6.1 Areas between Curves

Area between Curves

Calculating the area between curves is a core skill in Calculus I. Instead of finding the area under a single function, you're finding the area of a region trapped between two functions. The key idea: integrate the difference between the "top" and "bottom" (or "right" and "left") functions across the interval where they bound a region.

Choosing the right variable to integrate with respect to matters a lot. The wrong choice can turn a clean one-integral problem into a messy multi-part calculation.

Area between Two Curves

The fundamental idea is straightforward: area equals the integral of (bigger function minus smaller function).

Integrating with respect to x:

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

• $f(x)$ is the upper function (higher $y$-values) and $g(x)$ is the lower function on the interval $[a, b]$
• $a$ and $b$ are the $x$-coordinates where the curves intersect (or the endpoints you're given)

Integrating with respect to y:

$A = \int_{c}^{d} [h(y) - j(y)]\, dy$

• $h(y)$ is the right function (larger $x$-values) and $j(y)$ is the left function on the interval $[c, d]$
• $c$ and $d$ are the $y$-coordinates where the curves intersect

Steps to find the area:

1. Sketch the curves. Even a rough sketch helps you see which function is on top (or to the right) and where the region lives.
2. Find intersection points. Set the two functions equal and solve. These give you the limits of integration.
3. Identify which function is greater on the interval. Plug in a test point between the intersection values if you're unsure.
4. Set up and evaluate the integral of (greater function minus lesser function).

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

1. Set $x^2 = x + 2$, which gives $x^2 - x - 2 = 0$, so $(x-2)(x+1) = 0$. The curves intersect at $x = -1$ and $x = 2$.

2. Test a point: at $x = 0$, the line gives $y = 2$ and the parabola gives $y = 0$. So $y = x + 2$ is the upper function.

3. $A = \int_{-1}^{2} [(x + 2) - x^2]\, dx = \int_{-1}^{2} (x + 2 - x^2)\, dx$

4. Evaluate: $\left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2} = \left(2 + 4 - \frac{8}{3}\right) - \left(\frac{1}{2} - 2 + \frac{1}{3}\right) = \frac{9}{2}$

Compound Regions with Intersecting Curves

Sometimes the curves cross each other within the interval, so which function is "on top" switches partway through. When that happens, you can't use a single integral because the difference $f(x) - g(x)$ would change sign, and negative area would cancel out positive area.

How to handle this:

1. Find all intersection points of the curves within the region.
2. Split the interval at each intersection point, creating sub-regions.
3. For each sub-region, determine which function is greater (use a test point).
4. Set up a separate integral for each sub-region.
5. Add the areas together:

$A_{\text{total}} = A_1 + A_2 + \cdots + A_n$

Example: If $f(x)$ is above $g(x)$ on $[a, c]$ but below $g(x)$ on $[c, b]$, the total area is:

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

Notice the order flips in the second integral. An equivalent shortcut is to use absolute value:

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

This always works conceptually, but in practice you still need to split the integral at the crossing points to evaluate it.

Choosing the Variable of Integration

Picking the right variable can be the difference between a clean problem and a painful one.

• Integrate with respect to $x$ when the boundary curves are naturally written as $y = f(x)$ and you can clearly tell which is on top.
• Integrate with respect to $y$ when the boundary curves are naturally written as $x = h(y)$ and you can clearly tell which is to the right.

The big payoff of choosing wisely is that you may avoid splitting the region into multiple sub-regions. A region that requires two or three integrals in $x$ might need only one integral in $y$ (or vice versa).

Example: The region between $x = y^2$ and $x = y + 2$. These are already written as functions of $y$, so integrating with respect to $y$ is natural. If you rewrote them as functions of $x$, you'd need to split the parabola into upper and lower halves and use multiple integrals.