Integrals
Properties of definite integrals
Use properties P₀–P₇ to evaluate definite integrals.
Some definite integrals have no easy anti derivative, but the properties solve them. The main ones:
| Property | What it says |
|---|---|
| P₁ | ∫ₐᵇ f(x) dx = –∫ᵇₐ f(x) dx (swap limits, change sign) |
| P₂ | ∫ₐᵇ f = ∫ from a to c + ∫ from c to b (split the interval at a point c) |
| P₃ | ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b – x) dx |
| P₄ | ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a – x) dx |
| P₆ | ∫₀²ᵃ f(x) dx = 2∫₀ᵃ f(x) dx if f(2a – x) = f(x), and 0 if f(2a – x) = –f(x) |
| P₇ | ∫₋ₐᵃ f(x) dx = 2∫₀ᵃ f(x) dx if f is even, and 0 if f is odd |
P₂ for modulus: split where the inside of |...| changes sign. ∫₀² |x – 1| dx = ∫₀¹ (1 – x) dx + ∫₁² (x – 1) dx = ½ + ½ = 1.
P₇, even and odd: f is even if f(–x) = f(x) (like x², cos x) and odd if f(–x) = –f(x) (like x³, sin x). The two halves of an odd function cancel: ∫₋₂² x³ dx = 0, with no working needed.
P₃ / P₄, the 'replace and add' trick: call the integral I. Write it again with x replaced by (a + b – x). Often sin and cos swap places. Add the two forms: the awkward parts combine into something simple, like 1, giving 2I = an easy integral.
For ∫₀^(π/2), replacing x by π/2 – x changes sin x into cos x and cos x into sin x. That is why fractions like sinⁿx/(sinⁿx + cosⁿx) over [0, π/2] always give π/4.
When x itself appears in the top, as in ∫₀^π x f(sin x) dx, P₄ changes x into π – x; adding gives 2I = π∫₀^π f(sin x) dx, which removes the x.
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