Unit 1: Fundamental Integration Concepts

Integration of Rational Functions

A rational function is a function of the form f(x) = P(x) / Q(x), where P(x) and Q(x) are polynomials.

Proper vs. Improper Fractions

1. Proper Rational Function: The degree of the numerator P(x) is less than the degree of the denominator Q(x).
   Example: (2x + 1) / (x² + 3x - 4)
2. Improper Rational Function: The degree of the numerator P(x) is greater than or equal to the degree of the denominator Q(x).
   Example: (x³ + 2x) / (x² + 1)

Partial Fraction Decomposition

This is the technique for integrating proper rational functions. The goal is to break down a complex fraction into a sum of simpler fractions.

Case 1: Denominator has distinct linear factors.

Q(x) = (x - a)(x - b)...

Decomposition: A/(x - a) + B/(x - b) + ...

Example: ∫ (x + 1) / (x² - 4) dx = ∫ (x + 1) / ((x - 2)(x + 2)) dx
Set (x + 1) / ((x - 2)(x + 2)) = A/(x - 2) + B/(x + 2)
Multiply by the denominator: x + 1 = A(x + 2) + B(x - 2)
(Cover-up method)
Let x = 2: 3 = A(4) + B(0) → A = 3/4
Let x = -2: -1 = A(0) + B(-4) → B = 1/4
Integral becomes: ∫ (3/4)/(x - 2) dx + ∫ (1/4)/(x + 2) dx
Result: (3/4)ln|x - 2| + (1/4)ln|x + 2| + C

Case 2: Denominator has repeated linear factors.

Q(x) = (x - a)³...

Decomposition: A/(x - a) + B/(x - a)² + C/(x - a)³ + ...

Example: ∫ (2x) / (x - 1)² dx
Set (2x) / (x - 1)² = A/(x - 1) + B/(x - 1)²
2x = A(x - 1) + B
Let x = 1: 2 = A(0) + B → B = 2
Equate coefficients of x: 2 = A → A = 2
Integral becomes: ∫ 2/(x - 1) dx + ∫ 2/(x - 1)² dx
Result: 2ln|x - 1| - 2/(x - 1) + C

Case 3: Denominator has distinct irreducible quadratic factors.

Q(x) = (ax² + bx + c)... (where b² - 4ac < 0)

Decomposition: (Ax + B) / (ax² + bx + c) + ...

Example: ∫ (3) / (x(x² + 1)) dx
Set 3 / (x(x² + 1)) = A/x + (Bx + C)/(x² + 1)
3 = A(x² + 1) + (Bx + C)x
Let x = 0: 3 = A(1) → A = 3
Equate coefficients of x²: 0 = A + B → 0 = 3 + B → B = -3
Equate coefficients of x: 0 = C
Integral becomes: ∫ 3/x dx + ∫ (-3x)/(x² + 1) dx
Result: 3ln|x| - (3/2)ln|x² + 1| + C

Case 4: Denominator has repeated irreducible quadratic factors.

Decomposition: (Ax + B)/(ax² + bx + c) + (Cx + D)/(ax² + bx + c)² + ...

(These are more complex and involve trigonometric substitution).

Definite Integral as the Limit of a Sum

This is the fundamental definition of the definite integral. It represents the area under a curve f(x) from x = a to x = b by summing the areas of an infinite number of infinitesimally thin rectangles.

∫ [from a to b] f(x) dx = lim (n → ∞) Σ [from i=1 to n] h * f(a + ih)

Where:

• n is the number of rectangles.
• h = (b - a) / n is the width of each rectangle.
• a + ih is the x-coordinate of the right-hand side of the i-th rectangle.

Sometimes the sum is written from i=0 to n-1, representing the left-hand sum: lim (n → ∞) Σ [from i=0 to n-1] h * f(a + ih). Both definitions are equivalent for continuous functions.

Key Summation Formulas:

• Σ 1 = n
• Σ i = n(n + 1) / 2
• Σ i² = n(n + 1)(2n + 1) / 6
• Σ i³ = [n(n + 1) / 2]²
• Sum of a Geometric Series: a + ar + ... + ar^(n-1) = a(r^n - 1) / (r - 1)

Example: Evaluate ∫ [from 0 to 2] x² dx as a limit of a sum.

1. Identify: a = 0, b = 2, f(x) = x².
2. Find h: h = (2 - 0) / n = 2/n.
3. Find f(a + ih): f(0 + i*h) = f(ih) = (ih)² = i²h².
4. Set up the sum: lim (n → ∞) Σ [from i=1 to n] h * (i²h²)
5. = lim (n → ∞) Σ [from i=1 to n] h³i² = lim (n → ∞) h³ * Σ i²
6. Substitute formulas for h and Σ i²:
   = lim (n → ∞) (2/n)³ * [n(n + 1)(2n + 1) / 6]
7. = lim (n → ∞) (8/n³) * [n(n + 1)(2n + 1) / 6]
8. = (8/6) * lim (n → ∞) [n(n + 1)(2n + 1) / n³]
9. = (4/3) * lim (n → ∞) [(n/n) * ((n + 1)/n) * ((2n + 1)/n)]
10. = (4/3) * lim (n → ∞) [1 * (1 + 1/n) * (2 + 1/n)]
11. As n → ∞, 1/n → 0.
    = (4/3) * [1 * (1 + 0) * (2 + 0)] = (4/3) * 2 = 8/3.

Properties of Definite Integrals

These properties are essential for simplifying and evaluating definite integrals, often without finding the antiderivative.

• P-0: ∫[a, b] f(x) dx = ∫[a, b] f(t) dt
  (Changing the variable of integration doesn't change the value).
• P-1: ∫[a, b] f(x) dx = - ∫[b, a] f(x) dx
  (Swapping the limits of integration negates the value).
• P-2: ∫[a, b] f(x) dx = ∫[a, c] f(x) dx + ∫[c, b] f(x) dx
  (Splitting the interval. This is key for piecewise and modulus functions).
• P-3 ("King's Property"): ∫[a, b] f(x) dx = ∫[a, b] f(a + b - x) dx
  (The most powerful property for evaluation).
• P-4 (Special case of P-3): ∫[0, a] f(x) dx = ∫[0, a] f(a - x) dx
• P-5: ∫[0, 2a] f(x) dx = ∫[0, a] f(x) dx + ∫[0, a] f(2a - x) dx
• P-6 (from P-5): ∫[0, 2a] f(x) dx =
2 * ∫[0, a] f(x) dx, if f(2a - x) = f(x) (even symmetry about a)
  0, if f(2a - x) = -f(x) (odd symmetry about a)
• P-7 (Even/Odd Function Property): ∫[-a, a] f(x) dx =
2 * ∫[0, a] f(x) dx, if f(x) is an even function (i.e., f(-x) = f(x))
  0, if f(x) is an odd function (i.e., f(-x) = -f(x))

Example using P-4: Evaluate I = ∫[0, π/2] (sin x) / (sin x + cos x) dx

1. Apply P-4: f(x) = (sin x) / (sin x + cos x). a = π/2.
   f(a - x) = f(π/2 - x) = sin(π/2 - x) / (sin(π/2 - x) + cos(π/2 - x))
f(π/2 - x) = (cos x) / (cos x + sin x)
2. By P-4, I = ∫[0, π/2] (cos x) / (sin x + cos x) dx
3. Add the original integral and the new one:
   I + I = ∫[0, π/2] (sin x) / (sin x + cos x) dx + ∫[0, π/2] (cos x) / (sin x + cos x) dx
4. 2I = ∫[0, π/2] (sin x + cos x) / (sin x + cos x) dx
5. 2I = ∫[0, π/2] 1 dx = [x] from 0 to π/2 = π/2
6. I = π/4

Unit 1: Exam Quick Tips