Do I always integrate the top function minus the bottom function?

Yes, as long as the top function is always above the bottom on your entire interval. Always check by plugging in a test point. If one function is above in one part and below in another, split the integral at each intersection point.

What if I am not sure which function is on top?

Substitute any x-value strictly between a and b into both functions and compare the outputs. The function with the larger value is on top. If you remain unsure, sketch both curves or test several points.

Can the area ever be negative?

No; area is always positive. A negative result means the functions are reversed (you subtracted top from bottom) or the curves intersect within your interval and you did not split at the intersection. Reverse the order or use absolute value in each piece to fix it.

What is the difference between the area between curves and the definite integral of one function?

The definite integral \int_a^b f(x)\,dx gives the signed area between the curve and the x-axis (negative if the curve dips below). The area between two curves \int_a^b [f(x) - g(x)]\,dx measures the region trapped between both and is always positive.

Area Between Curves

Find the area of a region between two curves by computing the definite integral of their difference over a specified interval.

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

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What each symbol means

What Area Between Curves takes

Symbol	Meaning
$A$	The area enclosed between the two curves, measured in square units.
$f$	The upper curve (the function whose graph is on top in the region of interest).
$g$	The lower curve (the function whose graph is on the bottom in the region of interest).
$a$	The left endpoint of the interval on the $x$ -axis; the region of interest begins here.
$b$	The right endpoint of the interval on the $x$ -axis; the region of interest ends here.
$x$	The variable of integration, representing position along the horizontal axis.

When to use it

When you need to find the area trapped between two curves over a given interval.

Level

Usually taught in: Calculus I · Appears on: AP Calculus

Worked examples

1. Find the area between a horizontal line and a parabola

Problem

Find the area between

f(x) = 4

and

g(x) = x^2

from

x = 0

x = 2

$A = \int_0^2 [4 - x^2]\,dx$
Set up the integral with the upper function $f(x) = 4$ minus the lower function $g(x) = x^2$ , integrated from 0 to 2.
$\int [4 - x^2]\,dx = 4x - \frac{x^3}{3}$
Integrate term by term using the power rule: $\int 4\,dx = 4x$ and $\int x^2\,dx = \frac{x^3}{3}$ .
$A = \left[4x - \frac{x^3}{3}\right]_0^2 = \left(8 - \frac{8}{3}\right) - (0) = 8 - \frac{8}{3}$
Evaluate the antiderivative at the bounds: substitute $x = 2$ to get $8 - \frac{8}{3}$ , and at $x = 0$ to get 0.
$A = \frac{24}{3} - \frac{8}{3} = \frac{16}{3}$
Simplify by converting 8 to thirds, then combine fractions.

Answer: $A = \frac{16}{3} \text{ square units}$

Both functions are given explicitly, and the horizontal line is clearly above the parabola on $[0, 2]$ , so we subtract the lower from the upper and integrate.

2. Find the area between a line and a quadratic with negative bounds and fractions

Problem

Find the area between

f(x) = x + 3

and

g(x) = x^2 - 1

from

x = -1

x = 2

$A = \int_{-1}^2 [(x + 3) - (x^2 - 1)]\,dx$
Set up with $f(x) = x + 3$ (upper) minus $g(x) = x^2 - 1$ (lower) over $[-1, 2]$ .
$A = \int_{-1}^2 [x + 3 - x^2 + 1]\,dx = \int_{-1}^2 [-x^2 + x + 4]\,dx$
Distribute the negative sign and combine like terms: $3 + 1 = 4$ and $30t - 5t = 25t$ .
$\int [-x^2 + x + 4]\,dx = -\frac{x^3}{3} + \frac{x^2}{2} + 4x$
Integrate each term using the power rule.
$\text{At } x = 2: -\frac{8}{3} + \frac{4}{2} + 8 = -\frac{8}{3} + 2 + 8 = \frac{-8 + 30}{3} = \frac{22}{3}$
Substitute $x = 2$ ; note that $(2)^3 = 8$ and $(2)^2 = 4$ .
$\text{At } x = -1: -\frac{-1}{3} + \frac{1}{2} - 4 = \frac{1}{3} + \frac{1}{2} - 4 = \frac{2 + 3 - 24}{6} = -\frac{19}{6}$
Substitute $x = -1$ ; note that $(-1)^3 = -1$ so $-\frac{(-1)^3}{3} = \frac{1}{3}$ ; convert to sixths.
$A = \frac{22}{3} - \left(-\frac{19}{6}\right) = \frac{44}{6} + \frac{19}{6} = \frac{63}{6} = \frac{21}{2}$
Subtract: upper bound minus lower bound. Convert $\frac{22}{3} = \frac{44}{6}$ ; subtracting a negative becomes addition.

Answer: $A = \frac{21}{2} = 10.5 \text{ square units}$

Negative bounds and fractions invite sign errors. Always evaluate at the upper bound first, subtract the lower bound value, and watch for the double negative when subtracting a negative result.

3. Calculate total net profit from revenue and cost functions over a time interval

Problem

At a school bake sale, hourly revenue is

R(t) = 30t

dollars and hourly costs are

C(t) = t^2 + 5t

dollars, where

t

is hours worked. Find the total net profit from

t = 1

t = 4

hours.

$P = \int_1^4 [R(t) - C(t)]\,dt = \int_1^4 [30t - (t^2 + 5t)]\,dt$
Profit equals revenue minus cost; set up the integral over the time interval from 1 to 4 hours.
$P = \int_1^4 [30t - t^2 - 5t]\,dt = \int_1^4 [-t^2 + 25t]\,dt$
Distribute the negative and combine like terms: $30t - 5t = 25t$ .
$\int [-t^2 + 25t]\,dt = -\frac{t^3}{3} + \frac{25t^2}{2}$
Integrate term by term: $\int t^2\,dt = \frac{t^3}{3}$ and $\int t\,dt = \frac{t^2}{2}$ .
$\text{At } t = 4: -\frac{64}{3} + \frac{25 \cdot 16}{2} = -\frac{64}{3} + \frac{400}{2} = -\frac{64}{3} + 200 = \frac{-64 + 600}{3} = \frac{536}{3}$
Substitute $t = 4$ : $(4)^3 = 64$ and $(4)^2 = 16$ ; convert 200 to thirds to combine.
$\text{At } t = 1: -\frac{1}{3} + \frac{25}{2} = \frac{-2 + 75}{6} = \frac{73}{6}$
Substitute $t = 1$ ; convert to sixths to combine fractions.
$P = \frac{536}{3} - \frac{73}{6} = \frac{1072}{6} - \frac{73}{6} = \frac{999}{6} = 166.5$
Subtract by converting both fractions to sixths, then simplify.

Answer: $P = 166.5 \text{ dollars}$

In word problems, identify which quantity dominates (revenue over cost here) and set up the integral as (top function) minus (bottom function). The definite integral measures the total accumulated difference over the entire time period.

Common mistakes

Where Area Between Curves usually goes wrong

The mistake	Why it is wrong	The fix
Writing $\int_a^b f(x)\,dx + \int_a^b g(x)\,dx$ to find the area between curves.	This adds the areas under both curves separately instead of finding the trapped region between them, yielding the wrong answer.	Use $A = \int_a^b [f(x) - g(x)]\,dx$ to subtract the lower function from the upper within a single integral.
Writing $\int_a^b f(x) - g(x)\,dx$ without brackets around the difference.	Without brackets, the integral distributes as $\int_a^b f(x)\,dx - \int_a^b g(x)\,dx$ (which is correct only by coincidence), and obscures the geometry of the region.	Always write $\int_a^b [f(x) - g(x)]\,dx$ with brackets to keep the subtraction inside and ensure the region is properly defined.
Failing to check which function is on top, or ignoring curve intersections within the interval.	If $g(x) > f(x)$ the integral is negative (not an area), and if curves cross, positive and negative contributions cancel incorrectly, giving the wrong net result.	Verify $f(x) \geq g(x)$ by testing a point. If curves intersect at $x = c$ , split the integral: $A = \int_a^c [f(x) - g(x)]\,dx + \int_c^b [g(x) - f(x)]\,dx$ or use absolute value.

Tips and when to use something else

Always determine which function is above the other by testing a sample point in your interval, especially if the curves have not been graphed.
If the curves intersect within $[a, b]$ , find the intersection points and split the integral at those $x$ -values, adjusting which function is subtracted in each piece.
If the curves are given as $x = f(y)$ instead of $y = f(x)$ (rotated sideways), use $A = \int_c^d [f(y) - g(y)]\,dy$ , integrating with respect to $y$ instead.
When the problem asks for the volume of a solid of revolution, use Volume by Disks or Volume by Shells instead of an area integral.

Related formulas

Fundamental Theorem of CalculusCompute the exact value of a definite integral using the antiderivative—connect area under a curve to the accumulation of change.Definite Integral PropertiesTwo key properties of definite integrals: reversing the bounds negates the integral, and integrating over a single point gives zero.Power Rule for IntegrationPower Rule for Integration instantly finds antiderivatives of polynomial terms; use it to integrate any power of x except x to the negative one.Volume by DisksVolume by Disks calculates the volume of a solid of revolution by integrating the areas of circular cross-sections perpendicular to the axis of rotation.

Frequently asked questions

Do I always integrate the top function minus the bottom function?: Yes, as long as the top function is always above the bottom on your entire interval. Always check by plugging in a test point. If one function is above in one part and below in another, split the integral at each intersection point.
What if I am not sure which function is on top?: Substitute any $x$ -value strictly between $a$ and $b$ into both functions and compare the outputs. The function with the larger value is on top. If you remain unsure, sketch both curves or test several points.
Can the area ever be negative?: No; area is always positive. A negative result means the functions are reversed (you subtracted top from bottom) or the curves intersect within your interval and you did not split at the intersection. Reverse the order or use absolute value in each piece to fix it.
What is the difference between the area between curves and the definite integral of one function?: The definite integral $\int_a^b f(x)\,dx$ gives the signed area between the curve and the $x$ -axis (negative if the curve dips below). The area between two curves $\int_a^b [f(x) - g(x)]\,dx$ measures the region trapped between both and is always positive.

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Reviewed 2026-09-18