Areas Between Curves

Say we have a pair of curves:

What's the area betwee ${a}$ and ${b}$ ? Well, we can chop it up into Riemann sums:

We'll denote the width of each rectangle as ${dx.}$ What's the height? It's the difference between the top point of the rectangle (where the rectangle touches ${f(x)}$ ) and the bottom point of the rectangle (where it touches ${g(x)}$ ). Thus, for a single rectangle we have the area:

A_i = (f(x) - g(x)) \cdot dx

Making ${dx}$ smaller:

and smaller:

and even smaller:

we eventually fill in the area bounded to the left by ${a,}$ to the right by ${b,}$ above by ${f(x),}$ and below by ${g(x).}$ What's the value of this area? It's the sum of all the areas of the individual rectangles:

A_T = A_0 + A_1 + A_2 + \ldots + A_n

Now that we've seen integrals, this is really:

\begin{aligned} A_T &= A_0 + A_1 + A_2 + \ldots + A_n \\[1em] &= \int_{a}^{b} (f(x) - g(x)) dx \end{aligned}

This is how we find the area between two curves. Importantly, whenever we want to know such an area we want to first identify the integrand (our rectangle's height):

\int_{a}^{b} {\color{blue}(f(x) - g(x))} dx

Once we have this information, we want to know the limits, or bounds:

{\color{red}\int_{a}^{b}} (f(x) - g(x)) dx

These are the key pieces of information. We want to know the lower bound ${a,}$ the upper bound ${b,}$ and the integrand ${f(x)-g(x).}$ While it may seem like a simple task, it turns out that for many areas, this can be difficult. For example, consider the following problem:

problem. What is the area between ${x = y^2}$ and ${y=x-2?}$

The first step is to always try and create a visualization of the curves.

At this point, we see a problem. ${x = y^2}$ is not a function. A function is a a relation from ${\reals \to \reals.}$ Thus, for ${x = y^2,}$ we've effectively specified the set of ordered pairs:

\{ (x,y) \in \reals^2 : x = y^2 \}

And in doing so, it's no longer clear whether ${x}$ is a function of ${y,}$ or whether ${y}$ is a function of ${x.}$ And because of that ambiguity, we have a situation where it's unclear where our rectangles' heights start and end. It could start at ${x = y^2}$ and end at ${y = x-2.}$ It could start at ${x = y^2}$ and end at ${x = y^2:}$

To solve this problem, we'll have to split the area into two halves and sum the two after. And to do that, we need the intersection points:

Since ${x = y^2}$ and ${y = x-2,}$ we have:

y = y^2 - 2

This is a quadratic. Solve for ${y:}$

\begin{aligned} y^2 - y - 2 &= 0 \\ (y+2)(y-2) &= 0 \end{aligned}

Thus, we have:

y \in \{ 2, -1 \}

Next, from the implicit function theorem, we know that the top curve can be rewritten as:

y = \sqrt{x}

and the bottom curve is:

y = - \sqrt{x}

Finally, the line between the ${(1,-1)}$ and ${(4,2)}$ is given by the equation:

y = x - 2

At this point, we can find the area:

\text{A} = \int_{0}^{1} \sqrt{x} - (-\sqrt{x}) dx + \int_{1}^{4} \sqrt{x} - (x - 2) dx

Integrating for a Different Variable

The procedure above is meticulous. There's a much faster way: Integrating with respect to ${y.}$ Visually, this means we compute the area by dividing the region into horizontal rectangles.

To do so, we rewrite all of the equations to render them as true functions. Thus:

\begin{aligned} x = y^2 \nc x &= y^2 \\ y = x - 2 \nc x &= y + 2 \end{aligned}

Now we integrate in the ${dy}$ variable:

\int_{-1}^{2} (y+2 - y^2) dy

The ${y+2}$ gives us the left endpoint of the rectangle, and the ${y^2}$ gives us the right endpoint of the rectangle.