8.3 Area between curves

Achievable AP Calculus AB

8. Applications of integrals

Our AP Calculus AB course is currently in development and is a work-in-progress.

Area between curves

8 min read

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What you’ll learn:

How to use definite integrals to find the area between two curves
When to integrate with respect to $x$ or $y$
Handling curves that intersect multiple times

Recall that a definite integral can represent area under a curve. To find the area between two curves, picture the region sliced into many thin rectangles.

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

If $f (x)$ is always above $g (x)$ for all $x$ on the interval $[a, b]$ , then

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

In other words, when you integrate with respect to $x$ :

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

Examples

Often, you’ll first find intersection points so you know the correct interval(s) of integration.

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

Solution

Find where the two curves intersect by setting them equal:

$x^{2} = x + 2$

$x^{2} - x - 2 = 0$

$(x + 1) (x - 2) = 0$

$x = - 1, 2$

On the interval $[- 1, 2]$ , the line $y = x + 2$ lies above the parabola $y = x^{2}$ . So the area is

$A = \int_{- 1}^{2} ((x + 2) - x^{2}) d x$

Evaluating,

$A = (\frac{x ^{2}}{2} + 2 x - \frac{x ^{3}}{3})_{- 1}^{2}$

$= \frac{9}{2}$

AP tip:

The area of a region bounded by two curves is always positive (it’s not a net signed area). If you get a negative value, re-check which function is on top.

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)

Make sure you’re using the region described in the problem:

Using vertical slices to find the area of a region

This region must be split into 2 integrals:

Blue: On the interval $[0, 1]$ , the top function is $y = x^{3}$ and the bottom function is $y = x^{2} - 2 x$ .
Green: On the interval $[1, 2]$ , the top function is $y = 2 - x$ and the bottom function is $y = x^{2} - 2 x$ .

So the total area is

$\int_{0}^{1} (x^{3} - (x^{2} - 2 x)) d x + \int_{1}^{2} ((2 - x) - (x^{2} - 2 x)) d x$

Integrate carefully (or use Desmos) to get

$A = \frac{25}{12}$

Horizontal slices

Sometimes it’s simpler to slice the region into horizontal rectangles and integrate with respect to $y$ . For example, consider the region bounded by

$x = y^{2}$

$y = - x + 2$

Using horizontal slices to find the area would be easier

If you try to use vertical slices here, the “top” and “bottom” functions change partway through:

From $x = 0$ to $x = 1$ , the region is bounded only by the parabola $x = y^{2}$ . Written as $y$ -functions, the top curve is $y = x$ and the bottom curve is $y = - x$ .
From $x = 1$ to $x = 4$ , the top curve is the line $y = - x + 2$ , while the bottom curve is still $y = - x$ .

You could set up the area using $x$ like this:

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

But that setup is messy (and as written, it’s easy to lose parentheses). Horizontal slices avoid the split.

Given two curves $x = f (y)$ and $x = g (y)$ where $f (y) \geq g (y)$ for all $y$ in $[c, d]$ ,

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

When you integrate with respect to $y$ , think:

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

Examples

Let’s redo the problem presented above.

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

Solution

Rewrite $y = - x + 2$ as a function of $y$ (solve for $x$ ):

$x = 2 - y$

From the graph, $x = 2 - y$ is the right-hand curve and $x = y^{2}$ is the left-hand curve. Now find the intersection points:

$2 - y = y^{2}$

$y^{2} + y - 2 = 0$

$(y - 1) (y + 2) = 0$

$y = 1, - 2$

So we integrate from $y = - 2$ to $y = 1$ :

$\int_{- 2}^{1} ((2 - y) - y^{2}) d y$

$= \frac{9}{2}$

Let $R$ be the region bounded above by the curves $y = \frac{1}{x} - 1$ and $y = 2 x$ and below by the $x$ -axis. Find the area of $R$ in two ways:

a) Vertical slices
b) Horizontal slices

Solutions

a) Vertical slices

(spoiler)

Region $R$ is as shown:

The curve $y = \frac{1}{x} - 1$ intersects the $x$ -axis when

$\frac{1}{x} - 1 = 0$

$\frac{1}{x} = 1$

$x = 1$

The line $y = 2 x$ crosses the $x$ -axis at $x = 0$ .

Now find where the curve and the line intersect:

$\frac{1}{x} - 1 = 2 x$

$1 - x = 2 x^{2}$

$2 x^{2} + x - 1 = 0$

$(x + 1) (2 x - 1) = 0$

$x = - 1, 0.5$

Only $x = 0.5$ is part of region $R$ .

Using vertical slices (integrating with respect to $x$ ), split the region at $x = 0.5$ :

$A = \int_{0}^{0.5} (2 x) d x + \int_{0.5}^{1} (\frac{1}{x} - 1) d x$

$= x^{2}_{0}^{0.5} + [ln (x) - x]_{0.5}^{1}$

$= [(0.5)^{2} - 0^{2}] + [(ln (1) - 1) - (ln (0.5) - 0.5)]$

$\approx 0.443$

b) Horizontal slices

(spoiler)

Rewrite both functions in terms of $y$ :

Line $y = 2 x$ becomes $x = \frac{1}{2} y$ (this is the left function).
Rational function $y = \frac{1}{x} - 1$ becomes

$y = \frac{1}{x} - 1$

$y + 1 = \frac{1}{x}$

$x (y + 1) = 1$

$x = \frac{1}{y + 1}$

The $x$ -axis is $y = 0$ , so it’s the lower bound. The upper bound is the $y$ -value where the line and curve meet.

From the vertical-slice work, the intersection in region $R$ occurs at $x = 0.5$ . Plugging into $y = 2 x$ gives $y = 1$ .

So the area is

$\int_{0}^{1} (\frac{1}{y + 1} - \frac{1}{2} y) d y$

$= ln ∣ y + 1∣ - \frac{1}{4} y^{2}_{0}^{1}$

$= [ln (2) - \frac{1}{4} (1)^{2})] - [ln (1) - \frac{1}{4} (0)^{2}]$

$\approx 0.443$

Multiple intersection points

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). When that happens, split the integral at each intersection point.

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

Solution

Here’s the region:

On $[0, π]$ , the curves meet at $x = \frac{π}{4}$ . On $[0, π /4]$ , $cos (x)$ is above $sin (x)$ . On $[π /4, π]$ , $sin (x)$ is above $cos (x)$ . So we split the area into two integrals:

$\int_{0}^{π /4} (cos (x) - sin (x)) d x + \int_{π /4}^{π} (sin (x) - cos (x)) d x$

$= (sin (x) + cos (x))_{0}^{π /4} + (- cos (x) - sin (x))_{π /4}^{π}$

$[(sin (π /4) + cos (π /4)) - (sin (0) + cos (0))] + [(- cos (π) - sin (π)) - (- cos (π /4) - sin (π /4))]$

$= 22$

On a calculator-allowed problem, you can also write this as a single integral using an absolute value:

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

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:

How to use definite integrals to find the area between two curves
When to integrate with respect to $x$ or $y$
Handling curves that intersect multiple times

Recall that a definite integral can represent area under a curve. To find the area between two curves, picture the region sliced into many thin rectangles.

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

If $f (x)$ is always above $g (x)$ for all $x$ on the interval $[a, b]$ , then

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

In other words, when you integrate with respect to $x$ :

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

Examples

Often, you’ll first find intersection points so you know the correct interval(s) of integration.

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

Solution

Find where the two curves intersect by setting them equal:

$x^{2} = x + 2$

$x^{2} - x - 2 = 0$

$(x + 1) (x - 2) = 0$

$x = - 1, 2$

On the interval $[- 1, 2]$ , the line $y = x + 2$ lies above the parabola $y = x^{2}$ . So the area is

$A = \int_{- 1}^{2} ((x + 2) - x^{2}) d x$

Evaluating,

$A = (\frac{x ^{2}}{2} + 2 x - \frac{x ^{3}}{3})_{- 1}^{2}$

$= \frac{9}{2}$

AP tip:

The area of a region bounded by two curves is always positive (it’s not a net signed area). If you get a negative value, re-check which function is on top.

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)

Make sure you’re using the region described in the problem:

This region must be split into 2 integrals:

Blue: On the interval $[0, 1]$ , the top function is $y = x^{3}$ and the bottom function is $y = x^{2} - 2 x$ .
Green: On the interval $[1, 2]$ , the top function is $y = 2 - x$ and the bottom function is $y = x^{2} - 2 x$ .

So the total area is

$\int_{0}^{1} (x^{3} - (x^{2} - 2 x)) d x + \int_{1}^{2} ((2 - x) - (x^{2} - 2 x)) d x$

Integrate carefully (or use Desmos) to get

$A = \frac{25}{12}$

Horizontal slices

Sometimes it’s simpler to slice the region into horizontal rectangles and integrate with respect to $y$ . For example, consider the region bounded by

$x = y^{2}$

$y = - x + 2$

If you try to use vertical slices here, the “top” and “bottom” functions change partway through:

From $x = 0$ to $x = 1$ , the region is bounded only by the parabola $x = y^{2}$ . Written as $y$ -functions, the top curve is $y = x$ and the bottom curve is $y = - x$ .
From $x = 1$ to $x = 4$ , the top curve is the line $y = - x + 2$ , while the bottom curve is still $y = - x$ .

You could set up the area using $x$ like this:

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

But that setup is messy (and as written, it’s easy to lose parentheses). Horizontal slices avoid the split.

Given two curves $x = f (y)$ and $x = g (y)$ where $f (y) \geq g (y)$ for all $y$ in $[c, d]$ ,

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

When you integrate with respect to $y$ , think:

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

Examples

Let’s redo the problem presented above.

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

Solution

Rewrite $y = - x + 2$ as a function of $y$ (solve for $x$ ):

$x = 2 - y$

From the graph, $x = 2 - y$ is the right-hand curve and $x = y^{2}$ is the left-hand curve. Now find the intersection points:

$2 - y = y^{2}$

$y^{2} + y - 2 = 0$

$(y - 1) (y + 2) = 0$

$y = 1, - 2$

So we integrate from $y = - 2$ to $y = 1$ :

$\int_{- 2}^{1} ((2 - y) - y^{2}) d y$

$= \frac{9}{2}$

Let $R$ be the region bounded above by the curves $y = \frac{1}{x} - 1$ and $y = 2 x$ and below by the $x$ -axis. Find the area of $R$ in two ways:

a) Vertical slices
b) Horizontal slices

Solutions

a) Vertical slices

(spoiler)

Region $R$ is as shown:

The curve $y = \frac{1}{x} - 1$ intersects the $x$ -axis when

$\frac{1}{x} - 1 = 0$

$\frac{1}{x} = 1$

$x = 1$

The line $y = 2 x$ crosses the $x$ -axis at $x = 0$ .

Now find where the curve and the line intersect:

$\frac{1}{x} - 1 = 2 x$

$1 - x = 2 x^{2}$

$2 x^{2} + x - 1 = 0$

$(x + 1) (2 x - 1) = 0$

$x = - 1, 0.5$

Only $x = 0.5$ is part of region $R$ .

Using vertical slices (integrating with respect to $x$ ), split the region at $x = 0.5$ :

$A = \int_{0}^{0.5} (2 x) d x + \int_{0.5}^{1} (\frac{1}{x} - 1) d x$

$= x^{2}_{0}^{0.5} + [ln (x) - x]_{0.5}^{1}$

$= [(0.5)^{2} - 0^{2}] + [(ln (1) - 1) - (ln (0.5) - 0.5)]$

$\approx 0.443$

b) Horizontal slices

(spoiler)

Rewrite both functions in terms of $y$ :

Line $y = 2 x$ becomes $x = \frac{1}{2} y$ (this is the left function).
Rational function $y = \frac{1}{x} - 1$ becomes

$y = \frac{1}{x} - 1$

$y + 1 = \frac{1}{x}$

$x (y + 1) = 1$

$x = \frac{1}{y + 1}$

The $x$ -axis is $y = 0$ , so it’s the lower bound. The upper bound is the $y$ -value where the line and curve meet.

From the vertical-slice work, the intersection in region $R$ occurs at $x = 0.5$ . Plugging into $y = 2 x$ gives $y = 1$ .

So the area is

$\int_{0}^{1} (\frac{1}{y + 1} - \frac{1}{2} y) d y$

$= ln ∣ y + 1∣ - \frac{1}{4} y^{2}_{0}^{1}$

$= [ln (2) - \frac{1}{4} (1)^{2})] - [ln (1) - \frac{1}{4} (0)^{2}]$

$\approx 0.443$

Multiple intersection points

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

Solution

Here’s the region:

On $[0, π]$ , the curves meet at $x = \frac{π}{4}$ . On $[0, π /4]$ , $cos (x)$ is above $sin (x)$ . On $[π /4, π]$ , $sin (x)$ is above $cos (x)$ . So we split the area into two integrals:

$\int_{0}^{π /4} (cos (x) - sin (x)) d x + \int_{π /4}^{π} (sin (x) - cos (x)) d x$

$= (sin (x) + cos (x))_{0}^{π /4} + (- cos (x) - sin (x))_{π /4}^{π}$

$[(sin (π /4) + cos (π /4)) - (sin (0) + cos (0))] + [(- cos (π) - sin (π)) - (- cos (π /4) - sin (π /4))]$

$= 22$

On a calculator-allowed problem, you can also write this as a single integral using an absolute value:

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

Achievable AP Calculus AB

8. Applications of integrals

Our AP Calculus AB course is currently in development and is a work-in-progress.

Area between curves

8 min read

Font

Discuss

Feedback

What you’ll learn:

How to use definite integrals to find the area between two curves
When to integrate with respect to $x$ or $y$
Handling curves that intersect multiple times

Recall that a definite integral can represent area under a curve. To find the area between two curves, picture the region sliced into many thin rectangles.

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

If $f (x)$ is always above $g (x)$ for all $x$ on the interval $[a, b]$ , then

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

In other words, when you integrate with respect to $x$ :

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

Examples

Often, you’ll first find intersection points so you know the correct interval(s) of integration.

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

Solution

Find where the two curves intersect by setting them equal:

$x^{2} = x + 2$

$x^{2} - x - 2 = 0$

$(x + 1) (x - 2) = 0$

$x = - 1, 2$

On the interval $[- 1, 2]$ , the line $y = x + 2$ lies above the parabola $y = x^{2}$ . So the area is

$A = \int_{- 1}^{2} ((x + 2) - x^{2}) d x$

Evaluating,

$A = (\frac{x ^{2}}{2} + 2 x - \frac{x ^{3}}{3})_{- 1}^{2}$

$= \frac{9}{2}$

AP tip:

The area of a region bounded by two curves is always positive (it’s not a net signed area). If you get a negative value, re-check which function is on top.

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)

Make sure you’re using the region described in the problem:

This region must be split into 2 integrals:

Blue: On the interval $[0, 1]$ , the top function is $y = x^{3}$ and the bottom function is $y = x^{2} - 2 x$ .
Green: On the interval $[1, 2]$ , the top function is $y = 2 - x$ and the bottom function is $y = x^{2} - 2 x$ .

So the total area is

$\int_{0}^{1} (x^{3} - (x^{2} - 2 x)) d x + \int_{1}^{2} ((2 - x) - (x^{2} - 2 x)) d x$

Integrate carefully (or use Desmos) to get

$A = \frac{25}{12}$

Horizontal slices

Sometimes it’s simpler to slice the region into horizontal rectangles and integrate with respect to $y$ . For example, consider the region bounded by

$x = y^{2}$

$y = - x + 2$

If you try to use vertical slices here, the “top” and “bottom” functions change partway through:

From $x = 0$ to $x = 1$ , the region is bounded only by the parabola $x = y^{2}$ . Written as $y$ -functions, the top curve is $y = x$ and the bottom curve is $y = - x$ .
From $x = 1$ to $x = 4$ , the top curve is the line $y = - x + 2$ , while the bottom curve is still $y = - x$ .

You could set up the area using $x$ like this:

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

But that setup is messy (and as written, it’s easy to lose parentheses). Horizontal slices avoid the split.

Given two curves $x = f (y)$ and $x = g (y)$ where $f (y) \geq g (y)$ for all $y$ in $[c, d]$ ,

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

When you integrate with respect to $y$ , think:

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

Examples

Let’s redo the problem presented above.

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

Solution

Rewrite $y = - x + 2$ as a function of $y$ (solve for $x$ ):

$x = 2 - y$

From the graph, $x = 2 - y$ is the right-hand curve and $x = y^{2}$ is the left-hand curve. Now find the intersection points:

$2 - y = y^{2}$

$y^{2} + y - 2 = 0$

$(y - 1) (y + 2) = 0$

$y = 1, - 2$

So we integrate from $y = - 2$ to $y = 1$ :

$\int_{- 2}^{1} ((2 - y) - y^{2}) d y$

$= \frac{9}{2}$

Let $R$ be the region bounded above by the curves $y = \frac{1}{x} - 1$ and $y = 2 x$ and below by the $x$ -axis. Find the area of $R$ in two ways:

a) Vertical slices
b) Horizontal slices

Solutions

a) Vertical slices

(spoiler)

Region $R$ is as shown:

The curve $y = \frac{1}{x} - 1$ intersects the $x$ -axis when

$\frac{1}{x} - 1 = 0$

$\frac{1}{x} = 1$

$x = 1$

The line $y = 2 x$ crosses the $x$ -axis at $x = 0$ .

Now find where the curve and the line intersect:

$\frac{1}{x} - 1 = 2 x$

$1 - x = 2 x^{2}$

$2 x^{2} + x - 1 = 0$

$(x + 1) (2 x - 1) = 0$

$x = - 1, 0.5$

Only $x = 0.5$ is part of region $R$ .

Using vertical slices (integrating with respect to $x$ ), split the region at $x = 0.5$ :

$A = \int_{0}^{0.5} (2 x) d x + \int_{0.5}^{1} (\frac{1}{x} - 1) d x$

$= x^{2}_{0}^{0.5} + [ln (x) - x]_{0.5}^{1}$

$= [(0.5)^{2} - 0^{2}] + [(ln (1) - 1) - (ln (0.5) - 0.5)]$

$\approx 0.443$

b) Horizontal slices

(spoiler)

Rewrite both functions in terms of $y$ :

Line $y = 2 x$ becomes $x = \frac{1}{2} y$ (this is the left function).
Rational function $y = \frac{1}{x} - 1$ becomes

$y = \frac{1}{x} - 1$

$y + 1 = \frac{1}{x}$

$x (y + 1) = 1$

$x = \frac{1}{y + 1}$

The $x$ -axis is $y = 0$ , so it’s the lower bound. The upper bound is the $y$ -value where the line and curve meet.

From the vertical-slice work, the intersection in region $R$ occurs at $x = 0.5$ . Plugging into $y = 2 x$ gives $y = 1$ .

So the area is

$\int_{0}^{1} (\frac{1}{y + 1} - \frac{1}{2} y) d y$

$= ln ∣ y + 1∣ - \frac{1}{4} y^{2}_{0}^{1}$

$= [ln (2) - \frac{1}{4} (1)^{2})] - [ln (1) - \frac{1}{4} (0)^{2}]$

$\approx 0.443$

Multiple intersection points

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

Solution

Here’s the region:

On $[0, π]$ , the curves meet at $x = \frac{π}{4}$ . On $[0, π /4]$ , $cos (x)$ is above $sin (x)$ . On $[π /4, π]$ , $sin (x)$ is above $cos (x)$ . So we split the area into two integrals:

$\int_{0}^{π /4} (cos (x) - sin (x)) d x + \int_{π /4}^{π} (sin (x) - cos (x)) d x$

$= (sin (x) + cos (x))_{0}^{π /4} + (- cos (x) - sin (x))_{π /4}^{π}$

$[(sin (π /4) + cos (π /4)) - (sin (0) + cos (0))] + [(- cos (π) - sin (π)) - (- cos (π /4) - sin (π /4))]$

$= 22$

On a calculator-allowed problem, you can also write this as a single integral using an absolute value:

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

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:

How to use definite integrals to find the area between two curves
When to integrate with respect to $x$ or $y$
Handling curves that intersect multiple times

Recall that a definite integral can represent area under a curve. To find the area between two curves, picture the region sliced into many thin rectangles.

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

If $f (x)$ is always above $g (x)$ for all $x$ on the interval $[a, b]$ , then

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

In other words, when you integrate with respect to $x$ :

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

Examples

Often, you’ll first find intersection points so you know the correct interval(s) of integration.

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

Solution

Find where the two curves intersect by setting them equal:

$x^{2} = x + 2$

$x^{2} - x - 2 = 0$

$(x + 1) (x - 2) = 0$

$x = - 1, 2$

On the interval $[- 1, 2]$ , the line $y = x + 2$ lies above the parabola $y = x^{2}$ . So the area is

$A = \int_{- 1}^{2} ((x + 2) - x^{2}) d x$

Evaluating,

$A = (\frac{x ^{2}}{2} + 2 x - \frac{x ^{3}}{3})_{- 1}^{2}$

$= \frac{9}{2}$

AP tip:

The area of a region bounded by two curves is always positive (it’s not a net signed area). If you get a negative value, re-check which function is on top.

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)

Make sure you’re using the region described in the problem:

This region must be split into 2 integrals:

Blue: On the interval $[0, 1]$ , the top function is $y = x^{3}$ and the bottom function is $y = x^{2} - 2 x$ .
Green: On the interval $[1, 2]$ , the top function is $y = 2 - x$ and the bottom function is $y = x^{2} - 2 x$ .

So the total area is

$\int_{0}^{1} (x^{3} - (x^{2} - 2 x)) d x + \int_{1}^{2} ((2 - x) - (x^{2} - 2 x)) d x$

Integrate carefully (or use Desmos) to get

$A = \frac{25}{12}$

Horizontal slices

Sometimes it’s simpler to slice the region into horizontal rectangles and integrate with respect to $y$ . For example, consider the region bounded by

$x = y^{2}$

$y = - x + 2$

If you try to use vertical slices here, the “top” and “bottom” functions change partway through:

From $x = 0$ to $x = 1$ , the region is bounded only by the parabola $x = y^{2}$ . Written as $y$ -functions, the top curve is $y = x$ and the bottom curve is $y = - x$ .
From $x = 1$ to $x = 4$ , the top curve is the line $y = - x + 2$ , while the bottom curve is still $y = - x$ .

You could set up the area using $x$ like this:

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

But that setup is messy (and as written, it’s easy to lose parentheses). Horizontal slices avoid the split.

Given two curves $x = f (y)$ and $x = g (y)$ where $f (y) \geq g (y)$ for all $y$ in $[c, d]$ ,

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

When you integrate with respect to $y$ , think:

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

Examples

Let’s redo the problem presented above.

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

Solution

Rewrite $y = - x + 2$ as a function of $y$ (solve for $x$ ):

$x = 2 - y$

From the graph, $x = 2 - y$ is the right-hand curve and $x = y^{2}$ is the left-hand curve. Now find the intersection points:

$2 - y = y^{2}$

$y^{2} + y - 2 = 0$

$(y - 1) (y + 2) = 0$

$y = 1, - 2$

So we integrate from $y = - 2$ to $y = 1$ :

$\int_{- 2}^{1} ((2 - y) - y^{2}) d y$

$= \frac{9}{2}$

Let $R$ be the region bounded above by the curves $y = \frac{1}{x} - 1$ and $y = 2 x$ and below by the $x$ -axis. Find the area of $R$ in two ways:

a) Vertical slices
b) Horizontal slices

Solutions

a) Vertical slices

(spoiler)

Region $R$ is as shown:

The curve $y = \frac{1}{x} - 1$ intersects the $x$ -axis when

$\frac{1}{x} - 1 = 0$

$\frac{1}{x} = 1$

$x = 1$

The line $y = 2 x$ crosses the $x$ -axis at $x = 0$ .

Now find where the curve and the line intersect:

$\frac{1}{x} - 1 = 2 x$

$1 - x = 2 x^{2}$

$2 x^{2} + x - 1 = 0$

$(x + 1) (2 x - 1) = 0$

$x = - 1, 0.5$

Only $x = 0.5$ is part of region $R$ .

Using vertical slices (integrating with respect to $x$ ), split the region at $x = 0.5$ :

$A = \int_{0}^{0.5} (2 x) d x + \int_{0.5}^{1} (\frac{1}{x} - 1) d x$

$= x^{2}_{0}^{0.5} + [ln (x) - x]_{0.5}^{1}$

$= [(0.5)^{2} - 0^{2}] + [(ln (1) - 1) - (ln (0.5) - 0.5)]$

$\approx 0.443$

b) Horizontal slices

(spoiler)

Rewrite both functions in terms of $y$ :

Line $y = 2 x$ becomes $x = \frac{1}{2} y$ (this is the left function).
Rational function $y = \frac{1}{x} - 1$ becomes

$y = \frac{1}{x} - 1$

$y + 1 = \frac{1}{x}$

$x (y + 1) = 1$

$x = \frac{1}{y + 1}$

The $x$ -axis is $y = 0$ , so it’s the lower bound. The upper bound is the $y$ -value where the line and curve meet.

From the vertical-slice work, the intersection in region $R$ occurs at $x = 0.5$ . Plugging into $y = 2 x$ gives $y = 1$ .

So the area is

$\int_{0}^{1} (\frac{1}{y + 1} - \frac{1}{2} y) d y$

$= ln ∣ y + 1∣ - \frac{1}{4} y^{2}_{0}^{1}$

$= [ln (2) - \frac{1}{4} (1)^{2})] - [ln (1) - \frac{1}{4} (0)^{2}]$

$\approx 0.443$

Multiple intersection points

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

Solution

Here’s the region:

On $[0, π]$ , the curves meet at $x = \frac{π}{4}$ . On $[0, π /4]$ , $cos (x)$ is above $sin (x)$ . On $[π /4, π]$ , $sin (x)$ is above $cos (x)$ . So we split the area into two integrals:

$\int_{0}^{π /4} (cos (x) - sin (x)) d x + \int_{π /4}^{π} (sin (x) - cos (x)) d x$

$= (sin (x) + cos (x))_{0}^{π /4} + (- cos (x) - sin (x))_{π /4}^{π}$

$[(sin (π /4) + cos (π /4)) - (sin (0) + cos (0))] + [(- cos (π) - sin (π)) - (- cos (π /4) - sin (π /4))]$

$= 22$

On a calculator-allowed problem, you can also write this as a single integral using an absolute value:

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