Integration — The Reverse of Differentiation

If differentiation is about finding the gradient (slope), integration is about going backwards — finding the original function from its derivative. It also lets us calculate areas under curves.

What Integration Actually Is

Differentiation says: “I have a function, what’s its gradient?” Integration says the opposite: “I know the gradient, what was the original function?”

Because many different functions could have the same derivative (they just differ by a constant), we always add + C (the “constant of integration”) for indefinite integrals.

Key Formulas

Power rule: ∫ xⁿ dx = xⁿ⁺¹ / (n+1) + C (where n ≠ −1)
Constant multiplier: ∫ kf(x) dx = k ∫ f(x) dx
Sum/difference: ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx

The Power Rule Reversed

In differentiation you “bring the power down and subtract one.” In integration you do the opposite:

Add 1 to the power.
Divide by the new power.
Don’t forget + C.

Worked Example 1

∫ x³ dx

Add 1 to power: 3 → 4. Divide by 4.

= x⁴/4 + C

Definite Integrals (With Limits)

A definite integral has an upper and lower limit. Integrate as normal (no + C needed), then substitute the upper limit into your answer, substitute the lower limit, and subtract.

Worked Example 2

Evaluate ∫₁³ 2x dx

Integrate: 2x → x²
Upper limit (x=3): 3² = 9
Lower limit (x=1): 1² = 1
Subtract: 9 − 1 = 8

Area Under a Curve

The definite integral between two x-values gives the area between the curve and the x-axis. If the curve dips below the x-axis, that part of the integral is negative — so you may need to split the region and take absolute values.

Worked Example 3

Find the area under y = x² between x = 0 and x = 2.

∫ x² dx = x³/3
Upper (x=2): 8/3
Lower (x=0): 0
Area = 8/3 − 0 = 8/3 ≈ 2.67 square units

Common Mistakes

Forgetting + C in indefinite integrals.
Subtracting limits the wrong way round (always upper minus lower).
Not rewriting terms before integrating (e.g. 1/x² should become x⁻² first).
Ignoring negative areas when the curve is below the x-axis.

Practice Questions

Integrate: ∫ 5x² dx
Integrate: ∫ (3x² + 4x − 1) dx
Evaluate: ∫₀² 3x² dx
Find the area under y = 4x between x = 1 and x = 3.
Integrate: ∫ x⁻² dx (rewrite as 1/x² to check)

Answers: 1) 5x³/3 + C 2) x³ + 2x² − x + C 3) 8 4) 16 5) −x⁻¹ + C (or −1/x + C)

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