Pure 1: Integration

Pure 1: Integration — CIE A Level Mathematics 9709

1. Standard Results

Integration is the reverse of differentiation (anti-differentiation).

| f(x) | ∫f(x)dx |

|---|---|

| xⁿ (n ≠ −1) | xⁿ⁺¹/(n+1) + C |

| 1/x | ln|x| + C |

| eˣ | eˣ + C |

| sin x | −cos x + C |

| cos x | sin x + C |

| sec²x | tan x + C |

| (ax+b)ⁿ | (ax+b)ⁿ⁺¹/[a(n+1)] + C |

Key: Always add the constant of integration C for indefinite integrals.

2. Integration by Substitution

Used when the integrand contains a composite function.

Method: Let u = g(x), find du/dx, replace dx = du/(du/dx), integrate in u, substitute back.

Example: ∫ x(x² + 1)³ dx

Let u = x² + 1 → du = 2x dx → x dx = du/2.

∫ u³ × du/2 = u⁴/8 + C = (x² + 1)⁴/8 + C

Example: ∫ sin(2x) dx

Let u = 2x → du = 2 dx → dx = du/2.

∫ sin u × du/2 = −cos u/2 + C = −cos(2x)/2 + C

3. Definite Integrals and Area

Definite integral: $int_a^b f(x) , dx = [F(x)]_a^b = F(b) - F(a)$

Area under a curve (between curve and x-axis, a ≤ x ≤ b):

• If f(x) ≥ 0: Area = ∫ₐᵇ f(x) dx
• If f(x) < 0 in part of [a,b]: split at roots and take |value| for negative regions

Area between two curves: Area = ∫ₐᵇ [f(x) − g(x)] dx, where f(x) ≥ g(x) in [a,b].

Find where curves meet (set equal) to determine limits.

Example: Find the area enclosed by y = x² and y = 2x − x².

Intersection: x² = 2x − x² → 2x² − 2x = 0 → x(x−1) = 0 → x = 0, 1.

In [0,1], 2x − x² > x² → Area = ∫₀¹ (2x − 2x²) dx = [x² − 2x³/3]₀¹ = 1 − 2/3 = 1/3 sq. units

4. Volumes of Revolution

About x-axis: V = π ∫ₐᵇ y² dx

About y-axis: V = π ∫_c^d x² dy

Example: The region bounded by y = √x, x = 0, x = 4, y = 0, is rotated about the x-axis. Find the volume.

V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π[x²/2]₀⁴ = π × 8 = 8π cubic units

5. Trapezium Rule (Numerical Integration)

When exact integration is impossible or impractical:

$$int_a^b f(x) , dx approx rac{h}{2}[y_0 + 2(y_1 + y_2 + ldots + y_{n-1}) + y_n]$$

where h = (b−a)/n (strip width) and y_i = f(a + ih).

Improving accuracy: Increase the number of strips (smaller h).

Overestimate or underestimate: depends on concavity of the curve.

6. Integration as the Reverse of Differentiation

Finding constants: If dy/dx = f(x) and a point (a, b) is given:

y = ∫f(x) dx + C, then substitute x = a, y = b to find C.

Example: dy/dx = 3x² − 2x and y = 5 when x = 1.

y = x³ − x² + C; 5 = 1 − 1 + C → C = 5; y = x³ − x² + 5