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
- Find the intersections Set the two function values equal to locate every crossing \(\displaystyle 6x=3x^2\Rightarrow x=0,2\)
- Split at every crossing Use a separate interval whenever the upper curve changes \(\displaystyle [-1,3]=[-1,0]\cup[0,2]\cup[2,3]\)
- Compare curve heights Use a test input in each piece to identify the upper curve \(\displaystyle f(1)=6>g(1)=3\)
- 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
- \(\displaystyle 6x=3x^2\Rightarrow 3x(2-x)=0\Rightarrow x=0,2\)Set the curves equal to find both intersection x-values
- \(\displaystyle g\left(-\frac{1}{2}\right)=\frac{3}{4}>f\left(-\frac{1}{2}\right)=-3,\quad f(1)=6>g(1)=3,\quad g\left(\frac{5}{2}\right)=\frac{75}{4}>f\left(\frac{5}{2}\right)=15\)Testing one input in each piece identifies the upper curve on [-1,0], [0,2] and [2,3]
- \(\displaystyle A=\int_{-1}^{0}\left(3x^2-6x\right)\,dx+\int_0^2\left(6x-3x^2\right)\,dx+\int_2^3\left(3x^2-6x\right)\,dx=4+4+4\)Integrate upper minus lower on each piece
- \(\displaystyle A=12\)Add the three positive area contributions
Common mistakes
Difficulty levels
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 a polynomial fit have a lower degree than the expression initially allowed?+
Yes. If the task permits degree at most three, a zero cubic coefficient is allowed. A constant fit can also satisfy particular common-height and common-slope conditions
Why keep both factors in the contact equation?+
Dividing by a factor can remove valid contacts, including a constant fitted curve. Solve each branch, enforce the original domain and check both original equations
Can I use a calculator?+
A calculator is allowed unless a question is marked no calculator
Which year level and chapter is this?+
Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 8: Integration