AP Calculus AB and BC

Area Between Two Curves: Setup, dx vs dy, Examples

Q: Do you always subtract top minus bottom?

When you integrate with respect to x, yes: the upper curve minus the lower curve. When you integrate with respect to y, the parallel rule is right minus left. In both cases you subtract so the height is positive, because area cannot be negative.

Q: When do you integrate with respect to y instead of x?

Switch to dy when the boundaries are given as x = f(y), or when a single vertical strip would hit different curves in different parts of the region. Horizontal strips often turn a two-integral problem into one.

To find the area between two curves, integrate top minus bottom with respect to x (or right minus left with respect to y) between the intersection points. If the curves cross inside the interval, split the integral where they swap and add the pieces so each difference stays positive.

The setup: upper boundary minus lower

The area of a region enclosed by two curves is the integral of the upper boundary minus the lower boundary. When both curves are written as functions of $x$ , that is top minus bottom, integrated across the interval where the region lives:

A = \int_a^b \bigl[\, f(x) - g(x) \,\bigr]\,dx, \qquad f(x) \ge g(x) \text{ on } [a,b]

Slice the region into thin vertical strips. Each strip has height $f(x) - g(x)$ , the gap between the curves, and width $dx$ ; multiplying gives the area of one strip, and the integral adds up every strip from $x = a$ to $x = b$ . That is the same Riemann-sum idea from Unit 6, now measuring the space between two graphs instead of the space under one. This is Topic 8.4.

One rule keeps you out of trouble: subtract in whatever order makes the height positive. Area is never negative, so the curve that is physically higher on the interval goes first. If you are unsure which is higher, plug a single test value from between the bounds into both functions and compare the outputs.

dx or dy: which way do the strips run?

The choice between integrating in $x$ or $y$ is a choice about strip direction, and it is the first real decision this topic asks you to make. Vertical strips give top minus bottom and integrate with respect to $x$ . Horizontal strips give right minus left and integrate with respect to $y$ (Topic 8.5):

A = \int_c^d \bigl[\, p(y) - q(y) \,\bigr]\,dy, \qquad p(y) \ge q(y) \text{ on } [c,d]

Two signals tell you to switch to $dy$ . The first is the form of the curves: if a boundary is given as $x = f(y)$ , such as $x = y^2$ , horizontal strips read the picture directly. The second is what a vertical cut hits: if a single vertical strip would meet one curve on part of the region and a different curve elsewhere, integrating in $x$ forces you to break the region into several integrals, while a horizontal strip might sweep the whole region in one. Pick the direction that keeps a single pair of boundaries.

Feature	Vertical strips (dx)	Horizontal strips (dy)
Strip runs	up and down	left and right
You subtract	top minus bottom	right minus left
Bounds are	x-values a to b	y-values c to d
Best when	curves are y = f(x)	curves are x = f(y)

Finding the limits: where the curves meet

The bounds of the integral are the points where the two curves intersect. Set the two expressions equal and solve. If you are integrating in $x$ , set $f(x) = g(x)$ and solve for $x$ ; those $x$ -values are $a$ and $b$ . If you are integrating in $y$ , set the two $x$ -expressions equal and solve for $y$ .

On the AP exam this splits by section. On the no-calculator parts the intersections are built to factor cleanly, so a quadratic like $x^2 - x - 2 = 0$ factors to $(x - 2)(x + 1)$ and hands you the bounds. On the calculator-active parts you are expected to use the intersect feature, carry several decimal places, and integrate numerically. Either way, never assume the region starts at the $y$ -axis or at $x = 0$ ; find the actual crossing points first.

When the curves cross (Topic 8.6)

If the two curves intersect at more than two points, the upper and lower boundaries trade places somewhere inside the region. A single subtraction no longer works for the whole interval, because on one piece $f$ is on top and on another piece $g$ is. Topic 8.6 handles exactly this case: split the interval at every interior crossing, subtract in the correct order on each piece, and add the results.

A = \int_a^b \left| f(x) - g(x) \right|\,dx

The absolute value is the compact way to say the same thing: it automatically takes the positive height on every piece. To actually evaluate it you still split at the crossings and drop the bars with the right sign on each subinterval, so the work is identical. Finding those crossings is the whole game, so solve $f(x) = g(x)$ across the full domain, not just near the ends: a factor like $x(x - 1)(x + 1)$ reveals all three roots rather than two.

\int_{-1}^{1} (x - x^3)\,dx = 0

Do not integrate straight through a crossing

Without the absolute value or a split, the signed areas partly cancel and your answer comes out too small, sometimes zero. For $y = x$ and $y = x^3$ on $[-1, 1]$ the honest area is $\frac{1}{2}$ , but integrating straight through gives $0$ (shown above) because the two halves cancel exactly.

Pitfalls that cost points

Subtracting in the wrong order. Bottom minus top gives a negative number; if your area comes out negative, you flipped the difference.
Skipping a crossing. If the curves meet three or more times, integrating straight through makes the signed areas cancel. Split at every interior intersection.
Forcing $dx$ on a sideways region. When the boundaries are $x = f(y)$ , horizontal strips and a $dy$ integral are usually one clean integral instead of two.
Using the wrong bounds. The limits are the intersection points of the two curves, not the $x$ -intercepts of either curve alone.
Rushing the calculator setup on Section II. On calculator-active free response, write the exact integral first, then evaluate numerically; you earn the setup point even if the arithmetic slips.

Worked examples

Worked example

Area with respect to x (Topic 8.4)

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

Find where the curves meet. Set $x + 2 = x^2$ , which gives $x^2 - x - 2 = 0$ , so $(x - 2)(x + 1) = 0$ and $x = -1$ or $x = 2$ . These are the limits of integration.
Decide which curve is on top. Test $x = 0$ : the line gives $0 + 2 = 2$ and the parabola gives $0$ , so the line $y = x + 2$ is the upper boundary. The strips are vertical, so integrate with respect to $x$ .
Set up top minus bottom: $A = \int_{-1}^{2} \bigl[(x + 2) - x^2\bigr]\,dx$ .
Antidifferentiate term by term: $\int (x + 2 - x^2)\,dx = \frac{x^2}{2} + 2x - \frac{x^3}{3}$ .
Evaluate at the upper limit $x = 2$ : $\frac{4}{2} + 4 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{10}{3}$ .
Evaluate at the lower limit $x = -1$ : $\frac{1}{2} - 2 + \frac{1}{3} = -\frac{7}{6}$ .
Subtract the two values: $\frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$ .

$A = \frac{9}{2}$ square units.

Worked example

Area with respect to y (Topic 8.5)

Find the area of the region enclosed by $x = y^2$ and $x = 2y + 3$ .

Find the intersections. Set $y^2 = 2y + 3$ , giving $y^2 - 2y - 3 = 0$ , so $(y - 3)(y + 1) = 0$ and $y = -1$ or $y = 3$ .
Look at the region. The parabola $x = y^2$ opens to the right and the line $x = 2y + 3$ sits to its right; test $y = 0$ , where the parabola gives $x = 0$ and the line gives $x = 3$ . Horizontal strips run from the parabola on the left to the line on the right.
Choose $dy$ . Solving $x = y^2$ for $y$ splits into $y = \pm\sqrt{x}$ , so vertical strips would need two integrals, while horizontal strips need only one. Integrate right minus left with respect to $y$ .
Set up the integral: $A = \int_{-1}^{3} \bigl[(2y + 3) - y^2\bigr]\,dy$ .
Antidifferentiate: $\int (2y + 3 - y^2)\,dy = y^2 + 3y - \frac{y^3}{3}$ .
Evaluate at $y = 3$ : $9 + 9 - 9 = 9$ .
Evaluate at $y = -1$ : $1 - 3 + \frac{1}{3} = -\frac{5}{3}$ .
Subtract: $9 - \left(-\frac{5}{3}\right) = 9 + \frac{5}{3} = \frac{32}{3}$ .

$A = \frac{32}{3}$ square units.

Worked example

Curves that cross (Topic 8.6)

Find the total area of the region enclosed between $y = x^3$ and $y = x$ .

Find every intersection. Set $x^3 = x$ , so $x^3 - x = 0$ and $x(x - 1)(x + 1) = 0$ , giving $x = -1$ , $x = 0$ , and $x = 1$ . Three intersection points mean the curves cross, so this is a Topic 8.6 region.
Determine the order on each piece. On $(-1, 0)$ test $x = -\frac{1}{2}$ : $x^3 = -\frac{1}{8}$ sits above $x = -\frac{1}{2}$ , so $y = x^3$ is on top. On $(0, 1)$ test $x = \frac{1}{2}$ : $x = \frac{1}{2}$ sits above $x^3 = \frac{1}{8}$ , so $y = x$ is on top.
Split the area at $x = 0$ and subtract in the correct order on each piece: $A = \int_{-1}^{0} (x^3 - x)\,dx + \int_{0}^{1} (x - x^3)\,dx$ .
First integral: $\int (x^3 - x)\,dx = \frac{x^4}{4} - \frac{x^2}{2}$ . At $x = 0$ this is $0$ ; at $x = -1$ it is $\frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$ . The piece equals $0 - \left(-\frac{1}{4}\right) = \frac{1}{4}$ .
Second integral: $\int (x - x^3)\,dx = \frac{x^2}{2} - \frac{x^4}{4}$ . At $x = 1$ this is $\frac{1}{2} - \frac{1}{4} = \frac{1}{4}$ ; at $x = 0$ it is $0$ . The piece equals $\frac{1}{4}$ .
Add the pieces: $A = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ .

$A = \frac{1}{2}$ square units.

Frequently asked questions

Do you always subtract top minus bottom?

When you integrate with respect to $x$ , yes: the upper curve minus the lower curve. When you integrate with respect to $y$ , the parallel rule is right minus left. In both cases you subtract so the height is positive, because area cannot be negative.

When do you integrate with respect to y instead of x?

Switch to $dy$ when the boundaries are given as $x = f(y)$ , or when a single vertical strip would hit different curves in different parts of the region. Horizontal strips often turn a two-integral problem into one.

What do you do when the curves cross in the middle?

Split the interval at every interior intersection. On each piece, subtract the curve that is on top there, then add the pieces. Equivalently, integrate the absolute value of the difference. This is Topic 8.6.

Is area between curves on the calculator or non-calculator part?

Both. Non-calculator problems use curves whose intersections factor cleanly. Calculator-active free response expects you to find intersections with the intersect feature and evaluate the integral numerically.