The definite integral and area

A definite integral gives the area enclosed by a curve and the $x$ -axis¹. We compute it for $f(x) = x^2$ from $0$ to $2$ .

Looking for an antiderivative

Looking for a function whose derivative is $x^2$ turns up $F(x) = \dfrac{x^3}{3}$ , since indeed $F'(x) = x^2$ .

\int_0^2 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} - 0 = \frac{8}{3}

The area is $\dfrac{8}{3} \approx 2.667$ . The rectangle $0 \leq x \leq 2$ , $0 \leq y \leq 4$ has area $8$ , so the region under the parabola takes up exactly one third of it.

The constant of integration cancels

An antiderivative is not unique: $\dfrac{x^3}{3} + 5$ also differentiates to $x^2$ . But a definite integral takes the difference between the upper and the lower end, so the constant cancels and the answer is unchanged. Any antiderivative will do.

Orientation and splitting

Property	Expression
Swapping the ends	$\int_2^0 x^2 dx = -\dfrac{8}{3}$
Cutting the interval	$\int_0^2 = \int_0^1 + \int_1^2$
The actual values	$\dfrac{1}{3} + \dfrac{7}{3} = \dfrac{8}{3}$

A definite integral is an area with an orientation, counted positive when moving from left to right.

Intervals where the value is negative

Where $f$ is negative the definite integral is negative too. To get an area, the part below the $x$ -axis has to have its sign put back. Here $x^2 \geq 0$ , so the question does not arise.

Turning on the integral in the settings panel with lower end $0$ and upper end $2$ shades the region just computed. The parabola on the graph is $y = x^2$ , and the large dots are the two ends of the integral.

Integral, Wikipedia