The integral as a signed area

The integral $\displaystyle\int_a^b f(x)\,\mathrm{d}x$ measures an area — but an algebraic area, that is, a signed one. This is the crucial point to understand so that you never get it wrong.

Where the curve is above the x-axis, when $f$ is positive, the area counts positively. Where it goes below, when $f$ is negative, the area counts negatively.

Position of the curve	Sign of $f$	Contribution to the integral
above the axis	$f(x) > 0$	positive
below the axis	$f(x) < 0$	negative

An important consequence: if one part is above and another below, the integral gives the difference of the two areas.

$$\int_a^b f(x)\,\mathrm{d}x = \text{area above} - \text{area below}$$

It can even be zero if the two cancel out, while the actually shaded surface is not zero.

A telling example: the integral of the sine function over a full period is $0$, because the positive hump and the negative trough cancel out exactly.

$$\int_0^{2\pi} \sin x\,\mathrm{d}x = \bigl[-\cos x\bigr]_0^{2\pi} = -1 + 1 = 0$$

Yet the surface between the curve and the axis is not zero.

Remember this distinction, the source of half of all mistakes: the integral gives a signed area. If you want the true geometric area, which is always positive, you must take the sign of $f$ into account — that is the subject of the next lesson.