Does it matter if the area is below the x-axis?

No. As long as you subtract the bottom function from the top function, the formula automatically handles negative $y$-values. You will still get a positive area without needing to adjust the signs.

What if the curves are given as x in terms of y?

If the curves are functions of $y$, such as $x = f(y)$ and $x = g(y)$, you integrate with respect to $y$. Instead of top minus bottom, you subtract the left curve from the right curve: $\int_c^d [\text{right} - \text{left}] dy$.

How do I know which curve is on top without graphing?

Find all intersection points to establish your intervals. Then, pick one test number inside each interval and plug it into both original equations. The equation that produces the larger value is the top curve for that entire interval.

How do you find the area between two curves?

To find the area between two curves, you subtract the bottom function from the top function and integrate the result between their intersection points.

Think of it like finding the area of a picture frame: you calculate the area of the entire outer rectangle and subtract the area of the empty space in the middle. In calculus, if you have a top curve $f(x)$ and a bottom curve $g(x)$ , subtracting $g(x)$ from $f(x)$ leaves you with exactly the height of the region between them. Adding up all those tiny heights using an integral gives you the total area.

The area formula

If a function $f(x)$ is greater than or equal to another function $g(x)$ on an interval from $x = a$ to $x = b$ , the area $A$ between them is calculated using a definite integral. The precise definition is: $A = \int_a^b [f(x) - g(x)] dx$ .

The expression $[f(x) - g(x)]$ represents the height of an infinitely thin rectangle at position $x$ , and $dx$ represents its width. The integral sums up all these rectangles from the starting point $a$ to the ending point $b$ .

How to set it up

First, you need to know where your region starts and ends. If the bounds aren't given, you find the intersection points by setting the two functions equal to each other ( $f(x) = g(x)$ ) and solving for $x$ .

Next, figure out which curve is on top. You can do this by sketching a quick graph or by picking a test $x$ -value between your intersection points and plugging it into both functions. The function that gives the higher $y$ -value is your $f(x)$ (the top curve).

Where students slip: Crossing curves

The most common mistake happens when the two curves cross each other somewhere inside the interval. If they cross, the curve that was on top is now on the bottom, and vice versa.

If you just integrate from start to finish without checking, the area after the crossing will come out negative, canceling out some of your true area. To fix this, you must split the integral at the intersection point. Calculate the area of the first section, calculate the area of the second section (with the new top curve minus the new bottom curve), and add them together.

Worked through

Find the area of the region enclosed by the curves $y = x^2$ and $y = 2x$ .

Step 1: Find the intersection points by setting the equations equal. $x^2 = 2x$ $x^2 - 2x = 0$ $x(x - 2) = 0$ So, $x = 0$ and $x = 2$ . These are our bounds $a$ and $b$ .

Step 2: Determine which curve is on top between $x = 0$ and $x = 2$ . Let's test $x = 1$ . For $y = 2x$ : $y = 2(1) = 2$ . For $y = x^2$ : $y = (1)^2 = 1$ . Since $2 > 1$ , the line $y = 2x$ is the top curve.

Step 3: Set up and evaluate the integral. $A = \int_0^2 [2x - x^2] dx$

Find the antiderivative: $[x^2 - \frac{x^3}{3}]$ evaluated from $0$ to $2$ .

Plug in the upper bound (2): $(2)^2 - \frac{(2)^3}{3} = 4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3}$ .

Plug in the lower bound (0): $(0)^2 - \frac{(0)^3}{3} = 0$ .

Subtract the lower bound result from the upper bound result: $A = \frac{4}{3} - 0 = \frac{4}{3}$ .

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Where this comes from: Stewart Calculus, 8th Edition · OpenStax Calculus Volume 1 · Khan Academy: Applications of Definite Integrals