly — or when you only have measured heights, not a formula.

A ≈ h2 [ y1 + yn + 2(y2 + y3 + … + yn−1) ]

Half the gap × [first + last + twice the rest].

  • y1 = first ordinate  ·  yn = last ordinate
  • The ones in the middle get doubled
  • h = strip width
Counting strips and ordinates. If you have n ordinates (vertical lines), you have n − 1 strips. Easy to mix up — count carefully.
If total width is W, then h = Wn−1.

It's an estimate. The curved top of each strip isn't exactly a straight line, so the answer is approximate — but close, and good enough for the exam.

Card 2

Motion — distance, velocity, acceleration

Velocity is the rate of change of distance. Acceleration is the rate of change of velocity. So integration takes you the other way — from acceleration back to velocity, from velocity back to distance.

The chain
  1. Differentiate down:   distance → velocity → acceleration
  2. Integrate up:   acceleration → velocity → distance
  • s = distance / displacement  ·  s = f(t)
  • v = dsdt = velocity / speed
  • a = dvdt = d2sdt2 = acceleration

Conditions:

  • Initial = at start = t = 0
  • At rest / max distance: v = 0
Don't forget + c — use the initial condition to find it.
Card 3

Average value of a function

A function changes as you move through its domain. What's its average value over an interval [a, b]?

Idea: total accumulated value divided by length of interval. The total is the integral.

f̄ = 1b − a ∫ba f(x) dx

Total area ÷ width = average height.

Think of it as: a rectangle of width (b − a) and height f̄ has the same area as the region under the curve.

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Integration · Need to Know