Properties of definite integrals
Properties evaluate integrals with no easy antiderivative. Key idea: replace x by a + b − x, then add the two forms of the integral.
The properties
| Property | Statement | Use it when |
|---|---|---|
| Change of letter | ∫ f(x) dx from a to b = ∫ f(t) dt from a to b | Renaming the variable after a substitution |
| Swap the limits | ∫ f(x) dx from a to b = −∫ f(x) dx from b to a; ∫ f(x) dx from a to a = 0 | A substitution turns the limits upside down |
| Split the interval | ∫ f(x) dx from a to b = ∫ from a to c + ∫ from c to b | Modulus or piecewise functions, e.g. |x − 2| |
| a + b − x | ∫ f(x) dx from a to b = ∫ f(a + b − x) dx from a to b | f(x) and f(a + b − x) add up to something simple |
| a − x | ∫ f(x) dx from 0 to a = ∫ f(a − x) dx from 0 to a | Limits 0 to π/2, where sin x and cos x swap places |
| Split at a | ∫ f(x) dx from 0 to 2a = ∫ f(x) dx from 0 to a + ∫ f(2a − x) dx from 0 to a | Proving the 0 to 2a property |
| 0 to 2a | = 2 ∫ f(x) dx from 0 to a if f(2a − x) = f(x); = 0 if f(2a − x) = −f(x) | Limits 0 to π or 0 to 2π |
| Even or odd | ∫ f(x) dx from −a to a = 2 ∫ f(x) dx from 0 to a (even), or 0 (odd) | Limits symmetric about 0 |
Proof of the a − x property
- 1Let t = a − x. Then dt = −dx.
- 2Limits: x = 0 gives t = a; x = a gives t = 0.
- 3∫ f(x) dx from 0 to a = ∫ f(a − t)(−dt) from a to 0.
- 4Swapping the limits removes the minus sign: = ∫ f(a − t) dt from 0 to a.
- 5Change the letter back to x. So, ∫ f(x) dx from 0 to a = ∫ f(a − x) dx from 0 to a.
Example 1: the a + b − x property
Evaluate ∫ √x/(√x + √(5 − x)) dx from 1 to 4.
- Let I be the integral. Here a + b = 1 + 4 = 5.
- Replace x by 5 − x: I = ∫ √(5 − x)/(√(5 − x) + √x) dx from 1 to 4.
- Add the two forms: 2I = ∫ (√x + √(5 − x))/(√x + √(5 − x)) dx from 1 to 4 = ∫ 1 dx from 1 to 4.
- 2I = 4 − 1 = 3.
I = 3/2
Example 2: odd and even parts
Evaluate ∫ (x⁵ + x² sin x + tan³ x + 3) dx from −π/4 to π/4.
- Test each term by replacing x with −x.
- x⁵, x² sin x and tan³ x change sign, so they are odd. Their integrals from −π/4 to π/4 are 0.
- 3 is even: ∫ 3 dx from −π/4 to π/4 = 2 × 3 × (π/4).
- So the integral = 0 + 3π/2.
3π/2
Example 3: a modulus
Evaluate ∫ |x − 2| dx from 0 to 5.
- x − 2 changes sign at x = 2. Split the interval there.
- For 0 ≤ x ≤ 2, |x − 2| = 2 − x. For 2 ≤ x ≤ 5, |x − 2| = x − 2.
- ∫ (2 − x) dx from 0 to 2 = [2x − x²/2] from 0 to 2 = 4 − 2 = 2.
- ∫ (x − 2) dx from 2 to 5 = [x²/2 − 2x] from 2 to 5 = (25/2 − 10) − (2 − 4) = 5/2 + 2 = 9/2.
- Total = 2 + 9/2.
13/2
Which property to try
- Limits −a to a: check first whether the function is even or odd.
- Limits 0 to π/2 with sin x and cos x: use the a − x property, which swaps sin x and cos x.
- A factor x times a function of sin x on 0 to π: replace x by π − x, add, and the x cancels.
- A modulus: split the interval where the expression inside is zero.
- After replacing x, add the old and new forms to get 2I.
Symmetry saves work in real design. An engineer checking a symmetric arch bridge or a ceiling fan blade works out one half and doubles it, just as the even-function property does.
1 marks: “Evaluate ∫ sin⁵ x cos² x dx from −π/2 to π/2.” Replace x by −x: sin⁵(−x) cos²(−x) = −sin⁵ x cos² x, so the integrand is odd. The integral is 0.
3 marks: “Evaluate ∫ |x³ − x| dx from −1 to 1.” x³ − x is odd, so |x³ − x| is even: the integral = 2 ∫ |x³ − x| dx from 0 to 1. On [0, 1], x³ ≤ x, so |x³ − x| = x − x³. 2[x²/2 − x⁴/4] from 0 to 1 = 2(1/4) = 1/2.
3 marks: “Evaluate ∫ (sin x − cos x)/(2 + sin 2x) dx from 0 to π/2.” Replace x by π/2 − x: sin x and cos x swap, and sin 2x stays the same. So I = −I, 2I = 0 and I = 0.
Swap and add. Replace x by (a + b − x), add the two forms, and the hard part cancels.