Definite Integrals — Moosa Academy

An integral measures area, and the area of a real shape is never negative. Nobody quotes the floor area of a room as $-5$ square metres. Yet a definite integral can come out negative, and it can come out as exactly zero even when the curve is far from flat. The sign is not recording size — it is recording direction of effect.

Definition The definite integral

The definite integral $\int_{a}^{b} f(x)\,dx$ is the signed area between the curve $y = f(x)$ and the $x$ -axis, taken from $x = a$ to $x = b$ .

The numbers

a

and

b

written on the sign are the limits. They turn the integral from a family of functions into a single number, so a definite integral has no

+ C

attached to it.

Where the curve sits above the axis the area counts as positive. Where it drops below, the same area counts as negative.

Note Why an area is allowed to be negative

Picture a tank being filled with water, and let $f$ be the flow function — how fast water is moving through the pipe at each moment.

While

f

is above the axis, water is going in and the level rises.

Where

f

crosses zero, the flow stops.

Below the axis the flow has reversed: water is being drawn back out.

The patch of area below the axis is not smaller than the one above it — it may well be exactly the same size. What differs is its effect on the tank, which is the opposite of the effect of the patch above. The minus sign is how the integral records that reversal.

Theorem Equal areas cancel

If the region above the axis has the same area as the region below it, the definite integral is zero.

Each triangle here has base

1

and height

1

, so each has area

\tfrac{1}{2}

. One counts as

+\tfrac{1}{2}

, the other as

-\tfrac{1}{2}

$\int_{0}^{2} (x - 1)\,dx = 0$

Zero does not mean nothing happened. The same water that went in was taken back out, so the tank finished where it started. A definite integral reports the net effect, not the amount of activity.

Theorem Evaluating between limits

If $F$ is any antiderivative of $f$ , then $\int_{a}^{b} f(x)\,dx = F(b) - F(a)$ .

Find an antiderivative

F

, so that

F'(x) = f(x)

Substitute the upper limit:

F(b)

Substitute the lower limit:

F(a)

Subtract, upper minus lower.

This is also where the constant goes. Carrying $+ C$ through gives $\big(F(b) + C\big) - \big(F(a) + C\big)$ , and the two constants cancel. That is why a definite integral never needs one.

Example A straightforward evaluation

Evaluate $\int_{1}^{3} 2x\,dx$ .

An antiderivative of

2x

F(x) = x^2

Upper limit:

F(3) = 3^2 = 9

Lower limit:

F(1) = 1^2 = 1

Subtract:

9 - 1

$\int_{1}^{3} 2x\,dx = 8$

The whole curve sits above the axis on $[1, 3]$ , so nothing is subtracted and the answer is the plain area of the region.

Example Net value against total area

For $y = x - 1$ on $[0, 2]$ , find the definite integral, then find the total area enclosed with the axis.

An antiderivative is

F(x) = \dfrac{x^2}{2} - x

F(2) = 2 - 2 = 0

and

F(0) = 0

$\int_{0}^{2} (x - 1)\,dx = 0$

For total area, split at the crossing point

x = 1

[0, 1]

the integral is

-\dfrac{1}{2}

, a region of size

\dfrac{1}{2}

[1, 2]

the integral is

+\dfrac{1}{2}

Total area $= \dfrac{1}{2} + \dfrac{1}{2} = 1$

Two different questions, two different answers from the same curve. If a question asks for area rather than for the integral, split at every point where the curve crosses the axis and add the sizes.

Summary

$\int_{a}^{b} f(x)\,dx$ is the signed area between the curve and the $x$ -axis.
Above the axis counts as positive, below the axis as negative.
The sign records direction of effect — filling against draining — not size.
Equal areas on opposite sides cancel, and the integral reads zero.
Evaluate with $F(b) - F(a)$ ; the $+ C$ cancels, so limits leave a single number.
For total area rather than net value, split at each crossing and add the sizes.