Areas Between Curves

Suppose that we have two curves $f$ and $g$ and that we want to find the area between the curves. If for all $x$ such that $a ≤ x ≤ b$, $f(x) ≥ g(x)$, then the area between $f$ and $g$ can be calculated with the formula:

(1)

\begin{align} A = \int_{a}^{b} [f(x) - g(x)] \: dx \end{align}

We note that this formula can be interpreted as the area from the "top curve $f$" subtracted from the area from the "bottom curve $g$".

For example, suppose that we want to find the area between the curves $f(x) = -x^2 -2$ and $g(x) = x^2$. We first need to find points where these functions intersect each other. The graph bellow illustrates our curves.

Now notice that the area is trapped on the interval $[-1, 1]$, or rather, where the graphs $f$ and $g$ intersect each other at $x = -1$ and $x = 1$. We can obtain this area by taking the antiderivative of the "top" function:

And subtracting the antiderivative of the "bottom" function $g$ on the interval $[-1, 1]$:

Hopefully you can see that when you subtract the green area from the red area, we will obtain the blue area.

Note: The definition for calculating areas between two curves defined by $f(y)$ and $g(y)$ is similar. If $f(y) ≥ g(y)$ for all $y$ such that $c ≤ y ≤ d$, then the area between the two curves is equal to $\int_{c}^{d} [f(y) - g(y)] \: dy$.

Example 1

Calculate the area trapped between the curves $f(x) = x^2$ and $g(x) = 2x - x^2$.

We first note that these two curves intersect at x = 0 and x = 1. Hence our limits of integration will be 0 and 1. Furthermore, we note that g(x) ≥ f(x) for 0 ≤ x ≤ 1 as seen when we graph the two curves on top of each other:

Hence it follows that by the formula we derived:

(2)

\begin{align} A = \int_0^1 g(x) - f(x) \: dx \\ A = \int_0^1 (2x - x^2) - (x^2) \: dx \\ A = \int_0^1 (2x - 2x^2) \: dx \\ A = (x^2 - \frac{2x^3}{3}) \bigg |_0^1 \\ A = 1 - \frac{2}{3} \\ A = \frac{1}{3} \end{align}