Area between curves | Applications of integrals

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 the definite integral can be used to represent area under a curve. When finding the area between curves, imagine slicing the region 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 interval $[a, b]$ , then

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

Just remember that $f (x)$ is the top function, or

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

Examples

Often times you’ll have to first find the intersection points to determine the interval.

1. 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:

$x^{2} = x + 2$

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

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

$x = - 1, 2$

Between $x = - 1$ and $x = 2$ , the curve $y = x + 2$ is always above/greater than the curve $y = x^{2}$ , so the area can be found by

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

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 (no net signed area) so if you arrive at a negative answer, re-check the integral set-up.

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)

Make sure to choose the correct region described in the problem:

Using vertical slices to find the area of a region

The region must be split into 2 integrals

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

Then 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 easier to find the area by slicing the region into horizontal rectangles and integrating with respect to $y$ . For example, take a look at the region bounded by the two curves

$x = y^{2}$

$y = - x + 2$

Using horizontal slices to find the area would be easier

Notice that from $x = 0$ to $x = 1$ , the region is bound by only the parabola $x = y^{2}$ . In terms of $x$ , the top function is $y = x$ and the bottom one is $y = - x$ .

On the other hand, from $x = 1$ to $x = 4$ , the top function is the line $y = - x + 2$ while the bottom function is the lower half of $x = y^{2}$ , or $y = - x$ .

We could set up the integral as such:

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

But this quickly becomes tedious for more complicated functions. Instead, if we choose to slice the region into horizontal rectangles, then the area can be found by integrating with respect to $y$ :

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

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

Just remember that $f (y)$ is the rightmost function, or

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

Examples

Let’s redo the problem presented above.

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

Solution

Rewrite $y = - x + 2$ to be a function of $y$ :

$x = 2 - y$

As shown on the graph, the line $x = 2 - y$ is the right function while the parabola $x = y^{2}$ is the left function. Finding the intersection points algebraically,

$2 - y = y^{2}$

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

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

$y = 1, - 2$

So the interval to integrate over is from $y = - 2$ to $y = 1$ . The area is

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

$= \frac{9}{2}$

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 at

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

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

$x = 1$

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

The curve and the line intersect at

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

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

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

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

$x = - 1, 0.5$

We only need the intersection point that’s involved in region $R$ , or $x = 0.5$ .

Using vertical slices and integrating with respect to $x$ , region $R$ has area

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

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

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

$\approx 0.443$

b) Horizontal slices

(spoiler)

Rewriting the two functions in terms of $y$ ,

Line $y = 2 x$ becomes $x = \frac{1}{2} y$ and is the left function.
Rational function $y = \frac{1}{x} - 1$ is the right function and becomes

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

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

$x (y + 1) = 1$

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

The $x$ -axis $y = 0$ becomes the lower bound of integration, while the intersection point between the line and the rational function is the upper bound.

Previously, we found the two curves to intersect at $x = 0.5$ for region $R$ . This corresponds to $y = 1$ . So the integral setup for 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 from being the top function to the bottom one or from left to right.

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 to find the area of:

Over the interval $[0, π], sin (x)$ and $cos (x)$ meet at $x = \frac{π}{4}$ . Before that intersection point, $cos (x)$ is the top function and after that, it becomes the bottom function. So the integral set-up is

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

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

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

$= 22$

On a calculator-allowed problem, you can also enter the split integral as a single one involving an absolute value function:

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

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 the definite integral can be used to represent area under a curve. When finding the area between curves, imagine slicing the region 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 interval $[a, b]$ , then

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

Just remember that $f (x)$ is the top function, or

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

Examples

Often times you’ll have to first find the intersection points to determine the interval.

1. 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:

$x^{2} = x + 2$

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

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

$x = - 1, 2$

Between $x = - 1$ and $x = 2$ , the curve $y = x + 2$ is always above/greater than the curve $y = x^{2}$ , so the area can be found by

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

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 (no net signed area) so if you arrive at a negative answer, re-check the integral set-up.

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)

Make sure to choose the correct region described in the problem:

The region must be split into 2 integrals

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

Then 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 easier to find the area by slicing the region into horizontal rectangles and integrating with respect to $y$ . For example, take a look at the region bounded by the two curves

$x = y^{2}$

$y = - x + 2$

Notice that from $x = 0$ to $x = 1$ , the region is bound by only the parabola $x = y^{2}$ . In terms of $x$ , the top function is $y = x$ and the bottom one is $y = - x$ .

On the other hand, from $x = 1$ to $x = 4$ , the top function is the line $y = - x + 2$ while the bottom function is the lower half of $x = y^{2}$ , or $y = - x$ .

We could set up the integral as such:

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

But this quickly becomes tedious for more complicated functions. Instead, if we choose to slice the region into horizontal rectangles, then the area can be found by integrating with respect to $y$ :

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

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

Just remember that $f (y)$ is the rightmost function, or

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

Examples

Let’s redo the problem presented above.

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

Solution

Rewrite $y = - x + 2$ to be a function of $y$ :

$x = 2 - y$

As shown on the graph, the line $x = 2 - y$ is the right function while the parabola $x = y^{2}$ is the left function. Finding the intersection points algebraically,

$2 - y = y^{2}$

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

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

$y = 1, - 2$

So the interval to integrate over is from $y = - 2$ to $y = 1$ . The area is

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

$= \frac{9}{2}$

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 at

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

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

$x = 1$

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

The curve and the line intersect at

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

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

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

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

$x = - 1, 0.5$

We only need the intersection point that’s involved in region $R$ , or $x = 0.5$ .

Using vertical slices and integrating with respect to $x$ , region $R$ has area

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

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

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

$\approx 0.443$

b) Horizontal slices

(spoiler)

Rewriting the two functions in terms of $y$ ,

Line $y = 2 x$ becomes $x = \frac{1}{2} y$ and is the left function.
Rational function $y = \frac{1}{x} - 1$ is the right function and becomes

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

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

$x (y + 1) = 1$

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

The $x$ -axis $y = 0$ becomes the lower bound of integration, while the intersection point between the line and the rational function is the upper bound.

Previously, we found the two curves to intersect at $x = 0.5$ for region $R$ . This corresponds to $y = 1$ . So the integral setup for 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 from being the top function to the bottom one or from left to right.

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 to find the area of:

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

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

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

$= 22$

On a calculator-allowed problem, you can also enter the split integral as a single one involving an absolute value function:

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