Integration as the reverse of differentiation
Integration undoes differentiation. If d/dx [F(x)] = f(x), then ∫ f(x) dx = F(x) + C, where C is any constant.
What integration means
Differentiation starts with a function and gives its derivative. Integration goes the other way. It starts with a derivative f(x) and finds a function F(x) whose derivative is f(x). F(x) is called an antiderivative (or primitive) of f(x). The derivative of a constant is zero, so x², x² + 5 and x² − 3 all have derivative 2x. That is why every answer carries an arbitrary constant C. The sign ∫ is the integral sign, f(x) is the integrand, and dx says that x is the variable of integration. Extra help (no longer in your textbook): y = F(x) + C is a family of curves, each shifted up or down. At any fixed x, all of them have parallel tangents, because their slope is the same f(x).
Standard integrals: read each derivative backwards
| Derivative you know | Integral it gives |
|---|---|
| d/dx [xⁿ⁺¹/(n + 1)] = xⁿ | ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C, n ≠ −1 |
| d/dx (x) = 1 | ∫ dx = x + C |
| d/dx (sin x) = cos x | ∫ cos x dx = sin x + C |
| d/dx (−cos x) = sin x | ∫ sin x dx = −cos x + C |
| d/dx (tan x) = sec² x | ∫ sec² x dx = tan x + C |
| d/dx (−cot x) = cosec² x | ∫ cosec² x dx = −cot x + C |
| d/dx (sec x) = sec x tan x | ∫ sec x tan x dx = sec x + C |
| d/dx (−cosec x) = cosec x cot x | ∫ cosec x cot x dx = −cosec x + C |
| d/dx (sin⁻¹ x) = 1/√(1 − x²) | ∫ dx/√(1 − x²) = sin⁻¹ x + C |
| d/dx (tan⁻¹ x) = 1/(1 + x²) | ∫ dx/(1 + x²) = tan⁻¹ x + C |
| Extra help, not in your textbook's list: d/dx (sec⁻¹ x) = 1/(x√(x² − 1)) for x > 1 | ∫ dx/(x√(x² − 1)) = sec⁻¹ x + C |
| d/dx (eˣ) = eˣ | ∫ eˣ dx = eˣ + C |
| d/dx (log|x|) = 1/x | ∫ (1/x) dx = log|x| + C |
| d/dx (aˣ/log a) = aˣ | ∫ aˣ dx = aˣ/log a + C |
| d/dx (cos⁻¹ x) = −1/√(1 − x²) | ∫ dx/√(1 − x²) = −cos⁻¹ x + C (the same pattern gives ∫ dx/(1 + x²) = −cot⁻¹ x + C) |
Differentiation vs integration
The two are inverse processes: d/dx [∫ f(x) dx] = f(x), and ∫ f′(x) dx = f(x) + C. Two antiderivatives of the same function differ only by a constant.
Example 1: split, then integrate
Find ∫ (2x³ − 3x² + 5)/x² dx.
- Split the fraction: (2x³ − 3x² + 5)/x² = 2x − 3 + 5x⁻².
- Use linearity: 2 ∫ x dx − 3 ∫ dx + 5 ∫ x⁻² dx.
- Power rule: ∫ x dx = x²/2, ∫ dx = x, ∫ x⁻² dx = x⁻¹/(−1) = −1/x.
- So the integral = 2(x²/2) − 3x + 5(−1/x) + C = x² − 3x − 5/x + C.
- Check: d/dx (x² − 3x − 5/x) = 2x − 3 + 5/x², which is the integrand.
x² − 3x − 5/x + C
Example 2: rewrite with trigonometric ratios
Find ∫ (3 + 4 cos x)/sin² x dx.
- Split: (3 + 4 cos x)/sin² x = 3/sin² x + 4 (cos x/sin x)(1/sin x).
- = 3 cosec² x + 4 cosec x cot x.
- Use ∫ cosec² x dx = −cot x and ∫ cosec x cot x dx = −cosec x.
- So the integral = −3 cot x − 4 cosec x + C.
- Check: d/dx (−3 cot x − 4 cosec x) = 3 cosec² x + 4 cosec x cot x, the integrand.
−3 cot x − 4 cosec x + C
Integration rebuilds a total from a rate. From the rate at which water flows into an overhead tank, or the speed shown on a bus's speedometer, integration gives the total water stored or the distance travelled.
1 marks: “Find ∫ (3√x + 1/√x) dx.” 3 × x^(3/2)/(3/2) + x^(1/2)/(1/2) = 2x^(3/2) + 2√x + C.
2 marks: “Find ∫ cos 2x/(cos² x sin² x) dx.” cos 2x = cos² x − sin² x, so the integrand = 1/sin² x − 1/cos² x = cosec² x − sec² x. Answer: −cot x − tan x + C.
2 marks: “Find F(x) if F′(x) = 3x² + 2x and F(1) = 4.” F(x) = x³ + x² + C. F(1) = 2 + C = 4, so C = 2. Answer: F(x) = x³ + x² + 2.
Integrate, then differentiate to check. If you get the integrand back, the answer is right. Never forget + C.