Integration

What you will learn

• See integration as the reverse of differentiation.
• Find areas under graphs with definite integrals.
• Connect integrals to distance and velocity.

Before you start: you should know derivatives and the power rule. See Derivatives.

The idea

Integration answers two questions that turn out to be the same question.

1. Reverse differentiation. If you know a function’s derivative, what was the original function?
2. Area under a curve. What is the area between a curve and the $x$-axis between two points?

Reversing differentiation

Differentiating $x^3$ gives $3x^2$. So integrating $3x^2$ should give back $x^3$. But $x^3 + 5$ and $x^3 - 100$ also differentiate to $3x^2$, because constants vanish. So we add an unknown constant of integration, $C$:

$\int 3x^2 \, dx = x^3 + C$

This is an indefinite integral: a whole family of functions, all with the same shape, shifted up or down.

Area as a sum of slices

To find the area under $y = f(x)$ from $x = a$ to $x = b$, chop the region into thin vertical rectangles of width $\Delta x$. Each has area about $f(x)\,\Delta x$. Add them up and let the width shrink to zero; the approximation becomes exact. The integral sign $\int$ is a stretched S, for “sum”.

The Fundamental Theorem of Calculus

The two ideas connect: the exact area is found by evaluating an antiderivative at the endpoints and subtracting.

$\int_a^b f(x)\, dx = F(b) - F(a)$

This is a definite integral, and the result is a number. The $+C$ cancels out in the subtraction.

Worked examples

Example 1: An indefinite integral

Find $\displaystyle\int (6x^2 - 4x + 3)\, dx$.

1. Integrate term by term with the power rule: add one to the power, then divide by the new power.
2. $6x^2 \to \dfrac{6x^3}{3} = 2x^3$; $-4x \to \dfrac{-4x^2}{2} = -2x^2$; $3 \to 3x$.
3. Add the constant.

Answer: $2x^3 - 2x^2 + 3x + C$. Check: differentiating gives back $6x^2 - 4x + 3$. ✓

Example 2: Area under a curve

Find the area under $y = x^2$ between $x = 1$ and $x = 3$.

1. Antiderivative: $F(x) = \dfrac{x^3}{3}$.
2. Evaluate at the limits: $F(3) = \dfrac{27}{3} = 9$ and $F(1) = \dfrac{1}{3}$.
3. Subtract: $9 - \dfrac{1}{3} = \dfrac{26}{3}$.

Answer: $\frac{26}{3} \approx 8.67$ square units.

Example 3: Distance from velocity

A cyclist’s velocity is $v = 2t + 1$ m/s. How far does she travel in the first $4$ seconds?

1. Distance is the integral of velocity: $\displaystyle\int_0^4 (2t + 1)\, dt$.
2. Antiderivative: $t^2 + t$.
3. Evaluate: $(16 + 4) - (0 + 0) = 20$.

Answer: $20$ metres

Areas below the $x$-axis come out negative in a definite integral. To find total area, integrate each part separately and add the absolute values.

Why it matters

Integration finds totals from rates. It calculates distance from speed, total rainfall from rainfall rate, energy used from power over time, and the volume of oddly shaped objects such as wine glasses or engine parts. In statistics, probabilities for continuous variables are areas under a probability curve, and in physics, work done is the integral of force over distance.