Area Between Two Curves (College Board AP® Calculus AB)

Study Guide

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Written by: Jamie Wood

Reviewed by: Dan Finlay

Area between two curves in terms of x

How do I find the area between two curves?

Consider the diagram below where
- The area between two curves, $y = f (x)$ and $y = g (x)$ is being found
  - bounded by the lines $x = a$ and $x = b$
- This region is labeled $R$

Graph showing two curves, y = f(x) and y = g(x), with a shaded region R between x = a and x = b. Integral expression for the area R is displayed.

The region $R$ is the difference between the two areas found by:
- Integrating $y = f (x)$ between $x = a$ and $x = b$
- Integrating $y = g (x)$ between $x = a$ and $x = b$
As $f (x) \geq g (x)$ for all of this interval, this area can be calculated as
- $\int_{a}^{b} f (x) d x - \int_{a}^{b} g (x) d x$
This is equivalent to
- $\int_{a}^{b} (f (x) - g (x)) d x$
It is essential to have the function which is "above" the other be first inside the integral
- If the curves overlap and form multiple regions, see the method outlined in the 'Multiple Areas' study guide
Also note that if the graph of $y = f (x)$ is above the graph of $y = g (x)$ on $[a, b]$
- then $f (x) - g (x) \geq 0$ everywhere on that interval
- This means that you don't need to worry about negative integrals when integrating $\int_{a}^{b} (f (x) - g (x)) d x$
  - The integral will give the correct area value whether the area is above or below the $x$ -axis

What if I am not told the limits?

If you are not told the limits, it is likely you are finding the area enclosed by two curves, $f (x)$ and $g (x)$ , which intersect each other
Find the $x$ -values of the points of intersection of the two curves by solving $f (x) = g (x)$
- These will be the limits for the integral

Worked Example

Find the area of the region enclosed by the two curves with equations $y = 2 x^{2} - 4 x + 2$ and $y = - 2 x^{2} + 8 x - 6$ .

The curves are shown on the graph below.

Two intersecting quadratic curves, one a negative quadratic, the other a positive quadratic. They intersect in two places.

Answer:

Find the points where the two curves intersect, by setting their equations equal to one another and solving

$\begin{array}{rcl} 2 x^{2} - 4 x + 2 & = & - 2 x^{2} + 8 x - 6 \\ 4 x^{2} - 12 x + 8 & = & 0 \\ x^{2} - 3 x + 2 & = & 0 \\ (x - 2) (x - 1) & = & 0 \end{array}$

$x = 2$ or $x = 1$

Between the intersections, the n-shaped graph (the negative quadratic) is above the u-shaped graph (the positive quadratic) so the area integral will be of the following form

$\int_{1}^{2} ((- 2 x^{2} + 8 x - 6) - (2 x^{2} - 4 x + 2)) d x$

Simplify and then find the value of the definite integral

$\begin{array}{rcl} \int_{1}^{2} - 4 x^{2} + 12 x - 8 d x & = & {[- \frac{4}{3} x^{3} + 6 x^{2} - 8 x]}_{1}^{2} \\ = & (- \frac{4}{3} {(2)}^{3} + 6 {(2)}^{2} - 8 (2)) - (- \frac{4}{3} {(1)}^{3} + 6 {(1)}^{2} - 8 (1)) \\ = & (- \frac{8}{3}) - (- \frac{10}{3}) \\ = & \frac{2}{3} \end{array}$

$\frac{2}{3}$ units squared

Area between two curves in terms of y

How do I find the area between two curves when the functions are in terms of y?

The same concepts apply as when the functions are in terms of $x$
- but the process is followed relative to the $y$ -axis instead of the $x$ -axis
Consider the diagram below where
- The area between two curves, $x = f (y)$ and $x = g (y)$ is being found
  - bounded by the lines $y = a$ and $y = b$
- This region is labeled $R$

Graph showing a shaded region R between curves x=f(y) and x=g(y) from y=a to y=b, with an integral formula for R: ∫[a to b] (f(y) - g(y)) dy.

The region $R$ is the difference between the two areas found by:
- Integrating $x = f (y)$ between $y = a$ and $y = b$
- Integrating $x = g (y)$ between $y = a$ and $y = b$
As $f (y) \geq g (y)$ for all of this interval, this area can be calculated as
- $\int_{a}^{b} f (y) d y - \int_{a}^{b} g (y) d y$
- I.e. $g (y)$ is closer to the $y$ -axis than $f (y)$ is on this interval
This is equivalent to
- $\int_{a}^{b} (f (y) - g (y)) d y$
It is essential to have the function which is further away from the $y$ -axis, be first inside the integral
- If the curves overlap and form multiple regions, see the method outlined in the 'Multiple Areas' study guide
In these scenarios we are integrating an equation for $x$ in terms of $y$
- If you are given an equation for $y$ in terms of $x$
  - you need to rearrange the equation for $x$ in terms of $y$

What if I am not told the limits?

If you are not told the limits, it is likely you are finding the area enclosed by the two curves, $f (y)$ and $g (y)$ , which intersect each other
Find the $y$ -values of the points of intersection of the two curves by solving $f (y) = g (y)$
- These will be the limits for the integral

Worked Example

The graph below shows two curves with the following equations

$y = \frac{1}{8} e^{x}$ and $y = 2 e^{- x}$

Graph of the functions y = 2e^(-x) and y = (1/8)e^(x) showing the area R bounded by the curves between x = 0 and x = 2, labeled and shaded in gray.

The region $R$ is bounded by the line $y = 2$ and the two curves. Find the area of region $R$ .

Answer:

Note that this area could be found by using areas between the curves and the $x$ -axis, and subtracting from the rectangular area underneath the line forming the upper boundary of region $R$

I.e.

$\begin{array}{rcl} Area of region R & = & (2 \times \ln 16) - \int_{0}^{\ln 4} 2 e^{- x} d x - \int_{\ln 4}^{\ln 16} \frac{1}{8} e^{x} d x \\ = & 2 \ln 16 - 3 \\ = & 2.54517744 . . . \end{array}$

But here we'll work it out using areas between the curves and the $y$ -axis, and integrating in terms of $y$

Start by working out the integral limits

The upper limit is $y = 2$ and the lower limit will be the $y$ value of the point of intersection of the two curves

As we are working in terms of $y$ , rewrite each equation as $x$ in terms of $y$

For $y = \frac{1}{8} e^{x}$

$\begin{array}{rcl} y & = & \frac{1}{8} e^{x} \\ 8 y & = & e^{x} \\ \ln (8 y) & = & x \end{array}$

For $y = 2 e^{- x}$

$\begin{array}{rcl} y & = & 2 e^{- x} \\ \frac{y}{2} & = & e^{- x} \\ \frac{y}{2} & = & \frac{1}{e^{x}} \\ \frac{2}{y} & = & e^{x} \\ \ln (\frac{2}{y}) & = & x \end{array}$

Find the $y$ -value of the point of intersection by setting these equations equal to each other; this will be the lower limit for the integral

You could also use your calculator to find this

$\begin{array}{rcl} \ln (8 y) & = & \ln (\frac{2}{y}) \\ 8 y & = & \frac{2}{y} \\ 8 y^{2} & = & 2 \\ y^{2} & = & \frac{1}{4} \\ y & = & \pm \frac{1}{2} \end{array}$

$y$ must be positive as neither graph has any negative $y$ values

$y = \frac{1}{2}$

Use an integral of the form $\int_{a}^{b} (f (y) - g (y)) d y$

The curve with equation $y = \frac{1}{8} e^{x}$ (or $\ln (8 y) = x$ ) is furthest away from the $y$ -axis, so will come first in the integral

$\int_{\frac{1}{2}}^{2} (\ln (8 y)) - (\ln (\frac{2}{y})) d y$

You could use your calculator at this point to evaluate the integral, or you can simplify first using laws of logarithms

$\begin{array}{rcl} \ln (8 y) - \ln (\frac{2}{y}) & = & \ln (\frac{8 y}{(\frac{2}{y})}) \\ = & \ln (4 y^{2}) \\ = & \ln {(2 y)}^{2} \\ = & 2 \ln (2 y) \end{array}$

Use your calculator to evaluate the integral

$\int_{\frac{1}{2}}^{2} 2 \ln (2 y) d y = 2.54517744 . . .$

Round to 3 decimal places

2.545 units squared

Last updated: 3 September 2024

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Area Between Two Curves (College Board AP® Calculus AB)

Study Guide

Area between two curves in terms of x

How do I find the area between two curves?

What if I am not told the limits?

Worked Example

Area between two curves in terms of y

How do I find the area between two curves when the functions are in terms of y?

What if I am not told the limits?

Worked Example

You've read 0 of your 5 free study guides this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Jamie Wood

Author: Dan Finlay