Area Between Curves

Applications of Integration

By the end of this lesson, you will be able to find areas between curves, solve differential equations by separation of variables, and apply integrals to motion and accumulation problems.

Area Between Two Curves

Area between $f(x)$ and $g(x)$ on $[a, b]$ (where $f(x) \geq g(x)$):

$$A = \int_a^b [f(x) - g(x)], dx$$

Step-by-step:

1. Find where the curves intersect (the limits of integration)
2. Determine which curve is on top on each interval
3. Integrate the difference (top minus bottom)

Example: Find the area between $f(x) = x^2$ and $g(x) = x$.

Intersections: $x^2 = x \Rightarrow x^2 - x = 0 \Rightarrow x(x-1) = 0 \Rightarrow x = 0, 1$

On $[0, 1]$: $x \geq x^2$ (check: at $x = 0.5$, $0.5 > 0.25$) → $g$ is on top.

Wait: $g(x) = x$ is on top, $f(x) = x^2$ is on bottom.

$A = \displaystyle\int_0^1 (x - x^2), dx = \left[\dfrac{x^2}{2} - \dfrac{x^3}{3}\right]_0^1 = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}$

Integrating with Respect to $y$

Sometimes it's easier to integrate horizontally (with respect to $y$).

Area between $x = g(y)$ and $x = h(y)$ on $[c, d]$ (where $h(y) \geq g(y)$):

$$A = \int_c^d [h(y) - g(y)], dy$$

Example: Find the area enclosed by $x = y^2 - 1$ and $x = 1 - y^2$.

Intersections: $y^2 - 1 = 1 - y^2 \Rightarrow 2y^2 = 2 \Rightarrow y = \pm 1$

On $[-1, 1]$: $1 - y^2 \geq y^2 - 1$ (right curve − left curve)

$A = \displaystyle\int_{-1}^1 [(1-y^2) - (y^2-1)], dy = \int_{-1}^1 (2 - 2y^2), dy$

$= \left[2y - \dfrac{2y^3}{3}\right]_{-1}^1 = \left(2 - \dfrac{2}{3}\right) - \left(-2 + \dfrac{2}{3}\right) = \dfrac{4}{3} + \dfrac{4}{3} = \dfrac{8}{3}$

Average Value of a Function

Average value of $f$ on $[a, b]$:

$$f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x), dx$$

Mean Value Theorem for Integrals: if $f$ is continuous on $[a, b]$, there exists $c \in [a, b]$ where $f(c) = f_{\text{avg}}$.

Example: Find the average value of $f(x) = x^2$ on $[0, 3]$.

$f_{\text{avg}} = \dfrac{1}{3-0}\displaystyle\int_0^3 x^2, dx = \dfrac{1}{3}\left[\dfrac{x^3}{3}\right]_0^3 = \dfrac{1}{3} \cdot 9 = 3$