Integration | Ronny Best Mathematics — CBC-Aligned Calculus Lessons

Foundations

What Is Integration?

All Levels

Integration answers two related questions: "What function, when differentiated, gives me this?" and "What is the total accumulated quantity represented by this rate of change?" The first question leads to the indefinite integral (a family of antiderivatives), and the second leads to the definite integral (a specific numerical value, often interpreted as an area).

Think of differentiation and integration as opposite operations, much like multiplication and division undo each other. If a car's velocity function is v(t), then differentiating its position function s(t) gives v(t). Working backwards — integrating v(t) — recovers s(t), up to an unknown constant representing the car's starting position. That unknown constant is precisely why every indefinite integral carries a "+ C".

The Fundamental Theorem of Calculus

This single theorem links the two branches of calculus. It states that if F is an antiderivative of a continuous function f on an interval, then the definite integral of f over that interval can be computed simply by evaluating F at the endpoints and subtracting. This means we rarely need to think about "area under a curve" as a sum of infinitely many rectangles in practice — we just find an antiderivative and substitute.

Fundamental Theorem of Calculus

              If F′(x) = f(x), then ∫ f(x) dx = F(x) + C   (indefinite integral)

              ∫ₐᵇ f(x) dx = F(b) − F(a)   (definite integral)

Why "+ C" matters: Because the derivative of any constant is zero, every antiderivative of f(x) differs only by a constant. Forgetting the constant of integration is one of the most common mistakes in KCSE — examiners specifically award a mark for including it in indefinite integrals (though it cancels out and is omitted in definite integrals).

Foundations

Notation & Terminology

All Levels

Before working through methods, it helps to be fluent in the language of integration. The table below summarises the key terms you will meet throughout this lesson and in examinations.

Term Meaning

∫ f(x) dx The indefinite integral — a family of antiderivatives of f, written as F(x) + C

∫ₐᵇ f(x) dx The definite integral — a single number representing net signed area between a and b

dx Indicates the variable of integration; "with respect to x"

C The constant of integration — represents the unknown initial value

Integrand The function being integrated, f(x)

Limits of integration The values a (lower) and b (upper) in a definite integral

Common confusion: Students often treat dx as decoration and drop it. In substitution and multiple integrals especially, dx (or dy, dθ, dr) is essential — it tells you which variable the integration is being performed with respect to, and it transforms correctly under substitution (du = g′(x) dx).

Secondary · KCSE

Standard Integrals

Secondary

Every integration technique eventually reduces to a handful of standard results. Memorising these — and recognising when an integrand matches one of them after simplification — is the single most useful skill for KCSE Paper 2 calculus questions.

The Power Rule

The power rule for integration is the reverse of the power rule for differentiation: increase the exponent by one, then divide by the new exponent. This works for any real exponent n except n = −1, where the result would involve dividing by zero — that special case integrates to a natural logarithm instead.

Power & Exponential Rules

Power rule (n ≠ −1)∫ xⁿ dx = xⁿ⁺¹ / (n+1) + C

Special case (n = −1)∫ x⁻¹ dx = ∫ (1/x) dx = ln|x| + C

Constant∫ k dx = kx + C

Natural exponential∫ eˣ dx = eˣ + C

General exponential∫ aˣ dx = aˣ / ln a + C (a > 0, a ≠ 1)

Trigonometric Integrals

Trigonometric Rules

Sine∫ sin x dx = −cos x + C

Cosine∫ cos x dx = sin x + C

Secant squared∫ sec²x dx = tan x + C

Cosecant squared∫ csc²x dx = −cot x + C

Sec × tan∫ sec x tan x dx = sec x + C

Csc × cot∫ csc x cot x dx = −csc x + C

Inverse Trigonometric Forms

Inverse Trig Rules

Arcsine∫ 1 / √(1 − x²) dx = arcsin x + C

Arctangent∫ 1 / (1 + x²) dx = arctan x + C

Worked Examples

Example 1 ∫ (3x² − 2x + 5) dx

Step 1 Integrate each term separately using the power rule.

Step 2 ∫3x² dx = x³, ∫−2x dx = −x², ∫5 dx = 5x

Check Differentiate the result: d/dx(x³ − x² + 5x) = 3x² − 2x + 5 ✔

= x³ − x² + 5x + C

Example 2 ∫ (4√x + 3/x²) dx

Step 1 Rewrite roots and reciprocals as powers: √x = x^(1/2), 1/x² = x⁻²

Step 2 ∫4x^(1/2) dx = 4 · x^(3/2)/(3/2) = (8/3)x^(3/2)

Step 3 ∫3x⁻² dx = 3 · x⁻¹/(−1) = −3x⁻¹ = −3/x

= (8/3)x^(3/2) − 3/x + C

Example 3 ∫ (2eˣ − 1/x) dx

Step 1 ∫2eˣ dx = 2eˣ (exponential rule)

Step 2 ∫−(1/x) dx = −ln|x| (the n = −1 special case)

= 2eˣ − ln|x| + C

Secondary · KCSE

Definite Integrals & Area Under a Curve

Secondary

A definite integral evaluates to a single real number. Geometrically, when f(x) ≥ 0 on [a, b], this number equals the area between the curve y = f(x), the x-axis, and the vertical lines x = a and x = b. If f(x) is negative on part of the interval, that portion contributes a negative value — the definite integral gives net signed area, not total area.

Evaluation Procedure

Find an antiderivative F(x) of f(x) (omit "+ C" — it cancels in the subtraction).
Write the result in square brackets with limits: [F(x)]ₐᵇ
Substitute the upper limit b into F(x), then subtract F(a) (lower limit substituted).
Simplify to a single numerical value.

Worked Examples

Example 1 Evaluate ∫₁³ (x² + 1) dx

Step 1 Antiderivative: F(x) = x³/3 + x

Step 2 F(3) = 27/3 + 3 = 9 + 3 = 12

Step 3 F(1) = 1/3 + 1 = 4/3

Step 4 F(3) − F(1) = 12 − 4/3 = 36/3 − 4/3 = 32/3

= 32/3

Example 2 — Negative area Evaluate ∫₀² (x² − 2x) dx and interpret the sign

Step 1 F(x) = x³/3 − x²

Step 2 F(2) = 8/3 − 4 = −4/3

Step 3 F(0) = 0

Step 4 F(2) − F(0) = −4/3

Note The curve lies below the x-axis on (0, 2), so the integral is negative. The actual area is |−4/3| = 4/3 square units.

Integral = −4/3 (Area = 4/3 sq. units)

Sketch first: When a question asks for "area" rather than "the value of the integral," always sketch the curve to identify where it crosses the x-axis. Split the integral at each crossing point and take the absolute value of each piece before summing — otherwise regions below the axis will cancel regions above it.

Secondary / A-Level

Integration by Substitution

Secondary / A-Level

Substitution is the integration analogue of the chain rule. Whenever the integrand is a composite function multiplied by the derivative of its "inner function" (possibly up to a constant), substitution simplifies it into a standard integral.

Method

Identify an inner function and let u equal it.
Differentiate to find du = g′(x) dx, and rearrange to express dx in terms of du.
Rewrite the entire integral in terms of u only — no x should remain.
Integrate using standard results, then substitute u back in terms of x.
For definite integrals, either convert the limits to u-values, or substitute back to x before evaluating.

Worked Examples

Example 1 ∫ 2x(x²+1)⁵ dx

Step 1 Let u = x² + 1, so du = 2x dx — this matches the 2x already present.

Step 2 The integral becomes ∫u⁵ du = u⁶/6 + C

Step 3 Substitute back u = x² + 1

= (x²+1)⁶ / 6 + C

Example 2 — Constant adjustment ∫ cos(3x+2) dx

Step 1 Let u = 3x + 2, so du = 3 dx, meaning dx = du/3

Step 2 ∫cos u · (du/3) = (1/3)∫cos u du = (1/3) sin u

Step 3 Substitute back u = 3x + 2

= (1/3) sin(3x+2) + C

Example 3 — Definite with limit change Evaluate ∫₀¹ x(x²+1)³ dx

Step 1 Let u = x² + 1, du = 2x dx, so x dx = du/2

Step 2 Change limits: when x = 0, u = 1; when x = 1, u = 2

Step 3 Integral becomes (1/2)∫₁² u³ du = (1/2)[u⁴/4]₁² = (1/8)(16 − 1)

= 15/8

Example 4 — Trigonometric substitution ∫ 1/√(4−x²) dx

Step 1 Let x = 2 sin θ, so dx = 2 cos θ dθ and √(4−x²) = 2 cos θ

Step 2 Integral becomes ∫ (2 cos θ)/(2 cos θ) dθ = ∫ dθ = θ

Step 3 Since x = 2 sin θ, θ = arcsin(x/2)

= arcsin(x/2) + C

Spotting substitution opportunities: Look for a function "inside" another function (e.g. inside a bracket, square root, exponent, or trig function) whose derivative — or a constant multiple of it — also appears in the integrand. If you can account for every factor of x this way, substitution will work.

A-Level

Integration by Parts

A-Level

Integration by parts is the reverse of the product rule for differentiation. It is used when the integrand is a product of two different types of function (e.g. a polynomial times an exponential, or a polynomial times a trig function) and substitution does not simplify it.

Integration by Parts Formula
∫ u dv = uv − ∫ v du

Choosing u and dv — the LIATE Rule

The hardest part of this method is deciding which factor is u (to be differentiated) and which is dv (to be integrated). The LIATE rule gives a priority order for choosing u: pick whichever type appears earliest in the list below.

L — Logarithmic

e.g. ln x, log x — choose these as u first

I — Inverse Trig

e.g. arctan x, arcsin x

A — Algebraic

e.g. x, x², x³ — polynomials

T — Trigonometric

e.g. sin x, cos x

E — Exponential

e.g. eˣ, aˣ — choose these as dv first

Worked Examples

Example 1 ∫ x eˣ dx

Step 1 By LIATE, x is Algebraic and eˣ is Exponential, so u = x (differentiate), dv = eˣ dx (integrate)

Step 2 du = dx, v = eˣ

Step 3 ∫x eˣ dx = x eˣ − ∫eˣ dx = x eˣ − eˣ

= (x − 1)eˣ + C

Example 2 — Applying twice ∫ x² sin x dx

Step 1 First application: u = x², dv = sin x dx → du = 2x dx, v = −cos x

Step 2 = −x² cos x + ∫2x cos x dx

Step 3 Second application on ∫2x cos x dx: u = 2x, dv = cos x dx → du = 2dx, v = sin x

Step 4 ∫2x cos x dx = 2x sin x − ∫2 sin x dx = 2x sin x + 2 cos x

= −x² cos x + 2x sin x + 2 cos x + C

Example 3 — The "invisible" dv ∫ ln x dx

Step 1 Write ln x as ln x · 1. Let u = ln x, dv = dx

Step 2 du = (1/x) dx, v = x

Step 3 ∫ln x dx = x ln x − ∫x · (1/x) dx = x ln x − ∫1 dx

= x ln x − x + C

When integration by parts seems to loop: For integrals like ∫eˣ sin x dx, applying the formula twice returns the original integral on the right-hand side. Treat the original integral as an unknown I, set up the equation I = (...) − I, and solve algebraically for I — this is a standard A-Level technique that surprises many students the first time they meet it.

A-Level

Integration by Partial Fractions

A-Level

Rational functions (one polynomial divided by another) can often be "split" into a sum of simpler fractions whose integrals are standard logarithmic forms. This technique connects directly to the Algebra strand's work on partial fraction decomposition.

Decomposition Patterns

Denominator Type Decomposition Form

(x−a)(x−b) A/(x−a) + B/(x−b)

(x−a)² A/(x−a) + B/(x−a)²

(x−a)(x²+bx+c) A/(x−a) + (Bx+C)/(x²+bx+c) irreducible quadratic

Worked Examples

Example 1 — Distinct linear factors ∫ (3x+2) / [(x−1)(x+2)] dx

Step 1 Write (3x+2)/[(x−1)(x+2)] = A/(x−1) + B/(x+2)

Step 2 Multiply through: 3x+2 = A(x+2) + B(x−1)

Step 3 Let x = 1: 5 = 3A → A = 5/3. Let x = −2: −4 = −3B → B = 4/3

Step 4 Integrate each term: (5/3)ln|x−1| + (4/3)ln|x+2|

= (5/3)ln|x−1| + (4/3)ln|x+2| + C

Example 2 — Repeated factor ∫ 1 / [x(x−1)²] dx

Step 1 Decompose: 1/[x(x−1)²] = A/x + B/(x−1) + C/(x−1)²

Step 2 Multiply through and compare coefficients/substitute convenient values: A = 1, B = −1, C = 1

Step 3 Integrate: ln|x| − ln|x−1| − 1/(x−1)

= ln|x| − ln|x−1| − 1/(x−1) + C

Check the degree first: Partial fractions only apply directly when the numerator's degree is less than the denominator's. If not, perform polynomial long division first to separate a polynomial part from a proper fraction remainder, then decompose the remainder.

Secondary / A-Level

Applications of Integration

Secondary / A-Level

Area Between Two Curves

When two curves intersect, the area enclosed between them on an interval [a, b] is found by integrating the difference of the "upper" and "lower" functions. The intersection points, found by solving f(x) = g(x), become the limits of integration.

Area Between Curves

A = ∫ₐᵇ [f(x) − g(x)] dx where f(x) ≥ g(x) on [a, b]

Example Find the area between y = x and y = x²

Step 1 Find intersections: x = x² → x² − x = 0 → x = 0 or x = 1

Step 2 On [0,1], y = x lies above y = x² (test x = 0.5: 0.5 > 0.25)

Step 3 A = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = (1/2 − 1/3) − 0

A = 1/6 square units

Volumes of Revolution

Rotating a region about an axis sweeps out a solid of revolution. The disc method treats the solid as a stack of thin circular discs (used when rotating about the x-axis), while the shell method treats it as a series of nested cylindrical shells (often more convenient when rotating about the y-axis).

Volumes of Revolution

Disc method (about x-axis)V = π ∫ₐᵇ [f(x)]² dx

Shell method (about y-axis)V = 2π ∫ₐᵇ x f(x) dx

Example Rotate y = √x, 0 ≤ x ≤ 4, about the x-axis. Find the volume.

Step 1 [f(x)]² = (√x)² = x

Step 2 V = π ∫₀⁴ x dx = π [x²/2]₀⁴ = π(16/2 − 0)

V = 8π cubic units

Arc Length and Surface Area

The length of a curve and the surface area generated by rotating it both arise from integrating expressions involving the derivative f′(x). These integrals are conceptually straightforward but often require substitution or numerical methods to evaluate in practice.

Arc Length & Surface of Revolution

Arc lengthL = ∫ₐᵇ √[1 + (f′(x))²] dx

Surface of revolutionS = 2π ∫ₐᵇ f(x) √[1 + (f′(x))²] dx

Kinematics: Displacement, Velocity and Acceleration

Integration reverses the differentiation chain used in motion problems. Given an acceleration function a(t), integrating once (with an initial velocity condition) recovers velocity v(t); integrating again (with an initial displacement condition) recovers displacement s(t). This is a recurring KCSE Paper 2 application.

Example A particle has acceleration a(t) = 6t − 4 m/s². If v(0) = 2 m/s and s(0) = 0, find s(t).

Step 1 v(t) = ∫(6t − 4) dt = 3t² − 4t + C₁. Since v(0) = 2, C₁ = 2 → v(t) = 3t² − 4t + 2

Step 2 s(t) = ∫(3t² − 4t + 2) dt = t³ − 2t² + 2t + C₂. Since s(0) = 0, C₂ = 0

s(t) = t³ − 2t² + 2t

University

Improper Integrals

University

An integral is "improper" if at least one limit of integration is infinite, or if the integrand becomes unbounded somewhere within the interval. Such integrals are evaluated as limits, and they may either converge (settle to a finite value) or diverge (grow without bound).

Definitions

Infinite upper limit∫ₐ^∞ f(x)dx = lim(t→∞) ∫ₐᵗ f(x)dx

Both limits infinite∫₋∞^∞ f(x)dx = ∫₋∞⁰ f(x)dx + ∫₀^∞ f(x)dx

Worked Examples

Example 1 — Converges ∫₁^∞ 1/x² dx

Step 1 Rewrite as a limit: lim(t→∞) ∫₁ᵗ x⁻² dx = lim(t→∞) [−1/x]₁ᵗ

Step 2 = lim(t→∞) (−1/t + 1) = 0 + 1

= 1 (the integral converges)

Example 2 — Diverges ∫₁^∞ 1/x dx

Step 1 lim(t→∞) ∫₁ᵗ x⁻¹ dx = lim(t→∞) [ln x]₁ᵗ = lim(t→∞) ln t

= ∞ (the integral diverges)

The p-test: ∫₁^∞ 1/xᵖ dx converges if p > 1 and diverges if p ≤ 1. This single result — compare the two examples above where p = 2 and p = 1 — is one of the most frequently used facts in introductory analysis.

University

Numerical Integration

University

Many real integrals have no elementary antiderivative, or the antiderivative is too complex to be practical. Numerical methods approximate the value of a definite integral by sampling the function at evenly spaced points and combining the values with appropriate weights.

The Trapezium Rule

The trapezium rule approximates the area under the curve by replacing each strip with a trapezium rather than a rectangle, giving a better estimate than simple rectangle sums for smooth functions.

Trapezium Rule

∫ₐᵇ f(x)dx ≈ (h/2)[f(x₀) + 2f(x₁) + 2f(x₂) + ... + 2f(x_{n−1}) + f(xₙ)], h = (b−a)/n

Simpson's Rule

Simpson's rule fits a parabola through each pair of strips, giving significantly higher accuracy for the same number of points — but it requires the number of strips n to be even.

Simpson's Rule (n even)

∫ₐᵇ f(x)dx ≈ (h/3)[f(x₀) + 4f(x₁) + 2f(x₂) + 4f(x₃) + ... + 4f(x_{n−1}) + f(xₙ)]

Worked Example

Example Use the trapezium rule with n = 4 strips to estimate ∫₀² x² dx

Step 1 h = (2−0)/4 = 0.5. Points: x₀=0, x₁=0.5, x₂=1, x₃=1.5, x₄=2

Step 2 f-values: f(0)=0, f(0.5)=0.25, f(1)=1, f(1.5)=2.25, f(2)=4

Step 3 Sum = f(x₀)+f(x₄)+2[f(x₁)+f(x₂)+f(x₃)] = (0+4)+2(0.25+1+2.25) = 4 + 7 = 11

Step 4 Estimate = (h/2) × Sum = (0.5/2) × 11 = 2.75

Check Exact value: ∫₀² x² dx = [x³/3]₀² = 8/3 ≈ 2.667 — the trapezium estimate is close, slightly overestimating because x² is convex.

≈ 2.75 (exact = 8/3 ≈ 2.667)

University

Multiple Integrals

University

Single integrals accumulate quantities along a line. Double and triple integrals extend the same idea to regions in the plane and solids in space — accumulating quantities such as area, volume, mass, or average value over two- or three-dimensional domains.

Double, Triple, Polar & Jacobian

Double integral∬_R f(x,y) dA = ∫∫ f(x,y) dy dx

Triple integral∭_V f(x,y,z) dV

Polar coordinates∬ f(r,θ) r dr dθ

Jacobian (change of variables)∂(x,y)/∂(u,v) — scaling factor for area under substitution

Evaluating Iterated Integrals

A double integral over a rectangular region is evaluated by integrating "inside-out": treat the inner variable as the variable of integration while holding the outer variable constant, producing a function of the outer variable alone, then integrate that result.

Worked Example

Example Evaluate ∫₀¹ ∫₀¹ (x + y) dy dx

Step 1 — Inner integral Treat x as constant: ∫₀¹ (x+y) dy = [xy + y²/2]₀¹ = x + 1/2

Step 2 — Outer integral ∫₀¹ (x + 1/2) dx = [x²/2 + x/2]₀¹ = (1/2 + 1/2) − 0

= 1

Why polar coordinates? When a region is circular or has radial symmetry (e.g. a disc x²+y² ≤ R²), converting to polar coordinates with the extra factor of r often turns an otherwise impossible Cartesian double integral into an elementary one.

Quick Reference

Integration Formula Sheet

All Levels

A consolidated reference of every result covered in this lesson — useful for revision before KCSE Paper 2 or university calculus assessments.

Power & Exponential

∫xⁿdx = xⁿ⁺¹/(n+1) + C (n≠−1)

∫(1/x)dx = ln|x| + C

∫eˣdx = eˣ + C

∫aˣdx = aˣ/ln a + C

Trigonometric

∫sin x dx = −cos x + C

∫cos x dx = sin x + C

∫sec²x dx = tan x + C

∫csc²x dx = −cot x + C

Inverse Trig

∫1/√(1−x²)dx = arcsin x + C

∫1/(1+x²)dx = arctan x + C

Key Techniques

Substitution: u = g(x), du = g′(x)dx

By parts: ∫u dv = uv − ∫v du

Partial fractions: split rational integrands

Applications

Area: A = ∫ₐᵇ[f(x)−g(x)]dx

Disc volume: V = π∫ₐᵇ[f(x)]²dx

Shell volume: V = 2π∫ₐᵇ x f(x)dx

Arc length: L = ∫ₐᵇ√[1+(f′(x))²]dx

Numerical & Multiple

Trapezium: (h/2)[f₀+2f₁+...+fₙ]

Simpson's: (h/3)[f₀+4f₁+2f₂+...+fₙ]

Polar area element: r dr dθ

Exam Tips

Common Errors to Avoid

All Levels

Wrong ∫x² dx = x³ (forgetting "+C" and the constant 1/3)

Correct ∫x² dx = x³/3 + C

Always divide by the new exponent, and always include the constant of integration in indefinite integrals.

Wrong ∫(1/x) dx = x⁰/0 (undefined application of the power rule)

Correct ∫(1/x) dx = ln|x| + C

The power rule fails when n = −1; this case is always a natural logarithm — and don't forget the modulus sign.

Wrong Area = ∫₀² (x²−2x) dx, reported as −4/3 sq. units

Correct Area = |∫₀² (x²−2x) dx| = 4/3 sq. units

Definite integrals give signed values. "Area" must always be reported as positive — split the integral at sign changes if necessary.

Wrong ∫cos(3x) dx = sin(3x) + C

Correct ∫cos(3x) dx = (1/3)sin(3x) + C

When the inner function is ax + b, divide by the coefficient a after integrating — a direct consequence of the chain rule in reverse.

Wrong ∫x eˣ dx with u = eˣ, dv = x dx (LIATE ignored)

Correct u = x, dv = eˣ dx

Choosing u and dv the wrong way around in integration by parts often produces an integral that is harder than the original — follow LIATE.

Wrong Forgetting to change the limits of integration after substitution in a definite integral

Correct Convert both limits to u-values (or back-substitute before evaluating)

Mixing x-limits with a u-integrand is one of the most common substitution errors in A-Level exams.