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CCEA GCE Mathematics (2210) · Integration
Mini-Lesson

Integration

This mini-lesson covers indefinite and definite integration, areas under and between curves, integration by substitution and by parts, integrating with partial fractions, and solving differential equations by separating the variables.

Where this sits: Unit AS 1: Pure Mathematics — integration as the reverse of differentiation, indefinite and definite integrals, area under a curve. Unit A2 1 — substitution, integration by parts, partial fractions, and differential equations.

Work through each screen, answer the questions as you go (some are reasoning, some are calculations) and collect ⭐ stars. Press Start when you are ready.

Foundations

Indefinite and definite integrals

∫ xⁿ dx = xⁿ⁺¹/(n + 1) + c, n ≠ −1∫ 1/x dx = ln|x| + c · ∫ eᵏˣ dx = eᵏˣ/k + c · ∫ sin x dx = −cos x + c · ∫ cos x dx = sin x + c

Integration reverses differentiation, so an indefinite integral always needs + c. A definite integral has limits and gives a number:

Worked example

∫₁³ (3x² + 2) dx = [x³ + 2x]₁³

= (27 + 6) − (1 + 2) = 33 − 3 = 30

Why ln|x|? The modulus lets the result cover negative x, where 1/x is still defined.

Calculate

Definite integral

1Evaluate ∫₁³ (3x² + 2) dx.
Hint: Integrate: [x³ + 2x] from 1 to 3 = (27 + 6) − (1 + 2) = 33 − 3.
Calculate

Exponential integral

2Evaluate ∫₀¹ e^(2x) dx, to 3 significant figures.
Hint: ∫e^(2x) dx = e^(2x)/2, so the value is (e² − e⁰)/2 = (7.389 − 1)/2 = 3.1945.
Applications

Area under and between curves

The definite integral ∫ₐᵇ y dx gives the signed area between the curve and the x-axis.

  • Area below the axis comes out negative. For a total area, split at the roots and take the modulus of each piece.
  • Area between two curves = ∫ (upper − lower) dx between the intersections.
Worked example — between curves

y = 4 − x² and y = x + 2 meet where 4 − x² = x + 2 ⇒ x² + x − 2 = 0 ⇒ x = −2, 1.

Area = ∫₋₂¹ [(4 − x²) − (x + 2)] dx = ∫₋₂¹ (2 − x − x²) dx

= [2x − x²/2 − x³/3] from −2 to 1 = (2 − 0.5 − 1/3) − (−4 − 2 + 8/3) = 1.1667 − (−3.3333) = 4.5

Quick check

Negative area

?∫₀² (x² − 4) dx comes out negative. Why?
A2 technique

Integration by substitution

Substitution reverses the chain rule. Change the variable, and change the limits if the integral is definite.

Worked example — ∫₀² 2x(x² + 1)³ dx

Let u = x² + 1 ⇒ du/dx = 2x ⇒ du = 2x dx

Limits: x = 0 ⇒ u = 1; x = 2 ⇒ u = 5

Integral becomes ∫₁⁵ u³ du = [u⁴/4]₁⁵ = (625 − 1)/4 = 156

Look for the pattern ∫ f′(x)[f(x)]ⁿ dx — the derivative of the inside is sitting outside, ready to be absorbed.

Calculate

Substitution

3Use the substitution u = x² + 1 to evaluate ∫₀² 2x(x² + 1)³ dx.
Hint: du = 2x dx and the limits become u = 1 to u = 5, so the integral is ∫₁⁵ u³ du = [u⁴/4] = (625 − 1)/4.
A2 technique

Integration by parts

∫ u (dv/dx) dx = uv − ∫ v (du/dx) dxchoose u to be the part that gets SIMPLER when differentiated (LATE: Logs, Algebra, Trig, Exponentials)
Worked example — ∫₀¹ x eˣ dx

u = x (so du/dx = 1); dv/dx = eˣ (so v = eˣ).

= [x eˣ]₀¹ − ∫₀¹ eˣ dx = [x eˣ − eˣ]₀¹

= (e − e) − (0 − 1) = 1

Trick: ∫ ln x dx is by parts with u = ln x and dv/dx = 1, giving x ln x − x + c.

Calculate

By parts

4Evaluate ∫₀¹ x eˣ dx exactly.
Hint: By parts with u = x, dv/dx = eˣ: [xeˣ − eˣ]₀¹ = (e − e) − (0 − 1) = 1.
A2 technique

Differential equations

Separate the variables, integrate both sides, then use the initial condition to find c.

Worked example — dy/dx = 2xy with y = 1 when x = 0

(1/y) dy = 2x dx

∫(1/y) dy = ∫2x dx ⇒ ln|y| = x² + c

x = 0, y = 1 ⇒ ln 1 = 0 + c ⇒ c = 0

So ln y = x² ⇒ y = e^(x²). At x = 2: y = e⁴ = 54.6 (3 s.f.)

Growth/decay models come straight from dy/dt = ky, whose solution is y = Ae^(kt) — the same exponential model as before.

Calculate

Differential equation

5Solve dy/dx = 2xy with y = 1 when x = 0, then find y when x = 2, to 3 significant figures.
Hint: Separating gives ln y = x² + c with c = 0, so y = e^(x²). At x = 2, y = e⁴ = 54.598.
Sort it

Which method of integration?

Tap an integral, then tap the technique it needs.

🔄 Substitution

🧩 By parts

📗 Standard integral

Quick check

Choosing u

?For ∫ x ln x dx, which choice for u is correct?
Match it

Match the standard integral

Tap the integral on the left, then its result on the right.

Statement
Answer
Quick check

Partial fractions first

?To integrate ∫ (7x − 4)/((x − 2)(2x + 1)) dx, what should you do first?
Quick check

Constant of integration

?Why does ∫ f(x) dx need + c?
Examiner traps

Calculus pitfalls (integration)

  • Missing + c in every indefinite integral.
  • ∫ sin x dx = −cos x + c — the minus sign is lost every year.
  • Substitution: change the limits to the new variable, or convert back before substituting.
  • Negative area: if the curve dips below the axis, split at the roots and take moduli — otherwise areas cancel.
  • Area between curves: ∫(upper − lower) dx, with the limits at the intersections.
  • Differential equations: use the initial condition to find c before answering the question.
Recap

The big ideas to know

Standard: ∫xⁿ dx = xⁿ⁺¹/(n+1) + c (n ≠ −1) · ∫1/x dx = ln|x| + c · ∫eᵏˣ dx = eᵏˣ/k + c

Definite integrals give signed area — split at roots for a true area

Area between curves = ∫(upper − lower) dx between the intersections

Substitution reverses the chain rule — change the limits too

By parts: ∫u(dv/dx) dx = uv − ∫v(du/dx) dx — choose u to simplify on differentiating

Differential equations: separate variables, integrate, use the initial condition to find c

Integration powers area, volume and — in mechanics — the link from acceleration back to velocity and displacement. Press Finish to see your score.

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