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

Jamie Wood

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on

College Board Calculus ABCollege Board Calculus BC

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 equals f open parentheses x close parentheses and y equals g open parentheses x close parentheses is being found

      • bounded by the lines x equals a and x equals 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 equals f open parentheses x close parentheses between x equals a and x equals b

    • Integrating y equals g open parentheses x close parentheses between x equals a and x equals b

  • As f open parentheses x close parentheses greater or equal than g open parentheses x close parentheses for all of this interval, this area can be calculated as

    • integral subscript a superscript b f open parentheses x close parentheses space italic d x space minus space integral subscript a superscript b g open parentheses x close parentheses space italic d x

  • This is equivalent to

    • integral subscript a superscript b open parentheses f open parentheses x close parentheses minus g open parentheses x close parentheses close parentheses space italic 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 equals f open parentheses x close parentheses is above the graph of y equals g open parentheses x close parentheses on open square brackets a comma space b close square brackets

    • then f open parentheses x close parentheses minus g open parentheses x close parentheses greater or equal than 0 everywhere on that interval

    • This means that you don't need to worry about negative integrals when integrating integral subscript a superscript b open parentheses f open parentheses x close parentheses minus g open parentheses x close parentheses close parentheses space italic 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 open parentheses x close parentheses and g open parentheses x close parentheses, which intersect each other

  • Find the bold italic x-values of the points of intersection of the two curves by solving f open parentheses x close parentheses equals g open parentheses x close parentheses

    • These will be the limits for the integral

Worked Example

Find the area of the region enclosed by the two curves with equations y equals 2 x squared minus 4 x plus 2 and y equals negative 2 x squared plus 8 x minus 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

table row cell 2 x squared minus 4 x plus 2 end cell equals cell negative 2 x squared plus 8 x minus 6 end cell row cell 4 x squared minus 12 x plus 8 end cell equals 0 row cell x squared minus 3 x plus 2 end cell equals 0 row cell open parentheses x minus 2 close parentheses open parentheses x minus 1 close parentheses end cell equals 0 end table

x equals 2 or x equals 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

integral subscript 1 superscript 2 open parentheses open parentheses negative 2 x squared plus 8 x minus 6 close parentheses minus open parentheses 2 x squared minus 4 x plus 2 close parentheses close parentheses space italic d x

Simplify and then find the value of the definite integral

table row cell integral subscript 1 superscript 2 minus 4 x squared plus 12 x minus 8 space italic d x end cell equals cell open square brackets negative 4 over 3 x cubed plus 6 x squared minus 8 x close square brackets subscript 1 superscript 2 end cell row blank equals cell open parentheses negative 4 over 3 open parentheses 2 close parentheses cubed plus 6 open parentheses 2 close parentheses squared minus 8 open parentheses 2 close parentheses close parentheses minus open parentheses negative 4 over 3 open parentheses 1 close parentheses cubed plus 6 open parentheses 1 close parentheses squared minus 8 open parentheses 1 close parentheses close parentheses end cell row blank equals cell open parentheses negative 8 over 3 close parentheses minus open parentheses negative 10 over 3 close parentheses end cell row blank equals cell 2 over 3 end cell end table

2 over 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 equals f open parentheses y close parentheses and x equals g open parentheses y close parentheses is being found

      • bounded by the lines y equals a and y equals 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 equals f open parentheses y close parentheses between y equals a and y equals b

    • Integrating x equals g open parentheses y close parentheses between y equals a and y equals b

  • As f open parentheses y close parentheses greater or equal than g open parentheses y close parentheses for all of this interval, this area can be calculated as

    • integral subscript a superscript b f open parentheses y close parentheses space italic d y space minus space integral subscript a superscript b g open parentheses y close parentheses space italic d y

    • I.e. g open parentheses y close parentheses is closer to the y-axis than f open parentheses y close parentheses is on this interval

  • This is equivalent to

    • integral subscript a superscript b open parentheses f open parentheses y close parentheses minus g open parentheses y close parentheses close parentheses space italic 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 xin 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 open parentheses y close parentheses and g open parentheses y close parentheses, which intersect each other

  • Find the y-values of the points of intersection of the two curves by solving f open parentheses y close parentheses equals g open parentheses y close parentheses

    • These will be the limits for the integral

Worked Example

The graph below shows two curves with the following equations

y equals 1 over 8 e to the power of x and y equals 2 e to the power of negative x end exponent

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 equals 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.

table row cell Area space of space region space R end cell equals cell open parentheses 2 cross times ln 16 close parentheses minus integral subscript 0 superscript ln 4 end superscript 2 e to the power of negative x end exponent space d x space minus integral subscript ln 4 end subscript superscript ln 16 end superscript 1 over 8 e to the power of x space d x end cell row blank equals cell 2 ln 16 minus 3 end cell row blank equals cell 2.54517744... end cell end table

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 equals 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 equals 1 over 8 e to the power of x

table row y equals cell 1 over 8 e to the power of x end cell row cell 8 y end cell equals cell e to the power of x end cell row cell ln open parentheses 8 y close parentheses end cell equals x end table

For y equals 2 e to the power of negative x end exponent

table row y equals cell 2 e to the power of negative x end exponent end cell row cell y over 2 end cell equals cell e to the power of negative x end exponent end cell row cell y over 2 end cell equals cell 1 over e to the power of x end cell row cell 2 over y end cell equals cell e to the power of x end cell row cell ln open parentheses 2 over y close parentheses end cell equals x end table

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

table row cell ln open parentheses 8 y close parentheses end cell equals cell ln open parentheses 2 over y close parentheses end cell row cell 8 y end cell equals cell 2 over y end cell row cell 8 y squared end cell equals 2 row cell y squared end cell equals cell 1 fourth end cell row y equals cell plus-or-minus 1 half end cell end table

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

y equals 1 half

Use an integral of the form integral subscript a superscript b open parentheses f open parentheses y close parentheses minus g open parentheses y close parentheses close parentheses space italic d y

The curve with equation y equals 1 over 8 e to the power of x (or ln open parentheses 8 y close parentheses equals x) is furthest away from the y-axis, so will come first in the integral

integral subscript 1 half end subscript superscript 2 space open parentheses ln open parentheses 8 y close parentheses close parentheses minus open parentheses ln open parentheses 2 over y close parentheses close parentheses space italic d y

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

table row cell ln open parentheses 8 y close parentheses minus ln open parentheses 2 over y close parentheses end cell equals cell ln open parentheses fraction numerator 8 y over denominator open parentheses 2 over y close parentheses end fraction close parentheses end cell row blank equals cell ln open parentheses 4 y squared close parentheses end cell row blank equals cell ln open parentheses 2 y close parentheses squared end cell row blank equals cell 2 ln open parentheses 2 y close parentheses end cell end table

Use your calculator to evaluate the integral

integral subscript 1 half end subscript superscript 2 space 2 ln open parentheses 2 y close parentheses space italic d y equals 2.54517744...

Round to 3 decimal places

2.545 units squared

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    Author
    Reviewer
    Jamie Wood

    Author: Jamie Wood

    Expertise: Maths

    Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.

    Dan Finlay

    Author: Dan Finlay

    Expertise: Maths Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.