Finding the Area Between Curves That Intersect at More Than Two Points
Введено| English | Значение |
|---|---|
| subintervals/ˌsʌbɪnˈtɜːvlz/ | подынтервалы |
When the curves keep crossing
- Sometimes two curves cross more than twice, and the "top" curve swaps along the way.
- A single top-minus-bottom integral would then get the sign wrong on part of the region.
- The fix: split the region at each crossing and integrate each piece separately.
- Add the pieces' (positive) areas to get the total.
Find every intersection first
- Solve $f(x)=g(x)$ completely — find all the crossing points on the interval, not just the outer two.
- These crossings are the boundaries of the subintervals 子区间 where the ordering is fixed.
- Order them left to right along the $x$-axis.
- Each adjacent pair of crossings frames one piece of the region.
When curves cross more than twice, split at each crossing so every piece is top minus bottom
Кубическая пересекает прямую 3 раза
y = ax³ + bx
$y=x^3$ и $y=x$ cross at $-1,0,1$; the top curve switches at $0$, so the area needs two pieces.
Where do $y=x^3$ и $y=x$ cross?
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
$x^3=x\Rightarrow x(x-1)(x+1)=0$.
The intersection points divide the region into ____ where the ordering is fixed.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
On each subinterval one curve is consistently on top.
Split where top and bottom switch
- On each subinterval, one curve is consistently on top — test a point to see which.
- Set up a separate integral of (upper $-$ lower) for each subinterval, using the correct top curve there.
- Because the top curve switches between pieces, the integrands differ from piece to piece.
- The full area is the sum of these separate integrals.
You split the region at each intersection because...
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Where curves swap, part of the region would count negative.
Using one integral of $f-g$ across all crossings (curves swapping) gives...
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Sign-cancellation understates the true area.
Add the absolute areas
- Each subinterval integral, done with the correct top-minus-bottom, is positive.
- Sum them: $A=\displaystyle\sum \int_{\text{piece}} (\text{upper}-\text{lower})\,dx$.
- Equivalently, $A=\displaystyle\int_a^b |f(x)-g(x)|\,dx$ — the absolute gap, integrated.
- Both viewpoints give the same total geometric area.
За $y=x^3$ и $y=x$ on $[-1,1]$, each piece has area $\tfrac14$. Find the total area.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
$\tfrac14+\tfrac14=\tfrac12$.
Integrating $|f(x)-g(x)|$ over the whole interval gives the same total area as splitting.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The absolute value handles the switching automatically.
Do not use one integral of $f-g$ across all crossings — where the curves swap, part of the region contributes a negative value and cancels real area. You must split at every intersection and take upper-minus-lower correctly on each piece (or integrate $|f-g|$). Find all the crossings, not just the endpoints.
Area between $y=x^3$ and $y=x$ on $[-1,1]$.
- Crossings: $x^3=x\Rightarrow x(x^2-1)=0\Rightarrow x=-1,0,1$ — three points.
- On $(-1,0)$: $x^3>x$ (test $-0.5$). On $(0,1)$: $x>x^3$ (test $0.5$). The top curve switches at $0$.
- $A=\displaystyle\int_{-1}^{0}(x^3-x)\,dx+\int_{0}^{1}(x-x^3)\,dx=\tfrac14+\tfrac14=\tfrac12$.
When curves intersect more than twice, find all intersection points, split the region into subintervals, and integrate upper minus lower on each (the top curve switches between them). Sum the pieces — equivalently, integrate $|f(x)-g(x)|$. One straight integral of $f-g$ would wrongly cancel area.
Write from an absolute gap
- Two curves cross at three points.
- Between the outer crossings, the upper curve switches.
How do you get the total area?
Each piece has its own top curve.
Curves trade which one is on top. The total area is found how?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
A single signed integral cancels the swapped pieces.