Area Between Curves

Basic Formula

For two curves $y = f(x)$ and $y = g(x)$ where $f(x) \geq g(x)$ on $[a, b]$:

$\text{Area} = \int_a^b [f(x) - g(x)] \, dx = \int_a^b [\text{top} - \text{bottom}] \, dx$

General Formula

When curves cross, split the integral at intersection points:

$\text{Area} = \int_a^c |f(x) - g(x)| \, dx + \int_c^b |f(x) - g(x)| \, dx$

Integrating with Respect to y

For curves $x = f(y)$ and $x = g(y)$ where $f(y) \geq g(y)$ on $[c, d]$:

$\text{Area} = \int_c^d [f(y) - g(y)] \, dy = \int_c^d [\text{right} - \text{left}] \, dy$

Steps

1. Sketch the region
2. Find intersection points
3. Determine which curve is on top (or right)
4. Set up the integral
5. Evaluate

Examples

Example 1: Simple area between curves

Find the area between $y = x^2$ and $y = x$ from $x = 0$ to $x = 1$.

Since $x \geq x^2$ on $[0, 1]$:

$\text{Area} = \int_0^1 (x - x^2) \, dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1$

$= \left(\frac{1}{2} - \frac{1}{3}\right) - 0 = \frac{1}{6}$

Example 2: Finding intersection points

Find the area enclosed by $y = x^2$ and $y = 2x$.

Intersection: $x^2 = 2x \Rightarrow x(x-2) = 0 \Rightarrow x = 0, 2$

Since $2x \geq x^2$ on $[0, 2]$:

$\text{Area} = \int_0^2 (2x - x^2) \, dx = \left[x^2 - \frac{x^3}{3}\right]_0^2$

$= \left(4 - \frac{8}{3}\right) - 0 = \frac{4}{3}$

Example 3: Curves that cross

Find area between $y = \sin(x)$ and $y = \cos(x)$ from $0$ to $\frac{\pi}{2}$.

Intersection: $\sin(x) = \cos(x) \Rightarrow x = \frac{\pi}{4}$

On $[0, \frac{\pi}{4}]$: $\cos(x) \geq \sin(x)$

On $[\frac{\pi}{4}, \frac{\pi}{2}]$: $\sin(x) \geq \cos(x)$

$\text{Area} = \int_0^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx$

$= [\sin x + \cos x]_0^{\pi/4} + [-\cos x - \sin x]_{\pi/4}^{\pi/2}$

$= \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} - 0 - 1\right) + \left(-0 - 1 + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right)$

$= (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2\sqrt{2} - 2 \approx 0.828$

Example 4: Integrating with respect to y

Find the area between $x = y^2$ and $x = y + 2$.

Intersection: $y^2 = y + 2 \Rightarrow y^2 - y - 2 = 0 \Rightarrow (y-2)(y+1) = 0$

$y = -1$, $y = 2$

Since $y + 2 \geq y^2$ on $[-1, 2]$ (right curve minus left):

$\text{Area} = \int_{-1}^2 [(y + 2) - y^2] \, dy = \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_{-1}^2$

$= \left(2 + 4 - \frac{8}{3}\right) - \left(\frac{1}{2} - 2 + \frac{1}{3}\right)$

$= \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$

Example 5: Area requiring absolute value

Find area between $y = x^3 - x$ and $y = 0$ from $x = -1$ to $x = 1$.

Roots of $x^3 - x = x(x-1)(x+1) = 0$: $x = -1, 0, 1$

On $[-1, 0]$: $x^3 - x \geq 0$

On $[0, 1]$: $x^3 - x \leq 0$

$\text{Area} = \int_{-1}^0 (x^3 - x) \, dx + \int_0^1 -(x^3 - x) \, dx$

$= \left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{-1}^0 - \left[\frac{x^4}{4} - \frac{x^2}{2}\right]_0^1$

$= \left(0 - \left(\frac{1}{4} - \frac{1}{2}\right)\right) - \left(\left(\frac{1}{4} - \frac{1}{2}\right) - 0\right)$

$= \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

Example 6: Enclosed by three curves

Find area in the first quadrant bounded by $y = \sqrt{x}$, $y = 0$, and $x = 4$.

$\text{Area} = \int_0^4 \sqrt{x} \, dx = \left[\frac{2x^{3/2}}{3}\right]_0^4 = \frac{2(8)}{3} - 0 = \frac{16}{3}$