IB Maths AA Analysis and Approaches How to Solve: IB AA HL/SL – Integration (Substitution, By Parts, Partial Fractions, Definite Integrals for Area/Volume)

How to Solve: IB AA HL/SL – Integration (Substitution, By Parts, Partial Fractions, Definite Integrals for Area/Volume)

Complete Guide

Introduction

Mastering integration unlocks 5–7 marks per IB AA HL/SL Paper 1 or 2 question—and it’s the key to calculating work in Physics, reaction rates in Chemistry, population growth in Biology, and consumer surplus in Economics. If you can integrate, you can solve real-world problems—not just pass the exam.

WHAT YOU NEED TO KNOW FIRST

1. Basic differentiation rules (power rule, chain rule, product rule).
2. Algebraic manipulation (expanding, factoring, solving for variables).
3. Definite integrals as area under a curve (trapezoidal rule is not enough—you need exact values).

KEY TERMS & FORMULAS

1. Integration by Substitution (Reverse Chain Rule)

Formula: ∫ f(g(x)) · g'(x) dx = ∫ f(u) du where u = g(x) - u = substitution variable (inner function) - du/dx = derivative of u (must be present in the integral) - MEMORISE THIS: You must adjust limits if solving a definite integral.

2. Integration by Parts

Formula: ∫ u dv = uv – ∫ v du - u = part to differentiate (use LIATE: Logs, Inverse trig, Algebraic, Trig, Exponential) - dv = part to integrate - MEMORISE THIS: The formula—it’s not on the IB formula sheet!

3. Partial Fractions (for Rational Functions)

When to use: When the integrand is a fraction with a polynomial denominator (degree of numerator < degree of denominator). Steps:
1. Factor the denominator (if possible).
2. Write as a sum of simpler fractions (e.g., A/(x+1) + B/(x-2)).
3. Solve for constants (A, B) by equating coefficients or plugging in values.
4. Integrate each term separately. - MEMORISE THIS: The decomposition forms (linear factors, repeated roots, irreducible quadratics).

4. Definite Integrals for Area & Volume

Area under a curve (between x=a and x=b): ∫_a^b f(x) dx = F(b) – F(a) where F'(x) = f(x) - MEMORISE THIS: If the curve is below the x-axis, the integral gives a negative area—take the absolute value if asked for "area."

Volume of revolution (around x-axis): V = π ∫_a^b [f(x)]² dx - MEMORISE THIS: The formula—it’s not on the IB formula sheet!

STEP-BY-STEP METHOD

1. Integration by Substitution

Steps:
1. Identify u: Look for a function inside another function (e.g., e^(3x+2), (2x+1)⁵).
2. Find du/dx: Differentiate u to get du/dx.
3. Rewrite the integral: Replace the inner function with u and dx with du/(du/dx).
4. Integrate with respect to u.
5. Substitute back to x (if indefinite).
6. Adjust limits (if definite): Plug original limits into u = g(x) to get new limits.

Worked Example: ∫ 2x e^(x²) dx (from x=0 to x=1)
1. Let u = x² (inner function).
2. du/dx = 2x → du = 2x dx → dx = du/(2x).
3. Rewrite: ∫ 2x e^u · (du/(2x)) = ∫ e^u du.
4. Integrate: e^u + C.
5. Substitute back: e^(x²) + C.
6. Definite integral: e^(1) – e^(0) = e – 1.

What we did and why: We spotted x² inside e^(x²) and used substitution to simplify the integral. The 2x term was already present, making it a perfect candidate for substitution.

2. Integration by Parts

Steps:
1. Choose u and dv: Use LIATE (Logs, Inverse trig, Algebraic, Trig, Exponential).
2. Differentiate u to get du.
3. Integrate dv to get v.
4. Apply the formula: ∫ u dv = uv – ∫ v du.
5. Simplify and integrate the remaining term.

Worked Example: ∫ x e^x dx
1. Let u = x (Algebraic), dv = e^x dx (Exponential).
2. du = dx, v = e^x.
3. Apply formula: x e^x – ∫ e^x dx.
4. Integrate: x e^x – e^x + C = e^x (x – 1) + C.

What we did and why: We used LIATE to pick u = x (higher priority than e^x). The remaining integral was simple to solve.

3. Partial Fractions

Steps:
1. Factor the denominator (if possible).
2. Write the partial fraction decomposition (e.g., A/(x+1) + B/(x-2)).
3. Multiply through by the denominator to eliminate fractions.
4. Solve for constants (A, B) by plugging in values or equating coefficients.
5. Integrate each term separately.

Worked Example: ∫ (3x + 5)/(x² – x – 6) dx
1. Factor denominator: (x – 3)(x + 2).
2. Decompose: (3x + 5)/[(x – 3)(x + 2)] = A/(x – 3) + B/(x + 2).
3. Multiply: 3x + 5 = A(x + 2) + B(x – 3).
4. Solve for A and B: - Let x = 3: 14 = 5A → A = 14/5. - Let x = -2: -1 = -5B → B = 1/5.
5. Rewrite integral: ∫ (14/5)/(x – 3) + (1/5)/(x + 2) dx.
6. Integrate: (14/5) ln|x – 3| + (1/5) ln|x + 2| + C.

What we did and why: We split the complex fraction into simpler terms, making integration straightforward.

4. Definite Integrals for Area & Volume

Steps (Area):
1. Sketch the curve (if possible) to see if it crosses the x-axis.
2. Find x-intercepts (if needed) to split the integral.
3. Integrate f(x) from a to b.
4. Take absolute value if the question asks for "area" and the curve is below the x-axis.

Worked Example (Area): Find the area between y = x² – 4 and the x-axis from x=0 to x=3.
1. Find x-intercepts: x² – 4 = 0 → x = ±2.
2. Split integral: ∫₀² (x² – 4) dx + ∫₂³ (x² – 4) dx.
3. Integrate: [x³/3 – 4x] from 0 to 2 + [x³/3 – 4x] from 2 to 3.
4. Evaluate: - First part: (8/3 – 8) – 0 = -16/3 (negative, so take absolute value: 16/3). - Second part: (9 – 12) – (8/3 – 8) = -3 – (-16/3) = 7/3.
5. Total area: 16/3 + 7/3 = 23/3.

What we did and why: We split the integral at the x-intercept because the curve crosses the x-axis, making part of the area negative.

Steps (Volume of Revolution):
1. Identify the function to rotate (e.g., y = f(x)).
2. Square the function and multiply by π.
3. Integrate from a to b.

Worked Example (Volume): Find the volume when y = √x is rotated around the x-axis from x=0 to x=4.
1. Square the function: (√x)² = x.
2. Multiply by π: π ∫₀⁴ x dx.
3. Integrate: π [x²/2] from 0 to 4 = π (8 – 0) = 8π.

What we did and why: We used the volume formula for rotation around the x-axis, squaring the function first.

WORKED EXAMPLES

Example 1 – Basic (Substitution)

∫ cos(3x) dx
1. Let u = 3x → du/dx = 3 → dx = du/3.
2. Rewrite: ∫ cos(u) · (du/3) = (1/3) sin(u) + C.
3. Substitute back: (1/3) sin(3x) + C.

What we did and why: We used substitution to simplify cos(3x) into cos(u), which is easier to integrate.

Example 2 – Medium (By Parts)

∫ x ln(x) dx
1. Let u = ln(x) (Logs), dv = x dx (Algebraic).
2. du = (1/x) dx, v = x²/2.
3. Apply formula: ln(x) · (x²/2) – ∫ (x²/2)(1/x) dx.
4. Simplify: (x²/2) ln(x) – (1/2) ∫ x dx.
5. Integrate: (x²/2) ln(x) – (1/4) x² + C.

What we did and why: We used LIATE to pick u = ln(x) and applied integration by parts to simplify the integral.

Example 3 – Exam-Style (Partial Fractions + Definite Integral)

Evaluate ∫₂³ (2x + 1)/(x² – 1) dx.
1. Factor denominator: (x – 1)(x + 1).
2. Decompose: (2x + 1)/[(x – 1)(x + 1)] = A/(x – 1) + B/(x + 1).
3. Multiply: 2x + 1 = A(x + 1) + B(x – 1).
4. Solve for A and B: - Let x = 1: 3 = 2A → A = 3/2. - Let x = -1: -1 = -2B → B = 1/2.
5. Rewrite integral: ∫₂³ (3/2)/(x – 1) + (1/2)/(x + 1) dx.
6. Integrate: (3/2) ln|x – 1| + (1/2) ln|x + 1| from 2 to 3.
7. Evaluate: (3/2 ln(2) + 1/2 ln(4)) – (3/2 ln(1) + 1/2 ln(3)) = (3/2 ln(2) + ln(2)) – (1/2 ln(3)) = (5/2) ln(2) – (1/2) ln(3).

What we did and why: We used partial fractions to split the integrand, then evaluated the definite integral using the natural logarithm.

COMMON MISTAKES

1. MISTAKE: Forgetting to adjust limits in substitution. WHY IT HAPPENS: Students substitute u but keep the original x limits. CORRECT APPROACH: Always change limits when substituting (e.g., if u = 2x, and x goes from 0 to 1, u goes from 0 to 2).

2. MISTAKE: Choosing u and dv incorrectly in by parts. WHY IT HAPPENS: Students pick u as the harder part (e.g., e^x instead of x). CORRECT APPROACH: Use LIATE (Logs > Inverse trig > Algebraic > Trig > Exponential).

3. MISTAKE: Not factoring the denominator in partial fractions. WHY IT HAPPENS: Students try to integrate without decomposing first. CORRECT APPROACH: Always factor the denominator before decomposing.

4. MISTAKE: Ignoring negative area in definite integrals. WHY IT HAPPENS: Students forget that area below the x-axis is negative. CORRECT APPROACH: Split the integral at x-intercepts and take absolute values if needed.

5. MISTAKE: Forgetting π in volume of revolution. WHY IT HAPPENS: Students confuse area and volume formulas. CORRECT APPROACH: Remember: Volume = π ∫ [f(x)]² dx.

EXAM TRAPS

1. TRAP: Disguised substitution (e.g., ∫ x e^(x²) dx instead of ∫ e^(x²) dx). HOW TO SPOT IT: Look for a function and its derivative in the integrand. HOW TO AVOID IT: Check if multiplying/dividing by a constant balances the integral.

2. TRAP: Partial fractions with repeated roots (e.g., 1/(x – 1)²). HOW TO SPOT IT: Denominator has a squared term. HOW TO AVOID IT: Use A/(x – 1) + B/(x – 1)² instead of just A/(x – 1).

3. TRAP: Volume of revolution around the y-axis (not x-axis). HOW TO SPOT IT: The question says "rotate around the y-axis." HOW TO AVOID IT: Use V = π ∫ [f(y)]² dy or switch to x = f(y).

1-MINUTE RECAP

Student, listen up—this is your last-minute lifeline.

1. Substitution? Spot the inner function and its derivative. If they’re both there, substitute!
2. By parts? Use LIATE to pick u. If you see x ln(x) or x e^x, by parts is your friend.
3. Partial fractions? Factor the denominator, split into simpler fractions, solve for constants, then integrate.
4. Area/volume? Sketch the curve first. If it’s below the x-axis, split the integral. For volume, don’t forget π!
5. Definite integrals? Adjust limits in substitution, and always check if the question wants "area" (absolute value) or "integral" (signed value).

Final tip: If you’re stuck, differentiate your answer—it should give you back the integrand. Now go ace that exam!