Area Between Curves
Compute the area between two curves as ∫[f(x)−g(x)]dx. Handle cases where the curves intersect (splitting at intersection points) and integrate with respect to y when horizontal slices are more efficient.
A lesson of Calculus, in Applications of Integration, about 10 minutes.
By the end of this lesson, you will be able to find the area between two curves by integrating with respect to $x$ or $y$, set up correct limits of integration, and handle cases where curves cross.
The Setup
Area between $f(x)$ and $g(x)$ on $[a,b]$ where $f(x) \geq g(x)$:
$$A = \int_a^b [f(x) - g(x)],dx$$
Always: top function minus bottom function, integrating from left to right.
When curves cross, split the integral at the crossing points (where $f = g$).
Finding the Integration Limits
Given: two curves, find their intersections by solving $f(x) = g(x)$.
The intersections give the natural limits $a$ and $b$.
Example 1 — Basic Area
Find the area between $y = x^2$ and $y = x + 2$.
Intersections: $x^2 = x + 2 \implies x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0$; $x = -1, 2$.
On $[-1, 2]$: line $y = x+2$ is above parabola.
$A = \int_{-1}^2 [(x+2) - x^2],dx = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^2$
$= \left[2 + 4 - \frac{8}{3}\right] - \left[\frac{1}{2} - 2 + \frac{1}{3}\right] = \frac{10}{3} + \frac{7}{6} = \frac{20+7}{6} = \frac{27}{6} = \frac{9}{2}$
Example 2 — Curves That Cross
Area between $y = \sin x$ and $y = \cos x$ on $[0, \pi]$.
They cross at $x = \pi/4$ (where $\sin = \cos$).
On $[0, \pi/4]$: $\cos \geq \sin$; on $[\pi/4, \pi]$: $\sin \geq \cos$.
$A = \int_0^{\pi/4}(\cos x - \sin x),dx + \int_{\pi/4}^{\pi}(\sin x - \cos x),dx$
$= [\sin x + \cos x]0^{\pi/4} + [-\cos x - \sin x]{\pi/4}^{\pi}$
$= (\sqrt{2} - 1) + (1 + \sqrt{2}) = 2\sqrt{2}$
Integrating with Respect to $y$
When curves are described as $x = g(y)$, integrate horizontally:
$$A = \int_c^d [x_{\text{right}} - x_{\text{left}}],dy$$
When to use: when the region is more naturally described horizontally (e.g., parabolas opening left/right).
Example 3 — Horizontal Integration
Area between $x = y^2$ and $x = y + 2$.
Intersections: $y^2 = y + 2 \implies y = 2, -1$.
On $[-1, 2]$: $x = y+2$ is to the right of $x = y^2$.
$A = \int_{-1}^2 [(y+2) - y^2],dy = \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_{-1}^2 = \frac{9}{2}$
Same answer as Example 1 (same region, different orientation).
Example 4 — Three Curves
Area enclosed by $y = x^2$, $y = 4x$, and $y = 4$.
Find all pairwise intersections: