Sec 4 A-Math Integration & Areas

Overview

What you’ll learn

Integrate as the reverse of differentiation, including the constant of integration.
Integrate xⁿ (any rational n), sin x, cos x, sec²x and eˣ.
Integrate (ax + b)ⁿ, sin(ax + b), cos(ax + b) and e^(ax+b).
Evaluate definite integrals.
Find areas between a curve and the x-axis, handling regions below the axis.
Find areas bounded by a curve and a line, using intersection points as the limits.

In 30 seconds

1Integration reverses differentiation: ∫f(x) dx is any function whose derivative is f(x), so an indefinite integral carries + C.
2Power rule backwards: ∫xⁿ dx = x^(n+1)/(n+1) + C for any rational n ≠ −1 — raise the power, divide by the new power.
3Standard integrals: cos x → sin x + C; sin x → −cos x + C; sec²x → tan x + C; eˣ → eˣ + C.
4Shifted arguments: ∫(ax + b)ⁿ dx = (ax + b)^(n+1)/(a(n+1)) + C — divide by the inner derivative: ∫f(ax + b) dx = (1/a)F(ax + b) + C.
5Definite integral ∫ₐᵇ f(x) dx = [F(b) − F(a)] — a number, no + C.
6Area under a curve above the x-axis = the definite integral; below the axis the integral is negative, so split and take |·|.
7Region between a curve y = f(x) and a line y = g(x): integrate f(x) − g(x) between the intersection points.
8Odd-symmetry shortcut: ∫₋ₐᵃ (odd function) dx = 0, e.g. ∫₋₁¹ x³ dx = 0.

Notes · 1 of 5

Integration as reverse differentiation

Ask "what differentiates to this?", and never forget + C.

To integrate a function, ask what it was before differentiation. Because differentiation destroys any added constant, every indefinite integral is a family of functions: ∫2x dx = x² + C, where C could be any number. Omitting + C in an indefinite integral costs a mark every time; definite integrals need no C because it cancels in F(b) − F(a).

The power rule runs backwards with one extra step: differentiate x^(n+1) to get (n+1)xⁿ, so ∫xⁿ dx = x^(n+1)/(n+1) + C. Fractions and roots go through the same door as powers: ∫3x⁻⁴ dx = −x⁻³ + C and ∫√x dx = (2/3)x^(3/2) + C. The one forbidden value is n = −1: x⁻¹ is 1/x, whose integral is ln|x| — not required in this syllabus, which is why every set question keeps n ≠ −1.

Notes · 2 of 5

Standard integrals and the (ax + b) forms

The table, and the divide-by-the-inner-derivative rule.

Standard integrals (k, a, b constants; a ≠ 0)
f(x)	∫f(x) dx
xⁿ (n ≠ −1)	x^(n+1)/(n+1) + C
eˣ	eˣ + C
cos x	sin x + C
sin x	−cos x + C
sec²x	tan x + C
(ax + b)ⁿ	(ax + b)^(n+1)/(a(n + 1)) + C
e^(ax+b)	e^(ax+b)/a + C
cos(ax+b)	sin(ax+b)/a + C
sin(ax+b)	−cos(ax+b)/a + C

The last four rows share one idea: integrating f(ax + b) divides by a, because the chain rule would multiply by a going the other way. So ∫(2x + 3)⁵ dx = (2x + 3)⁶/12 + C (2 × 6 in the denominator), and ∫e^(5x) dx = e^(5x)/5 + C. Note the sign trap on sine: ∫sin 2x dx = −½cos 2x + C — differentiate −½cos 2x to see (−½)(−2 sin 2x) = sin 2x ✓. And sec²x integrates to tan x only because tan′ = sec².

Notes · 3 of 5

Definite integrals

Substitute the limits into the antiderivative and subtract — the constant disappears.

The definite integral ∫ₐᵇ f(x) dx = [F(b) − F(a)]: an antiderivative evaluated at the top limit minus the same at the bottom. No + C survives. It measures signed area: positive where the curve is above the x-axis, negative below. A useful property: for an odd function (like x³ or sin x), ∫₋ₐᵃ = 0 because the two halves cancel.

Notes · 4 of 5

Areas under curves — and below the axis

The integral is signed; real area may need a split and a modulus.

When the curve stays above the x-axis on the interval, the area under it is just the definite integral. When it dips below, the integral there is negative and undercounts: split the interval at the x-axis crossings, integrate each piece, and take the absolute value of the negative parts before adding.

Between a curve y = f(x) and a line y = g(x), find the intersection points (solve f(x) = g(x)) and integrate the difference: area = ∫ₐᵇ (f(x) − g(x)) dx with the upper curve first. If you are unsure which is upper, subtract in either order and take the absolute value of the whole integral (valid when the curves do not cross inside the interval).

Exam tips

Worked example: crossing the axis twice. Find the total area between y = x² − 4 and the x-axis from x = 0 to x = 3. The curve crosses at x = 2 (from x² − 4 = 0). On [0, 2] the integral is [x³/3 − 4x]₀² = 8/3 − 8 = −16/3; the area is 16/3. On [2, 3]: (9 − 12) − (8/3 − 8) = 7/3. Total area = 16/3 + 7/3 = 23/3 ≈ 7.67. Quoting −16/3 + 7/3 as "the area" loses the point — the negative piece must be flipped.
Worked example: curve over a line. Find the area enclosed by y = 9 − x² and y = x + 3. Intersections: 9 − x² = x + 3 gives x² + x − 6 = 0, so (x + 3)(x − 2) = 0 and x = −3 or 2. The parabola is above the line between them: area = ∫₋₃²[(9 − x²) − (x + 3)] dx = ∫₋₃²(6 − x − x²) dx = [6x − x²/2 − x³/3]₋₃² = (12 − 2 − 8/3) − (−18 − 9/2 + 9) = 22/3 + 27/2 = 125/6 ≈ 20.8.

Notes · 5 of 5

Checks and shortcuts for area answers

Five-second sanity checks that catch sign slips before the examiner does.

Three habits catch most area errors. First, a negative final answer is a red flag: real area is positive, so somewhere a below-axis piece was not flipped. Second, sketch before integrating — the drawing reveals how many crossings split the region and which curve is upper. Third, for symmetric parabola arches, the two-thirds rule (area = ⅔ × base × height) verifies the integral: y = x(6 − x) from 0 to 6 has base 6, height 9, area ⅔ × 6 × 9 = 36, matching ∫₀⁶(6x − x²) dx = 36.

Revise

Key terms

The words examiners expect you to use precisely.

Constant of integration: The arbitrary constant + C attached to every indefinite integral, since any vertical shift differentiates away.
Indefinite integral: A family of functions differing by a constant; written ∫f(x) dx = F(x) + C.
Definite integral: The number ∫ₐᵇ f(x) dx = F(b) − F(a); interpreted as signed area under the curve.
Signed area: The definite integral counts area above the x-axis positively and below it negatively.
Limits of integration: The lower and upper values a and b at which the antiderivative is evaluated.
Antiderivative: A function F with F′ = f; any two antiderivatives of f differ by a constant.

Show 2 more key terms

Upper and lower curve: In an area-between-curves question, the boundary with the larger (upper) or smaller (lower) y-value on the interval.
Parabola arch: The area of a parabola arch equals two-thirds of the base times the height — a useful check on definite integrals.

Revise

Common mistakes

Where students most often lose marks in this chapter — and what to do instead.

Mistake: Omitting + C in an indefinite integral.
A missing constant of integration means the answer is one member of the family, not the family.
Instead: Every ∫…dx answer ends with + C; definite integrals need none.
Mistake: Forgetting to divide by the new power: ∫x³ dx = 3x⁴ + C.
The reverse power rule raises then divides; raising alone overcounts.
Instead: ∫x³ dx = x⁴/4 + C.

Show 8 more mistakes

Mistake: Integrating sin x as cos x + C.
The reverse of cos x is sine with a plus, so the reverse of sine needs a minus.
Instead: ∫sin x dx = −cos x + C; check by differentiating.
Mistake: Not dividing by the inner coefficient: ∫e^(5x) dx = e^(5x) + C.
Differentiating e^(5x) would multiply by 5, so integration must divide by 5.
Instead: ∫e^(5x) dx = e^(5x)/5 + C; likewise ∫cos 4x dx = sin 4x/4 + C.
Mistake: Writing the definite integral with + C.
The constant cancels in F(b) − F(a); carrying it suggests confusion about what a definite integral is.
Instead: ∫ₐᵇ f(x) dx = [F(x)]ₐᵇ = F(b) − F(a), a pure number.
Mistake: Quoting a negative definite integral as an area when the curve is below the axis.
Signed area is negative below the axis; real area cannot be negative.
Instead: Split at the x-axis crossings, integrate each part, take absolute values of the below-axis pieces.
Mistake: Using the y-axis crossing instead of solving f(x) = g(x) for area-between-curves limits.
The region between the curves begins and ends where the curves intersect, not where one of them meets an axis.
Instead: Solve f(x) = g(x) first; those roots are the limits.
Mistake: Subtracting in the wrong order between a curve and a line.
A wrong order just flips the sign, so the error shows up as a negative area — an immediate red flag.
Instead: Identify the upper curve on the interval; integrate upper − lower (or take |·| of the whole integral).
Mistake: Integrating each term of a fraction separately without rewriting as powers.
Terms like 1/x² hide a negative index; term-by-term integration on the displayed form invites errors.
Instead: Rewrite 1/x² as x⁻² (and √x as x^½) before integrating.
Mistake: Dropping brackets when substituting the limits.
Evaluating F(b) − F(a) needs the whole antiderivative at each limit; wrong subtraction of negatives is the classic slip.
Instead: Bracket each substitution: [F(b)] − [F(a)], keeping negatives inside the brackets.

Check yourself

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Apply the reverse power rule

∫x⁵ dx equals

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5 multiple-choice questions and structured questions with answer lines. Chapter 5 (C1 Integration: standard integrals, definite integrals, areas).

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Frequently asked questions

What does Sec 4 A-Math Chapter 5 cover?

The integration half of syllabus sub-topic C1 of O-Level 4049 / SEC G3 K341: reverse differentiation, standard integrals including eˣ, sin x, cos x and sec²x, the (ax + b) shifted forms, definite integrals, areas under and below curves, and areas between curves and lines. Differentiation is Chapter 4 and kinematics is Chapter 6.

When do I need + C?

Only on indefinite integrals — answers to ∫f(x) dx. A definite integral ∫ₐᵇ evaluates to a number, so no constant appears.

Why does the answer to ∫(2x + 3)⁵ dx divide by 12?

Two factors: the new power 6 (one more than 5) gives a 6 in the denominator, and the inner coefficient 2 contributes another (differentiating (2x+3)⁶/12 would multiply by 6 × 2 = 12). Always divide by the product of the new power and the inner coefficient.

What is the rule for areas below the x-axis?

The definite integral is negative there. Split the region at the x-axis crossings, integrate each piece separately, take absolute values of the negative parts, and add the areas.

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