F4 Math | Integration | Masomotele Free Learning

📌 Learning Outcomes

By the end of this lesson, you will be able to:

🔄

Interpret integration as the reverse process of differentiation

📏

Integrate a polynomial using the power rule

∞

Distinguish between indefinite and definite integrals

📊

Calculate area under a curve using definite integration

🚗

Apply integration in kinematics (distance from velocity)

📐

Use the trapezium rule and mid-ordinate rule to approximate areas

🔄 Integration as Reverse Differentiation

The Concept

Integration is the reverse of differentiation. If differentiation gives the gradient function, integration finds the original function from its gradient.

Differentiate: x² → 2x
Integrate: 2x → x² + c

The constant c (called the constant of integration) is added because when we differentiate a constant it becomes zero — so we can't know what constant was there originally.

The Integration Rule (Power Rule)

∫xⁿ dx = xⁿ⁺¹/(n+1) + c (n ≠ −1)

∫axⁿ dx = axⁿ⁺¹/(n+1) + c

In words: add 1 to the power, divide by the new power, add c

📝 Integrating Polynomials

Examples — Indefinite Integrals

a) ∫2x⁵ dx
= 2x⁶/6 + c = x⁶/3 + c

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

c) Find y if dy/dx = 3x² and y=5 when x=1
y = ∫3x² dx = x³ + c
Substitute: 5 = 1 + c → c = 4
Particular solution: y = x³ + 4

d) Find y if dy/dx = 2x + 3 and passes through (0,1)
y = x² + 3x + c
At (0,1): 1 = 0 + 0 + c → c = 1
y = x² + 3x + 1

🔢 Definite Integrals

What is a Definite Integral?

A definite integral has upper and lower limits — it gives a specific numerical answer (no c).

∫ₐᵇ f(x) dx = [F(x)]ₐᵇ = F(b) − F(a)
where F(x) is the integral of f(x)

Examples

Evaluate ∫₁³ (x² + 2) dx

= [x³/3 + 2x]₁³
= (27/3 + 6) − (1/3 + 2)
= (9 + 6) − (1/3 + 2)
= 15 − 7⁄3
= 38/3 ≈ 12.67

Evaluate ∫₀² (3x² − 4x) dx

= [x³ − 2x²]₀²
= (8 − 8) − (0 − 0)
= 0

📊 Area Under a Curve

Finding Area Using Integration

      The area bounded by the curve y = f(x), the x-axis, and the lines x = a and x = b is:

      Area = ∫ₐᵇ f(x) dx

Note: Areas below the x-axis give negative values when integrated, so take the absolute value.

Worked Example

Find the area bounded by y = x² + 1, the x-axis, x = 0 and x = 3

Area = ∫₀³ (x² + 1) dx
= [x³/3 + x]₀³
= (27/3 + 3) − 0
= 9 + 3 = 12 square units

Area between y = 4 − x² and the x-axis (where y ≥ 0):
Curve crosses x-axis when x² = 4 → x = ±2
Area = ∫₋₂² (4 − x²) dx = [4x − x³/3]₋₂²
= (8 − 8/3) − (−8 + 8/3)
= 16 − 16/3 = 32/3 ≈ 10.67 sq units

🚗 Integration in Kinematics

Reversing the Kinematics Formulas

s = ∫v dt (integrate velocity to get displacement)
v = ∫a dt (integrate acceleration to get velocity)

Worked Example

A particle has velocity v = 3t² − 6t m/s. Find:

(a) Displacement after 4 seconds (starts from rest at origin):
s = ∫(3t² − 6t) dt = t³ − 3t² + c
At t=0, s=0: 0 = 0 + c → c = 0
At t=4: s = 64 − 48 = 16 m

(b) Distance travelled between t=0 and t=4:
v=0 when 3t(t−2)=0 → t=0 or t=2
s(2) = 8 − 12 = −4 m (moved 4 m in negative direction)
Total distance = 4 + (4+16) = 24 m

📐 Trapezium Rule & Mid-Ordinate Rule

Trapezium Rule (Area Approximation)

Trapezium Rule:
Area ≈ h/2 [(y₀ + yₙ) + 2(y₁ + y₂ + ... + yₙ₋₁)]
where h = uniform width of each strip

Example: A car's velocity (m/s) is measured every second:
t: 0, 1, 2, 3, 4, 5, 6
v: 0, 12, 24, 35, 41, 45, 47

Area (distance) ≈ 1/2[(0+47) + 2(12+24+35+41+45)]
= 1/2[47 + 2(157)]
= 1/2[47 + 314] = 1/2 × 361 = 180.5 m

Mid-Ordinate Rule

Area ≈ h × (sum of ordinates at midpoints of each strip)
Use midpoints of intervals instead of edges

📋 Summary

Power Rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) + c

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

Area under curve: A = ∫ₐᵇ f(x) dx

Trapezium Rule:
A ≈ h/2[(y₀+yₙ) + 2(y₁+y₂+...+yₙ₋₁)]

Kinematics: s = ∫v dt, v = ∫a dt

Key Reminders

Always add +c for indefinite integrals
Use given conditions to find the value of c
Definite integrals: substitute upper limit minus lower limit
Area below x-axis gives negative value — take absolute value

✏️ Practice Questions

1. Integrate: ∫(4x³ + 2x) dx

2. Evaluate: ∫₀² x² dx

3. If v = 6t + 2, find displacement s from t=0 to t=3 (s=0 at t=0)

🎯 Quiz — Integration

Fill in the Blanks (1 mark each)

1. Integration is the _____ of differentiation.

2. ∫xⁿ dx = xⁿ⁺¹/(n+1) + _____ (for indefinite integrals).

3. The area under a curve between x=a and x=b = ∫_a^b _____ dx.

4. In kinematics, displacement s = ∫_____ dt.

5. ∫6x² dx = _____ + c.

Short Answer (2 marks each)

6. Integrate: ∫(3x² − 4x + 5) dx

7. Evaluate ∫₁⁴ (2x + 1) dx

8. A velocity v = 4t − 2. Find the displacement from t=1 to t=3.