8.3 Area between curves

Achievable AP Calculus AB

8. Applications of integrals

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Area between curves

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What you’ll learn

Area between curves: Use definite integrals to calculate the area bounded by two functions.
Variable of integration: Determine whether to integrate with respect to $x$ (Top − Bottom) or $y$ (Right − Left).
Multiple intersections: Split the area into separate integrals when curves cross each other.

Recall that a definite integral represents the signed area between a curve and the $x$ -axis. To find the area between two curves, imagine slicing the region into infinitely thin rectangular strips. For each rectangle, the distance between the two functions represents its length, while $d x$ represents its infinitesimal width.

If the slices are vertical (parallel to the $y$ -axis), integrate with respect to $x$ .
If the slices are horizontal (parallel to the $x$ -axis), integrate with respect to $y$ .

Vertical slices ( $x$ -limits)

If a function $f (x)$ is always above $g (x)$ across the entire interval $[a, b]$ , the area of the region is defined as:

$Area = \int_{a}^{b} (Top function - bottom function) d x$

$or$

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

Example 1

Find the area of the region bounded by $y = x^{2}$ and $y = x + 2$ .

Solution

(spoiler)

1. Find the boundaries:

Set the equations equal to find their intersection points.

$x^{2} x^{2} - x - 2 (x + 1) (x - 2) x = x + 2 = 0 = 0 = - 1, 2$

The curves intersect at $x = - 1$ and $x = 2$ .

2. Identify Top vs. Bottom:

Test a point between $x = - 1$ and $x = 2$ . At $x = 0$ , the line is at $y = 2$ and the parabola is at $y = 0$ . So the line is the top function.

3. Set up and integrate:

$A = \int_{- 1}^{2} ((x + 2) - x^{2}) d x = [\frac{1}{2} x^{2} + 2 x - \frac{1}{3} x^{3}]_{- 1}^{2} = \frac{9}{2}$

AP tip:

The area of a region bounded by two curves is always positive (not a net signed area). If your calculation yields a negative value, re-check which function is on top.

Example 2

Find the area of the region bounded above by the curves $y = x^{3}$ and $y = 2 - x$ and below by $y = x^{2} - 2 x$ .

Solution

(spoiler)

Shown is the region described in the problem:

Using vertical slices to find the area of a region

Because the upper boundary changes its formula halfway through the region, the area must be split into two separate integrals at the intersection point $x = 1$ :

Blue region $(0, 1)$ : $y = x^{3}$ sits above $y = x^{2} - 2 x$ .
Green region $(1, 2)$ : $y = 2 - x$ sits above $y = x^{2} - 2 x$ .

$Total Area = \int_{0}^{1} (x^{3} - (x^{2} - 2 x)) d x + \int_{1}^{2} ((2 - x) - (x^{2} - 2 x)) d x = \frac{7}{12} + \frac{3}{2} = \frac{25}{12}$

Horizontal slices ( $y$ -limits)

Sometimes it’s simpler to slice the region horizontally and integrate with respect to $y$ . For example:

Using horizontal slices to find the area would be easier

The “top” and “bottom” functions change partway through:

On $(0, 1)$ , the region is bounded only by the parabola $x = y^{2}$ . Written as $y$ -functions:
- Top: $y = x$
- Bottom: $y = - x$
On $(1, 4)$ , the line $y = - x + 2$ sits above $y = - x$ .

The area can be found with the integral:

$\int_{0}^{1} (x - (- x)) d x + \int_{1}^{4} (- x + 2 - (- x)) d x$

But instead of looking top-to-bottom, orient your perspective from right-to-left. Given two curves expressed as functions of $y$ , where $f (y) \geq g (y)$ across the interval $[c, d]$ :

$A = \int_{c}^{d} (Right function - Left function) d y$

$or$

$Area = \int_{c}^{d} (f (y) - g (y)) d y$

Example 3

Let’s redo the problem presented above.

Find the area of the region bounded by $x = y^{2}$ and $y = - x + 2$ .

Solution

(spoiler)

1. Convert to functions of $y$ :

Solve the linear equation for $x$ .

$y = - x + 2 ⟹ x = 2 - y$

2. Find the $y$ -boundaries:

Set the functions equal to locate the intersection points.

$2 - y y^{2} + y - 2 (y - 1) (y + 2) y = y^{2} = 0 = 0 = 1, - 2$

The region is bounded from $y = - 2$ to $y = 1$ .

3. Identify Right vs. Left:

When a graph is not given, test points - on the interval $(- 2, 1)$ , testing $y = 0$ gives $x = 2$ for the line and $x = 0$ for the parabola. So the line is on the right.

4. Set up and integrate:

$A = \int_{- 2}^{1} ((2 - y) - y^{2}) d y = [2 y - \frac{1}{2} y^{2} - \frac{1}{3} y^{3}]_{- 2}^{1} = \frac{9}{2}$

Example 4

Let $R$ be the region bounded below by the $x$ -axis and above by $y = \frac{1}{x} - 1$ and $y = 2 x$ . Find the area of $R$ using both slicing orientations.

Solutions

a) Vertical slices ( $d x$ )

(spoiler)

Region $R$ is as shown:

We must split the region at the intersection point $x = \frac{1}{2}$ :

On $(0, \frac{1}{2})$ , the top boundary is $y = 2 x$ .
On $(\frac{1}{2}, 1)$ , the top boundary is $y = \frac{1}{x} - 1$ .
- For both, the “bottom function” is $y = 0$ (the $x$ -axis).

$A = \int_{0}^{0.5} 2 x d x + \int_{0.5}^{1} (\frac{1}{x} - 1) d x = [x^{2}]_{0}^{0.5} + [ln (x) - x]_{0.5}^{1} \approx 0.443 (calculator)$

b) Horizontal slices ( $d y$ )

(spoiler)

Solve both equations for $x$ to look right-to-left:

$y = 2 x ⟹ x = \frac{1}{2} y$ (Left curve)
$y = \frac{1}{x} - 1 ⟹ x = \frac{1}{y + 1}$ (Right curve)

The $y$ -boundaries span from the $x$ -axis ( $y = 0$ ) up to the intersection point ( $y = 1$ ).

$A = \int_{0}^{1} (\frac{1}{y + 1} - \frac{1}{2} y) d y = [ln ∣ y + 1∣ - \frac{1}{4} y^{2}]_{0}^{1} = (ln (2) - \frac{1}{4}) - 0 \approx 0.443$

Multiple intersections

Over an interval, functions can switch roles: a curve that starts on top may end up on the bottom (or a left curve may become the right curve). To calculate the area, split the integral at each intersection point.

Example 5

Find the area of the region bounded by $y = sin (x)$ and $y = cos (x)$ from $x = 0$ to $x = π$ .

Solution

(spoiler)

On the interval $[0, π]$ , the curves meet at $x = \frac{π}{4}$ .

On $(0, \frac{π}{4})$ , $cos (x)$ is on top.
On $(\frac{π}{4}, π)$ , $sin (x)$ is on top.

Set up two distinct integrals:

$A = \int_{0}^{π /4} (cos (x) - sin (x)) d x + \int_{π /4}^{π} (sin (x) - cos (x)) d x = [sin (x) + cos (x)]_{0}^{π /4} + [- cos (x) - sin (x)]_{π /4}^{π} = 22$

AP calculator tip:

On calculator-active sections, you can skip figuring out which goes above or below by wrapping the functions in an absolute value block inside a single integral expression:

$\int_{0}^{π} ∣ sin (x) - cos (x) ∣ d x$

Area between two curves: basics

Area = definite integral of (top function - bottom function) over interval
For vertical slices: integrate with respect to $x$
For horizontal slices: integrate with respect to $y$

Vertical slices

Area formula: $\int_{a}^{b} (f (x) - g (x)) d x$
“Top” function minus “bottom” function on $[a, b]$
Always use positive area (if negative, switch order)

Finding intersection points

Set curves equal to find limits of integration
Split integral if curves intersect multiple times

Horizontal slices

Use when region is easier to describe with $y$
Area formula: $\int_{c}^{d} (f (y) - g (y)) d y$
- $f (y)$ = right function, $g (y)$ = left function

Multiple intersection points

Functions may switch roles (top/bottom or left/right)
Split integral at each intersection point
Alternatively, use $\int ∣ difference ∣ d x$ for total area

General strategies

Choose vertical or horizontal slices based on region shape
Always check which function is on top/right in each interval
For complicated regions, break into sub-intervals and sum areas

Area between curves

What you’ll learn

Area between curves: Use definite integrals to calculate the area bounded by two functions.
Variable of integration: Determine whether to integrate with respect to $x$ (Top − Bottom) or $y$ (Right − Left).
Multiple intersections: Split the area into separate integrals when curves cross each other.

If the slices are vertical (parallel to the $y$ -axis), integrate with respect to $x$ .
If the slices are horizontal (parallel to the $x$ -axis), integrate with respect to $y$ .

Vertical slices ( $x$ -limits)

If a function $f (x)$ is always above $g (x)$ across the entire interval $[a, b]$ , the area of the region is defined as:

$Area = \int_{a}^{b} (Top function - bottom function) d x$

$or$

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

Example 1

Find the area of the region bounded by $y = x^{2}$ and $y = x + 2$ .

Solution

(spoiler)

1. Find the boundaries:

Set the equations equal to find their intersection points.

$x^{2} x^{2} - x - 2 (x + 1) (x - 2) x = x + 2 = 0 = 0 = - 1, 2$

The curves intersect at $x = - 1$ and $x = 2$ .

2. Identify Top vs. Bottom:

Test a point between $x = - 1$ and $x = 2$ . At $x = 0$ , the line is at $y = 2$ and the parabola is at $y = 0$ . So the line is the top function.

3. Set up and integrate:

$A = \int_{- 1}^{2} ((x + 2) - x^{2}) d x = [\frac{1}{2} x^{2} + 2 x - \frac{1}{3} x^{3}]_{- 1}^{2} = \frac{9}{2}$

AP tip:

The area of a region bounded by two curves is always positive (not a net signed area). If your calculation yields a negative value, re-check which function is on top.

Example 2

Find the area of the region bounded above by the curves $y = x^{3}$ and $y = 2 - x$ and below by $y = x^{2} - 2 x$ .

Solution

(spoiler)

Shown is the region described in the problem:

Because the upper boundary changes its formula halfway through the region, the area must be split into two separate integrals at the intersection point $x = 1$ :

Blue region $(0, 1)$ : $y = x^{3}$ sits above $y = x^{2} - 2 x$ .
Green region $(1, 2)$ : $y = 2 - x$ sits above $y = x^{2} - 2 x$ .

$Total Area = \int_{0}^{1} (x^{3} - (x^{2} - 2 x)) d x + \int_{1}^{2} ((2 - x) - (x^{2} - 2 x)) d x = \frac{7}{12} + \frac{3}{2} = \frac{25}{12}$

Horizontal slices ( $y$ -limits)

Sometimes it’s simpler to slice the region horizontally and integrate with respect to $y$ . For example:

The “top” and “bottom” functions change partway through:

On $(0, 1)$ , the region is bounded only by the parabola $x = y^{2}$ . Written as $y$ -functions:
- Top: $y = x$
- Bottom: $y = - x$
On $(1, 4)$ , the line $y = - x + 2$ sits above $y = - x$ .

The area can be found with the integral:

$\int_{0}^{1} (x - (- x)) d x + \int_{1}^{4} (- x + 2 - (- x)) d x$

But instead of looking top-to-bottom, orient your perspective from right-to-left. Given two curves expressed as functions of $y$ , where $f (y) \geq g (y)$ across the interval $[c, d]$ :

$A = \int_{c}^{d} (Right function - Left function) d y$

$or$

$Area = \int_{c}^{d} (f (y) - g (y)) d y$

Example 3

Let’s redo the problem presented above.

Find the area of the region bounded by $x = y^{2}$ and $y = - x + 2$ .

Solution

(spoiler)

1. Convert to functions of $y$ :

Solve the linear equation for $x$ .

$y = - x + 2 ⟹ x = 2 - y$

2. Find the $y$ -boundaries:

Set the functions equal to locate the intersection points.

$2 - y y^{2} + y - 2 (y - 1) (y + 2) y = y^{2} = 0 = 0 = 1, - 2$

The region is bounded from $y = - 2$ to $y = 1$ .

3. Identify Right vs. Left:

When a graph is not given, test points - on the interval $(- 2, 1)$ , testing $y = 0$ gives $x = 2$ for the line and $x = 0$ for the parabola. So the line is on the right.

4. Set up and integrate:

$A = \int_{- 2}^{1} ((2 - y) - y^{2}) d y = [2 y - \frac{1}{2} y^{2} - \frac{1}{3} y^{3}]_{- 2}^{1} = \frac{9}{2}$

Example 4

Let $R$ be the region bounded below by the $x$ -axis and above by $y = \frac{1}{x} - 1$ and $y = 2 x$ . Find the area of $R$ using both slicing orientations.

Solutions

a) Vertical slices ( $d x$ )

(spoiler)

Region $R$ is as shown:

We must split the region at the intersection point $x = \frac{1}{2}$ :

On $(0, \frac{1}{2})$ , the top boundary is $y = 2 x$ .
On $(\frac{1}{2}, 1)$ , the top boundary is $y = \frac{1}{x} - 1$ .
- For both, the “bottom function” is $y = 0$ (the $x$ -axis).

$A = \int_{0}^{0.5} 2 x d x + \int_{0.5}^{1} (\frac{1}{x} - 1) d x = [x^{2}]_{0}^{0.5} + [ln (x) - x]_{0.5}^{1} \approx 0.443 (calculator)$

b) Horizontal slices ( $d y$ )

(spoiler)

Solve both equations for $x$ to look right-to-left:

$y = 2 x ⟹ x = \frac{1}{2} y$ (Left curve)
$y = \frac{1}{x} - 1 ⟹ x = \frac{1}{y + 1}$ (Right curve)

The $y$ -boundaries span from the $x$ -axis ( $y = 0$ ) up to the intersection point ( $y = 1$ ).

$A = \int_{0}^{1} (\frac{1}{y + 1} - \frac{1}{2} y) d y = [ln ∣ y + 1∣ - \frac{1}{4} y^{2}]_{0}^{1} = (ln (2) - \frac{1}{4}) - 0 \approx 0.443$

Multiple intersections

Example 5

Find the area of the region bounded by $y = sin (x)$ and $y = cos (x)$ from $x = 0$ to $x = π$ .

Solution

(spoiler)

On the interval $[0, π]$ , the curves meet at $x = \frac{π}{4}$ .

On $(0, \frac{π}{4})$ , $cos (x)$ is on top.
On $(\frac{π}{4}, π)$ , $sin (x)$ is on top.

Set up two distinct integrals:

$A = \int_{0}^{π /4} (cos (x) - sin (x)) d x + \int_{π /4}^{π} (sin (x) - cos (x)) d x = [sin (x) + cos (x)]_{0}^{π /4} + [- cos (x) - sin (x)]_{π /4}^{π} = 22$

AP calculator tip:

On calculator-active sections, you can skip figuring out which goes above or below by wrapping the functions in an absolute value block inside a single integral expression:

$\int_{0}^{π} ∣ sin (x) - cos (x) ∣ d x$

Achievable AP Calculus AB

8. Applications of integrals

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.