Area Between Curves

Finding the area between two curves is a direct application of definite integrals. Instead of finding the area under a single curve down to the x-axis, we calculate the area of the region bounded between an upper curve and a lower curve.

The Formula

If a function $f(x)$ is greater than or equal to another function $g(x)$ on an interval $[a, b]$ (meaning $f(x)$ is the "top" curve and $g(x)$ is the "bottom" curve), the area $A$ between them is:

$A = \int_{a}^{b} [f(x) - g(x)] \, dx$

Steps to Find the Area

1. Find the intersection points: Set the two equations equal to each other to find where the curves intersect. These $x$-values will usually be your limits of integration, $a$ and $b$.
2. Determine the upper and lower functions: Pick a test point between $a$ and $b$ to see which function has the higher $y$-value.
3. Set up the integral: Subtract the lower function from the upper function.
4. Integrate and evaluate: Compute the definite integral.

Example 1: Bounded by Two Functions

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

Step 1: Find intersections. Set the equations equal to each other: $x^2 = x + 2$ $x^2 - x - 2 = 0$ $(x - 2)(x + 1) = 0$ The curves intersect at $x = -1$ and $x = 2$. These are our bounds.

Step 2: Determine upper and lower functions. Pick a test point in the interval $[-1, 2]$, such as $x = 0$:

• $y = (0)^2 = 0$
• $y = (0) + 2 = 2$

Since $2 > 0$, $y = x + 2$ is the upper function and $y = x^2$ is the lower function.

Step 3 & 4: Set up and evaluate the integral. $A = \int_{-1}^{2} [(x + 2) - x^2] \, dx$ $A = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2}$ Evaluate at the upper bound ($x=2$): $\frac{4}{2} + 4 - \frac{8}{3} = 2 + 4 - \frac{8}{3} = \frac{10}{3}$ Evaluate at the lower bound ($x=-1$): $\frac{1}{2} - 2 - \left(-\frac{1}{3}\right) = \frac{1}{2} - 2 + \frac{1}{3} = -\frac{7}{6}$ Subtract the lower bound result from the upper bound result: $A = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$

Example 2: Functions that Cross Over

Problem: Find the area enclosed by $y = \sin x$ and $y = \cos x$ on the interval $[0, \pi]$.

Step 1: Find intersections. Set $\sin x = \cos x$. On the interval $[0, \pi]$, this occurs at $x = \frac{\pi}{4}$. Because the curves cross, the upper and lower functions will switch. We must split the integral into two parts: $[0, \frac{\pi}{4}]$ and $[\frac{\pi}{4}, \pi]$.

Step 2: Determine upper and lower functions for each interval.

• On $[0, \frac{\pi}{4}]$: $\cos x \ge \sin x$ (e.g., at $x=0$, $\cos(0)=1$, $\sin(0)=0$).
• On $[\frac{\pi}{4}, \pi]$: $\sin x \ge \cos x$ (e.g., at $x=\frac{\pi}{2}$, $\sin(\frac{\pi}{2})=1$, $\cos(\frac{\pi}{2})=0$).

Step 3 & 4: Set up and evaluate the integrals.

First Region: $A_1 = \int_{0}^{\pi/4} (\cos x - \sin x) \, dx = [\sin x + \cos x]_{0}^{\pi/4}$ $A_1 = \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) - (0 + 1) = \sqrt{2} - 1$

Second Region: $A_2 = \int_{\pi/4}^{\pi} (\sin x - \cos x) \, dx = [-\cos x - \sin x]_{\pi/4}^{\pi}$ $A_2 = (-(-1) - 0) - \left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\right) = 1 - (-\sqrt{2}) = 1 + \sqrt{2}$

Total Area: $A = A_1 + A_2 = (\sqrt{2} - 1) + (1 + \sqrt{2}) = 2\sqrt{2}$