Integration: Techniques & Applications
Integration is the reverse process of differentiation and is used to find original functions from their rates of change as well as exact areas under curves. At A-Level Maths, you combine standard integrals with integration by substitution, integration by parts, and partial fractions to evaluate both indefinite and definite integrals.
- Definite Integral
- An integral with upper and lower limits, used to calculate a specific value such as the exact area under a curve.
- Integration by Parts
- A technique based on the product rule for differentiation: ∫ u (dv/dx) dx = uv - ∫ v (du/dx) dx.
- Integration by Substitution
- A method where a variable change (e.g., let u = f(x)) simplifies the integral into a standard form.
Definite Integral & Trapezium Rule Step-by-Step Visualizer
Choose a function, slide the limits a and b to see the area under the curve update live, and compare Exact Integration vs the Trapezium Rule.
Step-by-Step Working for ∫03 (1x²) dx
- Step 1, Find the antiderivative F(x):F(x) = [(1/3)x³] from x = 0 to x = 3
- Step 2: Substitute upper limit (b = 3) minus lower limit (a = 0):F(3) - F(0) = (9.0000) - (0.0000) = 9.0000
- Step 3, Compare with Trapezium Rule (4 strips, h = 0.750):Area ≈ (h/2)[y₀ + yₙ + 2(y₁ + … + yₙ₋₁)] = 9.2813
Difference from exact integral: 3.13% (increase
nstrips to make the approximation more accurate!).
+ C on indefinite integrals, and use radians whenever integrating trigonometric functions like sin(x) or cos(x)!Exam notes: Which Integration Method to Choose + Worked Examples
- Before integrating any expression, check if it can be rewritten in index form: 1/xⁿ = x⁻ⁿ and √x = x^(1/2). Only apply the power rule ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C when n ≠ −1. When n = −1, ∫ (1/x) dx = ln|x| + C.
- Reverse Chain Rule (Inspection) is the fastest method in exams: whenever the numerator is a constant multiple of the derivative of the bracket or denominator, use ∫ f'(x)/f(x) dx = ln|f(x)| + C or ∫ f'(x)[f(x)]ⁿ dx = [f(x)]ⁿ⁺¹ / (n+1) + C.
- Integration by Parts (∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx): choose u using the LATE order: Logarithms (ln x) first, then Algebraic powers (x, x²), then Trig (sin x, cos x), then Exponentials (eˣ).
- When finding the area between a curve and the x-axis, integrate separately over any interval where y < 0 (below the x-axis), because a definite integral gives a signed area (negative below the axis).
| Integral Pattern | Method to Use | Result / Formula |
|---|---|---|
| ∫ (ax + b)ⁿ dx (n ≠ −1) | Reverse Chain Rule | (ax + b)ⁿ⁺¹ / [a(n + 1)] + C |
| ∫ f'(x) / f(x) dx (e.g. ∫ 2x/(x²+1) dx) | Logarithmic Inspection | ln|f(x)| + C |
| ∫ x·eˣ dx, ∫ x·sin(x) dx, ∫ ln(x) dx | Integration by Parts (LATE) | uv − ∫ v (du/dx) dx (for ∫ ln x dx, let u = ln x, dv/dx = 1) |
| ∫ sin²(x) dx or ∫ cos²(x) dx | Double-Angle Trig Identity | Use cos(2x) = 1 − 2sin²(x) = 2cos²(x) − 1 first |
| ∫ P(x) / [(x − a)(x − b)] dx | Partial Fractions | Split into A/(x−a) + B/(x−b), then integrate to ln terms |
Key Concepts & Exam Definitions (10)
- Definite Integral
- An integral with upper and lower limits, used to calculate a specific value such as the exact area under a curve.
- Integration by Parts
- A technique based on the product rule for differentiation: ∫ u (dv/dx) dx = uv - ∫ v (du/dx) dx.
- Integration by Substitution
- A method where a variable change (e.g., let u = f(x)) simplifies the integral into a standard form.
- Partial Fractions
- A method to decompose complex rational algebraic fractions into simpler fractions that can be integrated using ln(x).
- Trapezium Rule
- A numerical method used to estimate the value of a definite integral by dividing the area into trapeziums.
- Integrating 1/x and f'(x)/f(x)
- The power rule fails when n = −1. Instead, ∫ (1/x) dx = ln|x| + C. More generally, if the top of a fraction is the derivative of the bottom, ∫ [f'(x) / f(x)] dx = ln|f(x)| + C.
- Integrating sin²(x) and cos²(x) using Double-Angle Identities
- You cannot use the power rule on sin²(x). Always substitute sin²(x) = ½(1 − cos 2x) or cos²(x) = ½(1 + cos 2x), which integrates directly to ½x ∓ ¼sin(2x) + C.
- Area between two curves y = f(x) and y = g(x)
- 1) Solve f(x) = g(x) to find the x-coordinates of the points of intersection a and b. 2) Integrate the upper curve minus the lower curve: Area = ∫ₐᵇ [f(x) − g(x)] dx.
- Trapezium Rule Overestimate vs Underestimate
- If a curve is concave upwards (d²y/dx² > 0, like y = eˣ or y = x²), the straight tops of the trapezia sit above the curve, giving an overestimate. If the curve is concave downwards (d²y/dx² < 0, like y = ln x or y = sin x on [0, π]), it gives an underestimate.
- Separation of Variables (First-Order Differential Equations)
- To solve dy/dx = f(x)g(y), rearrange all y terms with dy and all x terms with dx: ∫ (1/g(y)) dy = ∫ f(x) dx. Integrate both sides and use the given boundary condition (x₀, y₀) to find the constant +C.
Quick-Fire Flashcards (10)
Click any card to check the answer, then mark it as Known to track your progress.
What is the integral of 1/x dx?
State the formula for integration by parts.
What is the integral of e^(kx) dx?
How do you integrate sin(x)?
What does the definite integral of f(x) from a to b represent?
How do you integrate ∫ ln(x) dx?
How do you integrate ∫ sin²(x) dx?
When does the Trapezium Rule give an overestimate vs an underestimate?
What is ∫ (6x) / (3x² + 5) dx?
Why can you not evaluate ∫₋₁² x² dx directly if a question asks for the area enclosed by a curve that crosses the x-axis?
Exam-Style Questions & Mark Schemes (7)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Evaluate ∫ (2x + 3) dx.
Click “Check Mark Scheme” to compare your answer, x² + 3x + C
- Q2
Use integration by parts to find ∫ x e^x dx.
Click “Check Mark Scheme” to compare your answer, Let u=x, dv/dx=e^x. Then du/dx=1, v=e^x. ∫ x e^x dx = x e^x - ∫ e^x dx = x e^x - e^x + C.
- Q3
Integrate 3/(2x-1) with respect to x.
Click “Check Mark Scheme” to compare your answer, (3/2) ln|2x - 1| + C.
- Q4
Find ∫ (4x³ − 6/x² + 3√x) dx, giving each term in its simplest form. [4 marks]
Click “Check Mark Scheme” to compare your answer, 1) Rewrite in index form: ∫ (4x³ − 6x⁻² + 3x^(1/2)) dx [1 mark]. 2) Integrate term by term: (4x⁴)/4 − (6x⁻¹)/(−1) + 3x^(3/2)/(3/2) + C [2 marks]. 3) Simplify: x⁴ + 6/x + 2x^(3/2) + C (don't forget + C) [1 mark].
- Q5
Use integration by parts to find the exact value of ∫₁ᵉ x² ln(x) dx. [5 marks]
Click “Check Mark Scheme” to compare your answer, 1) By LATE, let u = ln(x) ⇒ du/dx = 1/x, and dv/dx = x² ⇒ v = x³/3 [1 mark]. 2) Apply formula: [ (x³/3)ln(x) ]₁ᵉ − ∫₁ᵉ (x³/3)(1/x) dx = [ (x³/3)ln(x) − x³/9 ]₁ᵉ [2 marks]. 3) Substitute limits: Upper (x=e): (e³/3)(1) − e³/9 = (2/9)e³. Lower (x=1): (1/3)(0) − 1/9 = −1/9 [1 mark]. 4) Subtract: (2/9)e³ − (−1/9) = (2e³ + 1) / 9 [1 mark].
- Q6
Find the exact area enclosed between the curve y = 6x − x² and the line y = 2x. [4 marks]
Click “Check Mark Scheme” to compare your answer, 1) Find intersections: 6x − x² = 2x ⇒ x² − 4x = 0 ⇒ x(x − 4) = 0, so x = 0 and x = 4 [1 mark]. 2) Integrate (upper − lower): ∫₀⁴ ((6x − x²) − 2x) dx = ∫₀⁴ (4x − x²) dx [1 mark]. 3) [ 2x² − x³/3 ]₀⁴ = (2(16) − 64/3) − 0 = 32 − 21⅓ = 32/3 (or 10⅔) square units [2 marks].
- Q7
Solve the differential equation dy/dx = 3x²y given that y = 2 when x = 0, expressing y in terms of x. [4 marks]
Click “Check Mark Scheme” to compare your answer, 1) Separate variables: ∫ (1/y) dy = ∫ 3x² dx [1 mark]. 2) Integrate both sides: ln|y| = x³ + C [1 mark]. 3) Substitute x = 0, y = 2: ln(2) = 0 + C ⇒ C = ln(2) [1 mark]. 4) Rearrange: ln(y) = x³ + ln(2) ⇒ y = e^(x³ + ln 2) = 2e^(x³) [1 mark].
Common Exam Questions (3)
When should I use integration by parts?
Use integration by parts, ∫u(dv/dx)dx = uv - ∫v(du/dx)dx, when integrating the product of two functions of different types that cannot be expanded directly, such as x cos(x), x e^x, or ln(x).
What is the + C in integration?
The '+ C' is the constant of integration included in every indefinite integral. Because any constant term differentiates to zero, differentiating f(x) + C always gives f'(x), so C represents the family of vertical translations of the antiderivative.
How do I choose the substitution for integration by substitution?
Look for an inner function u = g(x) whose derivative g'(x) (up to a constant factor) also appears in the integrand, or substitute u for the expression inside a bracket, root, or denominator.
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