Lesson 7: Integration | Additional Mathematics

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1. Integration — The Reverse of Differentiation

Integration (Indefinite): Finding a function F(x) such that F'(x) = f(x). Written ∫f(x) dx = F(x) + c, where c is the constant of integration — always included in indefinite integrals. Integration is also called antidifferentiation.

Standard Integration Results — Must Know All

∫xⁿ dx = xⁿ⁺¹/(n+1) + c (n ≠ −1)

∫1/x dx = ln|x| + c (n = −1 case)

∫eˣ dx = eˣ + c ∫eᵃˣ dx = (1/a)eᵃˣ + c

∫sin x dx = −cos x + c ∫sin ax dx = −(1/a)cos ax + c

∫cos x dx = sin x + c ∫cos ax dx = (1/a)sin ax + c

∫sec²x dx = tan x + c ∫sec²ax dx = (1/a)tan ax + c

∫(ax+b)ⁿ dx = (ax+b)ⁿ⁺¹/[a(n+1)] + c (n ≠ −1)

∫1/(ax+b) dx = (1/a)ln|ax+b| + c

The constant factor 1/a appears because the chain rule would have multiplied by a on differentiation.

📐 Worked Example 1 — Basic Integration

Integrate: (a) ∫(4x³ − 3x + 5) dx (b) ∫(2/x + 3√x − 1/x²) dx (c) ∫(3e²ˣ + 2cos 4x − sec²3x) dx

(a) ∫(4x³−3x+5) dx = x⁴ − (3/2)x² + 5x + c

(b) Rewrite: 2x⁻¹ + 3x^(½) − x^(−2)
∫ = 2ln|x| + 3×(⅔)x^(3/2) − x^(−1)/(−1) + c
= 2ln|x| + 2x^(3/2) + 1/x + c

(c) ∫3e²ˣ dx = (3/2)e²ˣ. ∫2cos 4x dx = (2/4)sin 4x = (½)sin 4x. ∫sec²3x dx = (1/3)tan 3x
= (3/2)e²ˣ + (½)sin 4x − (1/3)tan 3x + c

⚠ Common Integration Errors:
1. Forgetting the constant of integration c in indefinite integrals.
2. Not dividing by the coefficient a when integrating (ax+b)ⁿ or sin(ax).
3. Trying to integrate products directly — product rule exists for differentiation, NOT integration. Use substitution or expand first.
4. Writing ∫1/x dx = ln x — must write ln|x| (absolute value), or ln x when x > 0 is stated.

2. Integrating Trigonometric Functions

Using Double Angle Formulae

Integrating sin²x and cos²x — use double angle identities first:
cos²x = ½(1 + cos 2x) → ∫cos²x dx = ½x + ¼sin 2x + c
sin²x = ½(1 − cos 2x) → ∫sin²x dx = ½x − ¼sin 2x + c
sin x cos x = ½sin 2x → ∫sin x cos x dx = −¼cos 2x + c (or ¼sin²2x method)

📐 Worked Example 2 — Integrating Trig Powers

Find: (a) ∫sin²3x dx (b) ∫cos²(x/2) dx (c) ∫sin 2x cos 2x dx

(a) sin²3x = ½(1−cos 6x)
∫sin²3x dx = ∫½(1−cos 6x) dx = ½x − (1/12)sin 6x + c
= ½x − (1/12)sin 6x + c

(b) cos²(x/2) = ½(1+cos x)
∫cos²(x/2) dx = ½x + ½sin x + c
= ½x + ½sin x + c

(c) sin 2x cos 2x = ½sin 4x
∫½sin 4x dx = −(1/8)cos 4x + c
= −(1/8)cos 4x + c

3. Integration by Substitution

Substitution: When an integral contains a composite function, substitute u = g(x) to simplify. The key is recognising that the integrand contains both g(x) and g'(x) (or a multiple of it) — making the substitution converts the integral to a standard form.

Method for Integration by Substitution:
1. Choose u = g(x) (usually the expression inside a bracket, under a root, or in an exponent).
2. Find du/dx → express dx in terms of du: dx = du/(du/dx)
3. Substitute u and dx into the integral — all x's must disappear.
4. Integrate with respect to u.
5. Substitute back to express in terms of x.
6. Add constant c (indefinite) or apply limits (definite).

📐 Worked Example 3 — Integration by Substitution (Indefinite)

Find: (a) ∫x(x²+3)⁵ dx (b) ∫x²√(x³−1) dx (c) ∫sin³x cos x dx

(a) Let u = x²+3 → du/dx = 2x → x dx = du/2
∫x(x²+3)⁵ dx = ∫u⁵ × du/2 = (1/2)u⁶/6 + c
= (x²+3)⁶/12 + c

(b) Let u = x³−1 → du = 3x² dx → x² dx = du/3
∫x²√(x³−1) dx = ∫√u × du/3 = (1/3)∫u^(½) du = (1/3)(⅔)u^(3/2) + c
= (2/9)(x³−1)^(3/2) + c

(c) Let u = sin x → du = cos x dx
∫sin³x cos x dx = ∫u³ du = u⁴/4 + c
= sin⁴x/4 + c

📐 Worked Example 4 — Substitution with ln and e

Find: (a) ∫2x/(x²+5) dx (b) ∫e^x/(1+eˣ) dx (c) ∫(ln x)/x dx

(a) Let u = x²+5 → du = 2x dx
∫2x/(x²+5) dx = ∫1/u du = ln|u| + c
= ln(x²+5) + c (x²+5 > 0 always, so no |)|)

(b) Let u = 1+eˣ → du = eˣ dx
∫eˣ/(1+eˣ) dx = ∫1/u du = ln|u| + c
= ln(1+eˣ) + c

(c) Let u = ln x → du = (1/x) dx
∫(ln x)/x dx = ∫u du = u²/2 + c
= (ln x)²/2 + c

Recognising Substitution Integrals:
Look for integrals of the form ∫f'(x)/f(x) dx → result is ln|f(x)| + c
Look for ∫f'(x)[f(x)]ⁿ dx → result is [f(x)]ⁿ⁺¹/[(n+1)] + c
These patterns arise directly from the chain rule — the integrand contains a function and its derivative (or a multiple).

4. Definite Integrals

Definite Integral

∫ₐᵇ f(x) dx = [F(x)]ₐᵇ = F(b) − F(a)

F(x) is any antiderivative of f(x). No constant of integration needed — it cancels out.
The definite integral gives a numerical value — the signed area between the curve and the x-axis from x=a to x=b.

📐 Worked Example 5 — Evaluating Definite Integrals

Evaluate: (a) ∫₁³ (3x²−2x+1) dx (b) ∫₀^(π/4) (sin 2x + cos x) dx (c) ∫₁² x/(x²+3) dx

(a) [x³−x²+x]₁³ = (27−9+3)−(1−1+1) = 21−1 = 20

(b) [−½cos 2x + sin x]₀^(π/4)
At x=π/4: −½cos(π/2)+sin(π/4) = 0+√2/2 = √2/2
At x=0: −½cos 0+sin 0 = −½+0 = −½
Result: √2/2−(−½) = √2/2 + ½

(c) Let u=x²+3, du=2x dx → x dx=du/2
When x=1: u=4. When x=2: u=7.
∫₄⁷ (1/u)(du/2) = [½ln u]₄⁷ = ½ln 7−½ln 4 = ½ln(7/4)
= ½ln(7/4)

📐 Worked Example 6 — Definite Integral by Substitution (Changing Limits)

Evaluate ∫₀² x√(x²+1) dx using substitution.

Let u = x²+1. When x=0: u=1. When x=2: u=5.
du = 2x dx → x dx = du/2

∫₁⁵ √u × du/2 = ½∫₁⁵ u^(½) du = ½[⅔u^(3/2)]₁⁵
= ⅓[u^(3/2)]₁⁵ = ⅓(5^(3/2)−1^(3/2)) = ⅓(5√5−1)
= (5√5−1)/3

5. Area Under and Between Curves

Area under a curve: The definite integral ∫ₐᵇ f(x) dx gives the signed area between y=f(x) and the x-axis from x=a to x=b. Areas below the x-axis give negative values — to find actual area, integrate sections separately and take the absolute value.

Area Rules:
Area above x-axis (f(x) ≥ 0): Area = ∫ₐᵇ f(x) dx (positive result)
Area below x-axis (f(x) ≤ 0): Area = |∫ₐᵇ f(x) dx| (take modulus)
Area spanning both sides: Split at x-intercepts. Total area = sum of |each section|
Area between two curves: Area = ∫ₐᵇ [f(x)−g(x)] dx where f(x) ≥ g(x) on [a,b]

📐 Worked Example 7 — Area Under a Curve

Find the area enclosed between y = x² − 4 and the x-axis.

Find x-intercepts: x²−4=0 → x=±2. The parabola dips below the x-axis between x=−2 and x=2.

∫₋₂² (x²−4) dx = [x³/3−4x]₋₂²
At x=2: 8/3−8 = −16/3
At x=−2: −8/3+8 = 16/3
Result: −16/3−16/3 = −32/3

Area = |−32/3| = 32/3 square units

📐 Worked Example 8 — Area Between Two Curves

Find the area enclosed between y = 6x − x² and y = x² − 2x.

Find intersection: 6x−x² = x²−2x → 8x−2x² = 0 → 2x(4−x) = 0
Intersect at x=0 and x=4.

Check which is on top: at x=2, y₁=8, y₂=0. So y₁ ≥ y₂ on [0,4].
Area = ∫₀⁴ [(6x−x²)−(x²−2x)] dx = ∫₀⁴ (8x−2x²) dx

[4x²−(2/3)x³]₀⁴ = 64−128/3 = (192−128)/3 = 64/3 square units

Area with Respect to the y-axis

Area between curve and y-axis: ∫_c^d x dy — integrate with respect to y, where x is expressed as a function of y. Limits c and d are y-values.

📐 Worked Example 9 — Area with y-axis

Find the area enclosed between x = y² and the y-axis between y=0 and y=3.

Area = ∫₀³ x dy = ∫₀³ y² dy

[y³/3]₀³ = 27/3−0 = 9 square units

6. Integrating Using Partial Fractions

Partial fractions (Lesson 2) decompose rational functions into simpler fractions that can be integrated directly using ∫1/(ax+b) dx = (1/a)ln|ax+b| + c.

📐 Worked Example 10 — Integration via Partial Fractions

Find ∫(5x+1)/[(x+1)(x−2)] dx.

From Lesson 2 (partial fractions): (5x+1)/[(x+1)(x−2)] = 4/[3(x+1)] + 11/[3(x−2)]

∫[4/(3(x+1)) + 11/(3(x−2))] dx
= (4/3)ln|x+1| + (11/3)ln|x−2| + c
= (1/3)[4ln|x+1| + 11ln|x−2|] + c

📐 Worked Example 11 — Definite Integral with Partial Fractions

Evaluate ∫₃⁵ 3/[(x−1)(x+2)] dx.

Partial fractions: 3/[(x−1)(x+2)] = A/(x−1)+B/(x+2)
3=A(x+2)+B(x−1). x=1: 3=3A → A=1. x=−2: 3=−3B → B=−1
= 1/(x−1) − 1/(x+2)

∫₃⁵ [1/(x−1)−1/(x+2)] dx = [ln|x−1|−ln|x+2|]₃⁵
= [ln|(x−1)/(x+2)|]₃⁵
= ln(4/7)−ln(2/5) = ln(4/7 ÷ 2/5) = ln(4/7×5/2) = ln(10/7)

7. Kinematics — Integration Applications

Kinematics Integration Results

s = ∫v dt (displacement from velocity)

v = ∫a dt (velocity from acceleration)

The constant of integration is found from initial conditions (e.g. s=s₀ when t=0, or v=v₀ when t=0).
Displacement = ∫ₜ₁ᵗ² v dt (signed — can be negative)
Distance = ∫ₜ₁ᵗ² |v| dt (always positive — split at v=0)

📐 Worked Example 12 — Kinematics Integration

A particle starts from rest at origin O. Its acceleration at time t is a = 6t − 4 m/s². Find expressions for v and s. Find the total distance in the first 3 seconds.

Velocity: v = ∫(6t−4) dt = 3t²−4t + c
v=0 when t=0 (starts from rest): 0=0+c → c=0
v = 3t²−4t

Displacement: s = ∫(3t²−4t) dt = t³−2t² + c
s=0 when t=0: c=0 → s = t³−2t²

v=0: t(3t−4)=0 → t=0 or t=4/3. At t=4/3: s=(4/3)³−2(4/3)²=64/27−32/9=64/27−96/27=−32/27
At t=3: s=27−18=9
Distance: t=0 to t=4/3: moves 32/27 backwards. t=4/3 to t=3: moves from −32/27 to 9 = 9+32/27 = 275/27
Total = 32/27+275/27 = 307/27 ≈ 11.4 m

8. Finding the Constant of Integration

Whenever additional information is given (a point on the curve, initial conditions), use it to find the constant c after integrating.

📐 Worked Example 13 — Finding the Equation of a Curve

A curve passes through (2, 5). Its gradient function is dy/dx = 3x² − 4x + 1. Find the equation of the curve.

Integrate: y = ∫(3x²−4x+1) dx = x³−2x²+x+c

Use (2,5): 5 = 8−8+2+c → c = 3
y = x³−2x²+x+3

📐 Worked Example 14 — Given d²y/dx²

Given d²y/dx² = 6x−2. When x=1, dy/dx=4. When x=0, y=3. Find y in terms of x.

Integrate once: dy/dx = 3x²−2x+c₁
At x=1, dy/dx=4: 4=3−2+c₁ → c₁=3
dy/dx = 3x²−2x+3

Integrate again: y = x³−x²+3x+c₂
At x=0, y=3: 3=0+c₂ → c₂=3
y = x³−x²+3x+3

📝 Exam Practice Questions

Q1 [4 marks] — Find: (a) ∫(2x+3)⁶ dx (b) ∫3/(4x−1) dx (c) ∫cos²2x dx (d) ∫sin x cos³x dx

(a) ∫(2x+3)⁶ dx = (2x+3)⁷/(7×2) + c = (2x+3)⁷/14 + c

(b) ∫3/(4x−1) dx = 3×(1/4)ln|4x−1| + c = (3/4)ln|4x−1| + c

(c) cos²2x = ½(1+cos 4x). ∫½(1+cos 4x) dx = ½x + (1/8)sin 4x + c

(d) Let u=cos x, du=−sin x dx:
∫sin x cos³x dx = ∫−u³ du = −u⁴/4 + c = −cos⁴x/4 + c

Q2 [4 marks] — Evaluate: (a) ∫₀¹ xe^(x²) dx (b) ∫₀^(π/6) sin²3x dx

(a) u=x², du=2x dx → x dx=du/2. Limits: x=0→u=0, x=1→u=1
∫₀¹ ½eᵘ du = [½eᵘ]₀¹ = ½e−½ = ½(e−1)

(b) sin²3x = ½(1−cos 6x)
∫₀^(π/6) ½(1−cos 6x) dx = [½x−(1/12)sin 6x]₀^(π/6)
= (π/12−(1/12)sin π)−0 = π/12−0 = π/12

Q3 [5 marks] — Find the area enclosed between y = x³ − 3x and the x-axis.

x-intercepts: x(x²−3)=0 → x=0, x=√3, x=−√3
By symmetry, integrate from 0 to √3 and double.
At x=1: y=1−3=−2<0 → curve is BELOW x-axis on [0,√3]
∫₀^√3 (x³−3x) dx = [x⁴/4−(3/2)x²]₀^√3
= 9/4−9/2 = 9/4−18/4 = −9/4
Area from 0 to √3 = 9/4. Total area = 2×9/4 = 9/2 square units

Exam Tip: Always sketch the curve first to identify where it is above/below the x-axis. For odd functions like x³−3x, the area on each side of the origin is equal — use symmetry to simplify.

Q4 [5 marks] — Find the area enclosed between y = x² and y = 4x − x².

Intersection: x² = 4x−x² → 2x²−4x=0 → 2x(x−2)=0 → x=0, x=2
At x=1: y₁=1, y₂=3. Upper curve y₂=4x−x².
Area = ∫₀² [(4x−x²)−x²] dx = ∫₀² (4x−2x²) dx
= [2x²−(2/3)x³]₀²
= 8−16/3 = 24/3−16/3 = 8/3 square units

Q5 [4 marks] — Evaluate ∫₂⁴ (2x+1)/[(x−1)(x+1)] dx using partial fractions.

(2x+1)/[(x−1)(x+1)] = A/(x−1)+B/(x+1)
2x+1=A(x+1)+B(x−1)
x=1: 3=2A → A=3/2. x=−1: −1=−2B → B=1/2
∫₂⁴ [3/[2(x−1)] + 1/[2(x+1)]] dx = [(3/2)ln|x−1|+(1/2)ln|x+1|]₂⁴
At x=4: (3/2)ln3+(1/2)ln5
At x=2: (3/2)ln1+(1/2)ln3=0+(1/2)ln3
= (3/2)ln3+(1/2)ln5−(1/2)ln3 = ln3+(1/2)ln5
= ln3 + ½ln5 = ln(3√5)

Q6 [5 marks] — A curve has d²y/dx² = 2−6x. The curve has a stationary point at (2, 5). Find the equation of the curve and determine the nature of the stationary point.

Integrate once: dy/dx = 2x−3x²+c₁
Stationary at x=2: 0=4−12+c₁ → c₁=8
dy/dx = 2x−3x²+8
Integrate again: y = x²−x³+8x+c₂
At (2,5): 5=4−8+16+c₂ → c₂=−7
y = x²−x³+8x−7
Nature: d²y/dx² at x=2: 2−12=−10<0 → maximum

Exam Tip: When integrating twice with two sets of conditions, apply each condition immediately after its integration step — don't wait until the end. This avoids carrying two unknowns simultaneously.

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