Partial Fractions and the Trapezium Rule
Partial Fractions and the Trapezium Rule
Partial fractions decompose a complicated fraction into simpler pieces that can be integrated individually. The trapezium rule provides a numerical method for approximating definite integrals that are difficult or impossible to evaluate analytically.
Partial Fractions
A proper algebraic fraction (degree of numerator < degree of denominator) with a factored denominator can be split into partial fractions.
Type 1: Linear Factors
For a denominator with distinct linear factors:
(px + q) / ((ax + b)(cx + d)) = A/(ax + b) + B/(cx + d)
Worked Example: Express (5x + 1) / ((x + 1)(x - 2)) in partial fractions.
Write: (5x + 1) / ((x + 1)(x - 2)) = A/(x + 1) + B/(x - 2)
Multiply both sides by (x + 1)(x - 2):
5x + 1 = A(x - 2) + B(x + 1)
Substitution method: Set x = 2: 11 = 3B, so B = 11/3.
Set x = -1: -4 = -3A, so A = 4/3.
Result: (5x + 1) / ((x + 1)(x - 2)) = (4/3)/(x + 1) + (11/3)/(x - 2)
Type 2: Repeated Linear Factors
If a factor appears twice:
(px + q) / ((ax + b)²) = A/(ax + b) + B/(ax + b)²
Worked Example: Express (3x + 5) / ((x + 1)²) in partial fractions.
3x + 5 = A(x + 1) + B
Set x = -1: 2 = B, so B = 2.
Compare x coefficients: 3 = A.
Result: 3/(x + 1) + 2/(x + 1)²
Type 3: Irreducible Quadratic Factor
If the denominator contains a quadratic that does not factorise:
(px² + qx + r) / ((ax + b)(cx² + d)) = A/(ax + b) + (Bx + C)/(cx² + d)
Integrating Partial Fractions
This is the main application. Splitting into partial fractions gives terms of the form:
- A/(ax + b) integrates to (A/a)·ln|ax + b| + c
- B/(ax + b)² integrates to -B/(a(ax + b)) + c (use substitution or recognise as power rule)
Worked Example: Find ∫ (5x + 1) / ((x + 1)(x - 2)) dx.
From above: ∫ [(4/3)/(x + 1) + (11/3)/(x - 2)] dx
= (4/3)ln|x + 1| + (11/3)ln|x - 2| + c
Improper Fractions
If the degree of the numerator is greater than or equal to the degree of the denominator, perform polynomial long division first to obtain a polynomial plus a proper fraction, then decompose the proper fraction.
Example: (x³ + 2) / (x² - 1) = x + (x + 2)/(x² - 1) after dividing.
Then decompose (x + 2)/((x - 1)(x + 1)) = 3/2·1/(x - 1) + (-1/2)·1/(x + 1).
The Trapezium Rule
When an integral cannot be found analytically, the trapezium rule gives an approximation by dividing the area under the curve into trapeziums.
Formula: ∫ₐᵇ y dx ≈ (h/2)[y₀ + yₙ + 2(y₁ + y₂ + ... + yₙ₋₁)]
where:
- h = (b - a)/n is the strip width
- n is the number of strips
- y₀, y₁, ..., yₙ are the y-values at the equally spaced x-values
In words: half the strip width, multiplied by (first + last + twice all the middle values).
Worked Example: Estimate ∫₁³ (1/x) dx using 4 strips.
h = (3 - 1)/4 = 0.5. The x-values are 1, 1.5, 2, 2.5, 3.
| x | 1 | 1.5 | 2 | 2.5 | 3 |
|---|---|---|---|---|---|
| 1/x | 1 | 0.6667 | 0.5 | 0.4 | 0.3333 |
Estimate = (0.5/2)[1 + 0.3333 + 2(0.6667 + 0.5 + 0.4)]
= 0.25[1.3333 + 2(1.5667)]
= 0.25[1.3333 + 3.1334]
= 0.25 × 4.4667 = 1.1167
The exact value is ln3 ≈ 1.0986, so the estimate is slightly too large (because 1/x is concave upward on this interval).
Accuracy of the Trapezium Rule
- More strips (larger n) gives a better approximation
- For a convex curve (curves upward), the trapezium rule overestimates
- For a concave curve (curves downward), the trapezium rule underestimates
- The error decreases proportionally to 1/n² — doubling n roughly quarters the error
Exam Tips
- When decomposing partial fractions, substituting strategic x-values (the roots of each factor) is the fastest method.
- Always check your decomposition by recombining — add the fractions back together to verify you recover the original.
- For the trapezium rule, set up a table of values before substituting into the formula. This helps prevent arithmetic errors.
- State whether the trapezium rule gives an overestimate or underestimate, with a reason based on the shape of the curve.
- In partial fractions questions that lead to integration, do not forget to add the constant of integration.