The area enclosed by a parabola and a line

The area a parabola and a line enclose is decided by the gap between their intersections alone. We confirm it for $y = x^2$ and $y = 2x + 3$ .

Finding the intersections

From $x^2 = 2x + 3$ , that is $x^2 - 2x - 3 = (x - 3)(x + 1) = 0$ , we get $x = -1$ and $x = 3$ . The points are $(-1, 1)$ and $(3, 9)$ .

Integrating directly

On this interval the line is above, so the area is as follows.

\int_{-1}^{3} (2x + 3 - x^2) \, dx = \left[ x^2 + 3x - \frac{x^3}{3} \right]_{-1}^{3} = 9 - \left( -\frac{5}{3} \right) = \frac{32}{3}

Getting it from the gap alone

Writing $\alpha$ and $\beta$ with $\alpha < \beta$ for the intersections, the integrand factors as $-(x - \alpha)(x - \beta)$ .

\int_{\alpha}^{\beta} -(x - \alpha)(x - \beta) \, dx = \frac{(\beta - \alpha)^3}{6}

Here $\dfrac{(3 - (-1))^3}{6} = \dfrac{64}{6} = \dfrac{32}{3}$ , in agreement. This is known as the one-sixth formula.

Method	Computation	Result
From the antiderivative	$9 - \left( -\dfrac{5}{3} \right)$	$\dfrac{32}{3}$
The one-sixth formula	$\dfrac{4^3}{6}$	$\dfrac{32}{3}$

When there is a coefficient

If the coefficient of $x^2$ is not $1$ , the area is multiplied accordingly: the region enclosed by $y = ax^2 + bx + c$ and a line has area $\dfrac{|a| (\beta - \alpha)^3}{6}$ . For the region between two parabolas the same formula applies, with $a$ the difference of their coefficients of $x^2$ .

It grows with the cube

What makes it interesting is that the area grows as the cube of the gap.

Gap between the intersections	Area, with coefficient $1$
$1$	$\dfrac{1}{6}$
$2$	$\dfrac{8}{6}$
$4$	$\dfrac{64}{6}$

Doubling the gap multiplies the area by eight.

Since the answer follows the moment the intersections are known, no antiderivative has to be computed. Once the values $-1$ and $3$ are in hand, all that remains is $\dfrac{4^3}{6}$ .

The parabola on the graph is $y = x^2$ , the line is $y = 2x + 3$ , and the large dots are the intersections.