Introduction to Integrals

What is an Integral?

Integral: Accumulation of quantities or area under a curve

Two types:

1. Definite integral: Area under curve from a to b
2. Indefinite integral: Antiderivative (reverse of derivative)

Fundamental connection: Integration is inverse of differentiation

Antiderivatives (Indefinite Integrals)

Antiderivative: Function whose derivative gives original function

Notation: ∫f(x)dx (read "integral of f(x) with respect to x")

If F'(x) = f(x), then ∫f(x)dx = F(x) + C

C = constant of integration (because derivative of constant is 0)

Example 1: Find Antiderivative

Find ∫x²dx

Ask: What function has derivative x²?

Power rule (reverse): x³/3 has derivative x²

Answer: ∫x²dx = x³/3 + C

Example 2: Check Result

∫2xdx = x² + C

Check: d/dx(x² + C) = 2x ✓

Power Rule for Integration

For x^n (n ≠ -1):

∫x^n dx = x^(n+1)/(n+1) + C

"Add 1 to exponent, divide by new exponent"

Example 1: Power Rule

∫x⁵dx = x⁶/6 + C

∫x⁴dx = x⁵/5 + C

∫x dx = x²/2 + C

Example 2: Constant

∫5 dx = 5x + C

Constant k integrates to kx

Example 3: Polynomial

∫(3x² - 4x + 2)dx

= 3(x³/3) - 4(x²/2) + 2x + C = x³ - 2x² + 2x + C

Special Integrals

∫sin(x)dx = -cos(x) + C

∫cos(x)dx = sin(x) + C

∫e^x dx = e^x + C

∫(1/x)dx = ln|x| + C (special case for n = -1)

Example: Trig Integral

∫(2sin(x) + 3cos(x))dx

= 2(-cos(x)) + 3(sin(x)) + C = -2cos(x) + 3sin(x) + C

Area Under a Curve

Definite integral: ∫[from a to b] f(x)dx

Geometric meaning: Area between curve and x-axis from x = a to x = b

Notation: ∫ₐᵇ f(x)dx or ∫[a,b] f(x)dx

Note: Area below x-axis counts as negative

Example: Interpret Area

∫[0 to 3] 2dx = ?

This is rectangle: height 2, width 3

Area = 2 × 3 = 6

Fundamental Theorem of Calculus

Part 1: If F'(x) = f(x), then

∫[a to b] f(x)dx = F(b) - F(a)

Process:

1. Find antiderivative F(x)
2. Evaluate F(b) - F(a)

Notation: F(x)|ₐᵇ means F(b) - F(a)

Example 1: Evaluate Definite Integral

∫[1 to 3] x²dx

Antiderivative: F(x) = x³/3

Evaluate:

F(3) - F(1) = 3³/3 - 1³/3
             = 27/3 - 1/3
             = 26/3

Answer: 26/3 ≈ 8.67

Example 2: Polynomial

∫[0 to 2] (2x + 1)dx

Antiderivative: F(x) = x² + x

Evaluate:

F(2) - F(0) = (4 + 2) - (0 + 0)
             = 6

Area under curve: 6 square units

Riemann Sums (Approximating Area)

Riemann sum: Approximate area using rectangles

Types:

• Left sum: Use left endpoint of each interval
• Right sum: Use right endpoint
• Midpoint sum: Use middle of each interval

As rectangles get narrower, approximation improves

Example: Left Riemann Sum

Approximate ∫[0 to 2] x²dx using 4 rectangles

Width of each: Δx = (2-0)/4 = 0.5

Left endpoints: 0, 0.5, 1, 1.5

Heights:

• f(0) = 0
• f(0.5) = 0.25
• f(1) = 1
• f(1.5) = 2.25

Area ≈ 0.5(0 + 0.25 + 1 + 2.25) = 0.5(3.5) = 1.75

Exact answer: ∫[0 to 2] x²dx = 8/3 ≈ 2.67

(Left sum underestimates for increasing function)

Properties of Definite Integrals

∫[a to a] f(x)dx = 0 (zero width)

∫[a to b] f(x)dx = -∫[b to a] f(x)dx (reverse limits)

∫[a to b] [f(x) + g(x)]dx = ∫[a to b] f(x)dx + ∫[a to b] g(x)dx

∫[a to b] c·f(x)dx = c·∫[a to b] f(x)dx

Example: Using Properties

∫[1 to 3] 5x²dx

= 5·∫[1 to 3] x²dx = 5·(x³/3)|₁³ = 5·(9 - 1/3) = 5·(26/3) = 130/3

Negative Areas

When f(x) < 0: Integral gives negative value

Total area: Take absolute value or split into regions

Example: Area Below Axis

∫[-1 to 1] xdx

= (x²/2)|₋₁¹ = 1/2 - 1/2 = 0

Why 0? Positive area (x > 0) cancels negative area (x < 0)

Displacement vs Distance

If v(t) = velocity:

∫[a to b] v(t)dt = displacement (net change in position)

∫[a to b] |v(t)|dt = total distance (absolute value)

Example: Motion

v(t) = t - 2 for 0 ≤ t ≤ 4

Displacement:

∫[0 to 4] (t - 2)dt = (t²/2 - 2t)|₀⁴
                    = (8 - 8) - 0
                    = 0

Returns to starting point!

But total distance traveled ≠ 0 (moved away then back)

Real-World Applications

Physics: Work = ∫F(x)dx (force over distance)

Economics: Total profit from marginal profit function

Biology: Total growth from growth rate

Probability: Area under probability density function

Example: Work

Force F(x) = 10x newtons to compress spring x meters

Work to compress 2 meters:

W = ∫[0 to 2] 10x dx
  = 10(x²/2)|₀²
  = 10(2)
  = 20 joules

Average Value of Function

Average value on [a, b]:

f_avg = (1/(b-a))·∫[a to b] f(x)dx

Example: Average Value

Find average value of f(x) = x² on [0, 3]

Calculate:

f_avg = (1/3)·∫[0 to 3] x² dx
      = (1/3)·(x³/3)|₀³
      = (1/3)·9
      = 3

Average value: 3

Fundamental Theorem Part 2

If f is continuous:

d/dx[∫[a to x] f(t)dt] = f(x)

"Derivative of integral gives back original function"

Example: Derivative of Integral

Let F(x) = ∫[0 to x] t²dt

F'(x) = x² (by FTC Part 2)

Substitution Method

For integrals requiring chain rule in reverse

If u = g(x), then du = g'(x)dx

∫f(g(x))·g'(x)dx = ∫f(u)du

Example: u-Substitution

∫2x(x² + 1)³dx

Let u = x² + 1, then du = 2xdx

= ∫u³ du = u⁴/4 + C = (x² + 1)⁴/4 + C

Practice