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Calculus BC · Unit 8 · 8.6
AP Calculus · Applications of Integration
Curves That Intersect at More Than Two Points
Every crossing starts a new region — and integrating straight through one cancels it away.
key idea$\text{Three intersections} \Rightarrow \text{two regions} \Rightarrow \text{two integrals, added}$
Picture the region
When two curves cross more than twice they enclose several separate regions, and the curve on top is not the same one throughout. Each region needs its own integral, with its own top and bottom, and the areas are then added.
$y=x$ and $y=x^{3}$ meet at $x=-1$, $0$ and $1$, enclosing two lobes. The solid line traces whichever curve is on top: the cubic wins on the left lobe, the line on the right. The rose marker at $x=0$ is where they trade places.Setting up the integral
Handling several crossings
- Solve $f(x)=g(x)$ completely. Factor rather than divide — dividing by a common factor destroys the root $x=0$, which is exactly the crossing that gets lost.
- List every solution in increasing order. These become the boundaries of the sub-intervals.
- Test one point inside each sub-interval to see which curve is on top there. Do not assume it alternates, even though it usually does.
- Write one integral per sub-interval, each with its own $\text{top}-\text{bottom}$.
- Add the results. Every piece is already a positive area, so nothing should be subtracted at the end.
which curve is on top
| $x\lt-1$ | $-1\lt x\lt0$ | $0\lt x\lt1$ | $x\gt1$ |
|---|
| $x-x^{3}$ | + | − | + | − |
| verdict | outside the region | $x^{3}$ on top | $x$ on top | outside the region |
| Symptom | What it means | What to do |
|---|
| three or more solutions of $f=g$ | more than one enclosed region | one integral per gap |
| the answer comes out $0$ | the pieces cancelled | you integrated across a crossing |
| a piece is negative | top and bottom swapped there | reverse that piece only |
| the region is symmetric | the lobes have equal area | compute one, double it |
| you divided by $x$ to solve | a root was destroyed | factor instead and recover it |
Row 2 is the trap this topic exists to teach. Integrating $f-g$ straight across a crossing lets a positive lobe cancel a negative one — and $0$ is a plausible-looking wrong answer.
Worked examples
EX 1
Two lobes
$y=x$ and $y=x^{3}$Find the total area enclosed by $y=x$ and $y=x^{3}$.
1
Solve completely
$x=x^{3}$ gives $x^{3}-x=0$, so $x\left(x^{2}-1\right)=0$ and $x=-1,\,0,\,1$.
2
Test each sub-interval
At $x=-0.5$: $x^{3}=-0.125\gt x=-0.5$, so the cubic is on top. At $x=0.5$: $x=0.5\gt x^{3}=0.125$, so the line is on top.
3
Left piece
$\displaystyle\int_{-1}^{0}\left(x^{3}-x\right)dx=\left[\dfrac{x^{4}}{4}-\dfrac{x^{2}}{2}\right]_{-1}^{0}=0-\left(-\dfrac14\right)=\dfrac14$.
4
Right piece
$\displaystyle\int_{0}^{1}\left(x-x^{3}\right)dx=\left[\dfrac{x^{2}}{2}-\dfrac{x^{4}}{4}\right]_{0}^{1}=\dfrac14$.
5
Add
Total area $=\dfrac14+\dfrac14=\dfrac12$. Integrating $x-x^{3}$ from $-1$ to $1$ in one go would have given $0$.
EX 2
Exploiting symmetry
compute one lobe, double itUse symmetry to shorten the previous calculation.
1
Test for odd symmetry
$f(x)=x-x^{3}$ satisfies $f(-x)=-f(x)$, so the graph has rotational symmetry about the origin.
2
Translate that into areas
The two lobes are congruent, so they have equal area.
3
Compute one
$\displaystyle\int_{0}^{1}\left(x-x^{3}\right)dx=\dfrac14$.
4
Double it
Total $=2\cdot\dfrac14=\dfrac12$. ✓
5
Note the safeguard
Doubling requires the lobes to be genuinely congruent. Check the symmetry algebraically before relying on it.
EX 3
Three crossings, unequal lobes
no shortcut availableSet up the area enclosed by $y=x^{3}-4x$ and $y=0$.
1
Find the crossings
$x^{3}-4x=x\left(x^{2}-4\right)=0$ gives $x=-2,\,0,\,2$.
2
Test the sub-intervals
At $x=-1$: $-1+4=3\gt0$, so the cubic is above the axis. At $x=1$: $1-4=-3\lt0$, so it is below.
3
Write both integrals
$\displaystyle\int_{-2}^{0}\left(x^{3}-4x\right)dx+\int_{0}^{2}\left(0-\left(x^{3}-4x\right)\right)dx$.
4
Evaluate each
$\left[\dfrac{x^{4}}{4}-2x^{2}\right]_{-2}^{0}=0-(4-8)=4$, and the second gives $4$ as well.
5
Add
Total $=8$. The single integral $\displaystyle\int_{-2}^{2}\left(x^{3}-4x\right)dx$ is $0$ — the classic cancellation.
EX 4
The root that gets divided away
factor, never divideA student solves $x^{2}=x^{4}$ by dividing both sides by $x^{2}$, getting $x=\pm1$. What was lost?
1
Spot the illegal step
Dividing by $x^{2}$ assumes $x\neq0$.
2
Solve properly
$x^{4}-x^{2}=0$ gives $x^{2}\left(x^{2}-1\right)=0$, so $x=-1,\,0,\,1$.
3
Identify the consequence
The crossing at $x=0$ was destroyed, so the two lobes were treated as one.
4
Predict the wrong answer
A single integral from $-1$ to $1$ would let the lobes cancel.
5
State the habit
Move everything to one side and factor. Dividing by a variable expression loses roots every time.
Watch out
Never integrate across a crossing
The area on either side has opposite sign in a single integral, so the pieces cancel. Split at every intersection.
Factor to solve, do not divide
Dividing by $x$ or $x^{2}$ silently discards the root at zero — usually the very crossing the problem was built around.
Test each sub-interval separately
Which curve is on top can stay the same across a crossing if the crossing is a tangency. Substitute a point rather than assuming it alternates.
Add the pieces, do not subtract
Each integral is already set up as $\text{top}-\text{bottom}$, so each is positive. The total is a sum.