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📐 Area Between Curves

$\displaystyle \text{Area} = \int_a^b [f(x) - g(x)] dx$ — finding the area between two curves

One of the most common applications of definite integrals is finding the area between two curves. If $f(x) \ge g(x)$ on the interval $[a, b]$, the area between the curves is given by the integral of the top function minus the bottom function.

🎯 In this section you will learn

📌 The Basic Formula

$$ \text{Area} = \int_a^b [\text{top} - \text{bottom}] \, dx = \int_a^b [f(x) - g(x)] \, dx $$
where $f(x) \ge g(x)$ on $[a, b]$
💡 Step-by-Step Process
  1. Sketch the curves to see which is on top.
  2. Find the intersection points (set $f(x) = g(x)$).
  3. Set up the integral with the top function minus the bottom function.
  4. Integrate and evaluate.

📊 Visualizing Area Between Curves

Interactive graph below. The shaded region shows the area between the two curves:

Example 1Area Between $y=x^2$ and $y=x$

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

①
Find intersection points
$x^2 = x$ ⇒ $x^2 - x = 0$ ⇒ $x(x-1)=0$ ⇒ $x=0$ and $x=1$
②
Determine top and bottom
On $[0,1]$, $x \ge x^2$ (since $x^2 \le x$ for $0 \le x \le 1$). Top: $x$, Bottom: $x^2$
③
Set up and integrate
$\text{Area} = \int_0^1 (x - x^2) dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$
Example 2Area Between $y=x^2-4$ and $y=0$

Find the area between $y = x^2 - 4$ and the x-axis ($y=0$).

①
Find intersection points
$x^2 - 4 = 0$ ⇒ $x = \pm 2$
②
Determine top and bottom
On $[-2,2]$, $0 \ge x^2 - 4$ (x-axis is above the parabola). Top: $0$, Bottom: $x^2 - 4$
③
Set up and integrate
$\text{Area} = \int_{-2}^2 [0 - (x^2 - 4)] dx = \int_{-2}^2 (4 - x^2) dx = \left[4x - \frac{x^3}{3}\right]_{-2}^2 = (8 - \frac{8}{3}) - (-8 + \frac{8}{3}) = \frac{16}{3} + \frac{16}{3} = \frac{32}{3}$
Example 3When Top and Bottom Switch

Find the area between $y = x$ and $y = x^3$ on $[-1, 1]$.

①
Find intersection points
$x = x^3$ ⇒ $x^3 - x = 0$ ⇒ $x(x-1)(x+1)=0$ ⇒ $x=-1, 0, 1$
②
Determine top and bottom on each subinterval
On $[-1,0]$: $x^3 \ge x$ (since $x^3$ is less negative). Top: $x^3$, Bottom: $x$
On $[0,1]$: $x \ge x^3$. Top: $x$, Bottom: $x^3$
③
Set up and integrate
$\text{Area} = \int_{-1}^0 (x^3 - x) dx + \int_0^1 (x - x^3) dx$
$= \left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{-1}^0 + \left[\frac{x^2}{2} - \frac{x^4}{4}\right]_0^1$
$= (0 - (\frac{1}{4} - \frac{1}{2})) + ((\frac{1}{2} - \frac{1}{4}) - 0) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

📌 Integrating with Respect to $y$

Sometimes it's easier to integrate with respect to $y$, especially when curves are given as $x = f(y)$ or when the region is bounded by functions that are easier to express in terms of $y$.

$$ \text{Area} = \int_c^d [\text{right} - \text{left}] \, dy $$
Example 4Integrating with Respect to $y$

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

①
Find intersection points (in $y$)
$y^2 = y + 2$ ⇒ $y^2 - y - 2 = 0$ ⇒ $(y-2)(y+1)=0$ ⇒ $y = -1, 2$
②
Determine right and left
For $y$ between $-1$ and $2$, $y+2 \ge y^2$. Right: $y+2$, Left: $y^2$
③
Set up and integrate
$\text{Area} = \int_{-1}^2 [(y+2) - y^2] dy = \int_{-1}^2 (y + 2 - y^2) dy = \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_{-1}^2$
$= (2 + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3}) = (6 - \frac{8}{3}) - (-\frac{3}{2} + \frac{1}{3}) = \frac{10}{3} + \frac{7}{6} = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$
⚠️ Common Mistakes
🔍 Key Takeaways
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