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A-Level

Integration — The Reverse of Differentiation

Learn integration from scratch — the reverse of differentiation. Understand indefinite and definite integrals, the +C constant, and finding areas under curves.

If differentiation tells you the gradient, integration goes backwards— it undoes differentiation. It's also used to find the area under a curve.

The Power Rule (Reversed)

∫ xⁿ dx = xⁿ⁺¹ ÷ (n+1) + C

In English: add 1 to the power, then divide by the new power. Always add + C.

FunctionIntegral
x⁴/4 + C
2xx² + C
55x + C
6x²2x³ + C

Why + C?

When you differentiate, constants disappear (e.g. x² + 3 and x² + 99 both give 2x). So when you reverse it, you don't know what the constant was. +C is a placeholder for that unknown constant. This is called an indefinite integral.

Worked Example 1 — Indefinite Integral

Find ∫ (3x² + 4x − 1) dx

  1. 3x² → 3 × x³/3 = x³
  2. 4x → 4 × x²/2 = 2x²
  3. −1 → −x
  4. Answer: x³ + 2x² − x + C

Definite Integrals (Finding Areas)

A definite integral has limits (numbers at the top and bottom of the ∫ sign). It gives you the exact area under the curve between two x-values.

∫ₐᵇ f(x) dx = F(b) − F(a)

Integrate, then substitute the top limit, subtract the bottom limit. No + C needed.

Worked Example 2 — Definite Integral

Find ∫₁³ 2x dx

  1. Integrate: 2x → x²
  2. Upper limit: 3² = 9
  3. Lower limit: 1² = 1
  4. Area = 9 − 1 = 8 square units

Worked Example 3 — Finding Area Under a Curve

Find the area under y = x² + 1 between x = 0 and x = 3.

  1. ∫₀³ (x² + 1) dx
  2. Integrate: x³/3 + x
  3. At x = 3: 27/3 + 3 = 9 + 3 = 12
  4. At x = 0: 0 + 0 = 0
  5. Area = 12 − 0 = 12 square units

Real-World Connection

If you have a speed-time graph, the area under the curve gives you the total distance travelled. Integration turns a rate of change back into a total.

Try It Yourself

  1. Find ∫ (6x² − 4x + 3) dx. (Answer: 2x³ − 2x² + 3x + C)
  2. Find ∫₂⁴ 3x² dx. (Answer: 64 − 8 = 56)
  3. Find the area under y = 4x between x = 0 and x = 5. (Answer: 50)