Area bounded by a parabola and a line

The interaction between a parabola and a line is one of the most common settings for area problems. The geometry is friendly: a parabola is a single quadratic, a line is linear, their intersection is two points (when they meet), and the bounded region is a single connected piece.

Setting up the problem

Two arrangements appear repeatedly.

Parabola $y = x^2$ (or $y = ax^2 + bx + c$) and a chord $y = mx + k$.

Solve simultaneously: $x^2 = mx + k$, a quadratic in $x$ with roots $\alpha$ and $\beta$ (assume $\alpha < \beta$). On $[\alpha, \beta]$, the line is above the parabola if $\alpha$ and $\beta$ are real. So $A = \int_\alpha^\beta [(mx + k) - x^2]\,dx.$

There is a beautiful formula for this: $\int_\alpha^\beta [\,L(x) - P(x)\,]\,dx = \frac{|c|}{6}(\beta - \alpha)^3,$ where $L - P = -c(x - \alpha)(x - \beta)$ and $c$ is the leading coefficient of $P$ (with sign chosen so the difference is positive on the interval). For $P(x) = x^2$ with $c = 1$, the area is $\dfrac{(\beta - \alpha)^3}{6}$.

Parabola $y^2 = 4ax$ (opens to the right) and a line.

Use horizontal slices. The parabola gives $x = y^2/(4a)$; the line gives $x$ as a function of $y$. Integrate the difference over the relevant $y$-range.

The $\tfrac{1}{6}$ derivation

Expand the integral with $L(x) - P(x) = -(x - \alpha)(x - \beta) = -(x^2 - (\alpha+\beta)x + \alpha\beta)$: $\int_\alpha^\beta -(x - \alpha)(x - \beta)\,dx.$ Substitute $u = x - \alpha$, so the integral becomes $\int_0^{\beta-\alpha} -u(u - (\beta-\alpha))\,du = \int_0^h u(h - u)\,du$ where $h = \beta - \alpha$. That equals $h \cdot \dfrac{h^2}{2} - \dfrac{h^3}{3} = \dfrac{h^3}{6}$.

So when a unit-leading-coefficient parabola is cut by a chord, the bounded area is $\dfrac{1}{6}$ times the cube of the horizontal width.

Variants

If the parabola has leading coefficient $c$ ($y = cx^2 + \ldots$ with $c > 0$ and the line above), the formula becomes $A = \dfrac{c}{6}(\beta - \alpha)^3$.

If the line is replaced by a horizontal chord $y = h$ on $y = x^2$, then $\alpha = -\sqrt h$, $\beta = \sqrt h$, and area $= \dfrac{(2\sqrt h)^3}{6} = \dfrac{4 h^{3/2}}{3}$.

For $y^2 = 4ax$ cut by the latus rectum $x = a$, the chord meets the parabola at $y = \pm 2a$. Area between curve and chord using horizontal slices: $\int_{-2a}^{2a}(a - y^2/(4a))\,dy = \dfrac{8a^2}{3}$.

Worked examples

Example 1. Area enclosed by $y = x^2$ and $y = 4$.

Intersections: $x^2 = 4$, so $x = \pm 2$. By the formula, area $= \dfrac{(\beta - \alpha)^3}{6} = \dfrac{4^3}{6} = \dfrac{32}{3}$. Or directly: $\int_{-2}^2 (4 - x^2)\,dx = [4x - x^3/3]_{-2}^2 = (8 - 8/3) - (-8 + 8/3) = \dfrac{32}{3}$.

Example 2. Area enclosed by $y = x^2 - 4$ and $y = 2x - 1$.

Intersections: $x^2 - 4 = 2x - 1 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0$, so $\alpha = -1$, $\beta = 3$. By the formula (line above parabola between roots): area $= \dfrac{(3 - (-1))^3}{6} = \dfrac{64}{6} = \dfrac{32}{3}$.

Example 3. Area enclosed by $y^2 = 4x$ and $y = x - 2$ , using horizontal slices.

Line: $x = y + 2$. Parabola: $x = y^2/4$. Intersections: $y + 2 = y^2/4 \Rightarrow y^2 - 4y - 8 = 0$... wait, let me redo. $y^2 = 4(y + 2) \Rightarrow y^2 - 4y - 8 = 0 \Rightarrow y = 2 \pm 2\sqrt 3$. Hmm, that doesn't match the cleaner version. Let me change: take $y^2 = x$ and $y = x - 2$. Then $y^2 = y + 2 \Rightarrow y^2 - y - 2 = 0 \Rightarrow y = -1, 2$. Right side at $y$ is line $x = y + 2$; left is parabola $x = y^2$. Area $= \int_{-1}^2 [(y + 2) - y^2]\,dy = [y^2/2 + 2y - y^3/3]_{-1}^2 = (2 + 4 - 8/3) - (1/2 - 2 + 1/3) = \dfrac{10}{3} - (-\dfrac{7}{6}) = \dfrac{27}{6} = \dfrac{9}{2}$.

Example 4. Area cut off from $y = x^2$ by the chord $y = x + 2$.

Intersections: $x^2 = x + 2 \Rightarrow x^2 - x - 2 = 0 \Rightarrow x = -1, 2$. By the formula: $\dfrac{3^3}{6} = \dfrac{27}{6} = \dfrac{9}{2}$.

Example 5. Area enclosed by $y^2 = 4ax$ and its latus rectum $x = a$.

Use symmetry. Chord at $x = a$ meets the curve at $y = \pm 2a$. Area $= 2\int_0^a 2\sqrt{ax}\,dx = 4\sqrt a \cdot \dfrac{2}{3}x^{3/2}\Big|_0^a = \dfrac{8 a^2}{3}$.

Example 6. Area enclosed by $y = x^2$ and the chord joining $(-1, 1)$ and $(2, 4)$.

Line through $(-1, 1)$ and $(2, 4)$: slope $1$, equation $y = x + 2$. Intersections with $y = x^2$: $x = -1, 2$. Area by formula: $\dfrac{3^3}{6} = \dfrac{9}{2}$.

Try it yourself

1. Area enclosed by $y = x^2$ and $y = 2x$.
2. Area enclosed by $y = x^2$ and $y = x + 6$.
3. Area enclosed by $y = x^2 + 1$ and $y = 5$.
4. Area enclosed by $y^2 = 4x$ and the line $y = x$.
5. Area enclosed by $y^2 = x$ and $x = 4$.
6. Area enclosed by $y = -x^2 + 4$ and $y = 0$.
7. Area enclosed by $y = x^2 - 2x$ and $y = x$.
8. Area enclosed by $y^2 = 4ax$ and its latus rectum (general $a$).
9. Area enclosed by $y = x^2$ and the line through $(0, 0)$ and $(3, 9)$.
10. Area enclosed by $y^2 = 9x$ and $y = 3x$.
11. Area enclosed by $y = x^2$ and the chord joining $(\alpha, \alpha^2)$ and $(\beta, \beta^2)$ , prove $A = \dfrac{(\beta - \alpha)^3}{6}$.
12. Area enclosed by $4y = 3x^2$ and $2y = 3x + 12$.
13. Area enclosed by $x^2 = y$ and the line through $(0, 0)$ and $(4, 16)$.
14. Area enclosed by $x = y^2$ and $x = 2 - y^2$.

Pitfalls and tricks

• Memorise the $\dfrac{(\beta - \alpha)^3}{6}$ formula for unit-leading-coefficient parabolas , it saves time.
• Find both intersection points first; mistakes here propagate.
• Choose slice direction wisely , for $y^2 = 4ax$ (opens right), horizontal slices ($dy$) often win.
• Always check the sign of "top minus bottom" in your integral. Negative answer means you have it backwards.
• Symmetric situations (parabola and a chord symmetric about its axis) cut the work in half.