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Find the area between two curves

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Area between two curves

The area between two curves is found from vertical widths. Find all intersections, determine which curve is above on each piece, integrate upper minus lower, and add the positive contributions

\[A=\int_a^b\left(\text{upper curve}-\text{lower curve}\right)\,dx\]
  1. 1Find the intersections Set the two function values equal to locate every crossing \(\displaystyle 6x=3x^2\Rightarrow x=0,2\)
  2. 2Split at every crossing Use a separate interval whenever the upper curve changes \(\displaystyle [-1,3]=[-1,0]\cup[0,2]\cup[2,3]\)
  3. 3Compare curve heights Use a test input in each piece to identify the upper curve \(\displaystyle f(1)=6>g(1)=3\)
  4. 4Integrate positive widths Subtract the lower curve from the upper curve on each piece and add the areas \(\displaystyle A=\int_0^2\left(6x-3x^2\right)\,dx=4\)
Year 12, VCE Units 3 & 4, Chapter 5, 4 difficulty levels, No calculatorBuild a worksheet

Area between two curves

The area between two curves is found from vertical widths. Find all intersections, determine which curve is above on each piece, integrate upper minus lower, and add the positive contributions

\[A=\int_a^b\left(\text{upper curve}-\text{lower curve}\right)\,dx\]
  1. Find the intersections Set the two function values equal to locate every crossing \(\displaystyle 6x=3x^2\Rightarrow x=0,2\)
  2. Split at every crossing Use a separate interval whenever the upper curve changes \(\displaystyle [-1,3]=[-1,0]\cup[0,2]\cup[2,3]\)
  3. Compare curve heights Use a test input in each piece to identify the upper curve \(\displaystyle f(1)=6>g(1)=3\)
  4. Integrate positive widths Subtract the lower curve from the upper curve on each piece and add the areas \(\displaystyle A=\int_0^2\left(6x-3x^2\right)\,dx=4\)

Worked example

Find the area between the curves on the stated interval

\[f(x)=6x,\quad g(x)=3x^2,\quad -1\le x\le3\]
  1. \(\displaystyle 6x=3x^2\Rightarrow 3x(2-x)=0\Rightarrow x=0,2\)Set the curves equal to find both intersection x-values
  2. \(\displaystyle g\left(-\frac{1}{2}\right)=\frac{3}{4}>f\left(-\frac{1}{2}\right)=-3,\quad f(1)=6>g(1)=3,\quad g\left(\frac{5}{2}\right)=\frac{75}{4}>f\left(\frac{5}{2}\right)=15\)Test one input in each piece to identify the upper curve
  3. \(\displaystyle A=\int_{-1}^{0}\left(3x^2-6x\right)\,dx+\int_0^2\left(6x-3x^2\right)\,dx+\int_2^3\left(3x^2-6x\right)\,dx=4+4+4\)Integrate upper minus lower on each piece
  4. \(\displaystyle A=12\)Add the three positive area contributions

Common mistakes

Ignoring intersections
✕\(\displaystyle A=\int_{-1}^{3}\left(6x-3x^2\right)\,dx=-4\)✓\(\displaystyle A=\int_{-1}^{0}\left(3x^2-6x\right)\,dx+\int_0^2\left(6x-3x^2\right)\,dx+\int_2^3\left(3x^2-6x\right)\,dx=12\)
The interval must be split at both intersections so that the upper curve is used on every piece
Reversing upper and lower
✕\(\displaystyle \int_0^2\left(3x^2-6x\right)\,dx=-4\)✓\(\displaystyle \int_0^2\left(6x-3x^2\right)\,dx=4\)
Area uses upper curve minus lower curve, so the integrand must be non-negative on the piece
Renaming does not fix the region
✕Changing x to y removes the need to split the interval✓A dummy-variable rename preserves the integral; keep the bounds and split at each crossing
The unsplit signed integral is −4 in either variable, while geometric area is 12.

Difficulty levels

introFind the area between a polynomial and a line with one intersection
standardFind two intersections, choose the upper curve and split the interval
challengeUse exact circular or exponential curve pairs with two intersections
examChoose the area from values produced by reversed order, ignored intersections and confused axes

Questions about this topic

Why must the interval be split at an intersection?+

The upper curve can change when the curves cross. Splitting makes upper minus lower non-negative on each piece

How is the upper curve identified?+

Choose an input inside a piece, calculate both function values, and identify the larger value

Why is the area positive?+

Geometric area adds the magnitudes of the vertical widths, so the lower curve is subtracted from the upper curve

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 12, VCE Units 3 & 4, Specialist Mathematics, chapter 5: Integral calculus

In this chapter

Integrate inverse circular kernels and change variablesAnti-differentiate reciprocal and linear-reciprocal functionsIntegration by recognition with logarithmic componentsAnti-differentiate by recognition from a supplied derivativeIntegrate using partial fractions, identities and partsDecompose rational expressions into partial fractionsCompound, double-angle and reciprocal identitiesAnti-differentiate exponential and circular functionsFind exact areas and assess numerical integration resultsFind the area between two curvesApproximate area by an endpoint or trapezium ruleFind areas involving reciprocal curvesAudit numerical integration loops and error enclosuresSet up volumes, parametric lengths and surfaces of revolution