Why Crossings Require a Split
When one curve stays above another throughout an interval, the area between them is found by integrating upper curve minus lower curve across that interval. If the curves cross, however, their vertical order switches at the crossing. An integrand that keeps the same subtraction order across the whole interval then becomes negative on one side, so its integral records signed net change rather than the total geometric area.
The remedy is to divide the interval at every crossing where the curves exchange which one is higher. On each resulting subinterval, choose the upper curve and subtract the lower curve. Each integral then gives a nonnegative area, and the total area is the sum of those pieces.
Set Up the Partition
Suppose the graphs of \(y=f(x)\) and \(y=g(x)\) are considered on \([a,b]\). Their vertical order is determined by the difference \(f(x)-g(x)\): it is positive where \(f\) is above \(g\), and negative where \(g\) is above \(f\). First locate the intersections in the interval, then determine which ones actually change that order.
Write the order-changing crossing locations in increasing order as \(c_1,c_2,\dots,c_n\). Together with the endpoints, they divide the interval into consecutive pieces. On each piece, the sign of \(f(x)-g(x)\) tells which function is uppermost. A test point strictly inside each piece can confirm the sign. The area is the sum of the integrals of upper minus lower over those pieces.
The crossing points themselves have zero width, so assigning a boundary point to either adjacent piece does not change an integral. What matters is choosing the correct order on the interior of each piece. This method concerns crossings that switch which curve is above; deciding how to handle curves that meet without switching order is a separate issue.
Worked Example: Two Crossings and Three Area Pieces
Worked Example: Two Crossings and Three Area Pieces
Find the area between \(y=f(x)=x^2-x\) and \(y=g(x)=0\) on \([-1,2]\).
First find the intersections by setting the functions equal:
The crossings divide \([-1,2]\) into \([-1,0]\), \([0,1]\), and \([1,2]\). Check the sign of \(f(x)-g(x)=x^2-x\) at one point in each open piece. At \(x=-\tfrac12\), the difference is \(\tfrac14+\tfrac12=\tfrac34>0\), so \(f\) is above \(g\) on the first piece. At \(x=\tfrac12\), it is \(\tfrac14-\tfrac12=-\tfrac14<0\), so \(g\) is above \(f\) on the middle piece. At \(x=\tfrac32\), it is \(\tfrac94-\tfrac32=\tfrac34>0\), so \(f\) is above \(g\) on the last piece.
An antiderivative of \(x^2-x\) is \(F(x)=\tfrac{x^3}{3}-\tfrac{x^2}{2}\). Thus the three areas are:
For the last piece:
Add the nonnegative pieces:
Check: Differentiating \(F(x)=\tfrac{x^3}{3}-\tfrac{x^2}{2}\) gives \(F'(x)=x^2-x\), the required difference on the first and last pieces. On the middle piece, the integrand is its negative, \(x-x^2\), and the displayed antiderivative differentiates to \(x-x^2\). Each integrand matches the sign test and is nonnegative over its piece, as an area calculation requires.
Why Adding the Pieces Gives the Area
On any one of the subintervals, suppose \(f\) is above \(g\). A thin vertical slice has height \(f(x)-g(x)\) and width approximately \(\Delta x\), so its approximate area is \((f(x)-g(x))\Delta x\). Adding these slice areas and taking the limit gives the integral of \(f-g\) on that subinterval. If \(g\) is above \(f\), the slice height is instead \(g(x)-f(x)\), and that is the integrand to use.
The partition matters because the identity of the upper curve is fixed between consecutive crossings that switch the order. Each integral therefore accumulates the actual slice heights on one region of consistent ordering. Adjacent pieces meet only at their shared boundary, which contributes no area of its own. Adding their areas counts the full region without subtracting one side from another.
Worked Example: A Crossing That Changes the Sign of the Integral
Worked Example: A Crossing That Changes the Sign of the Integral
Find the area between \(y=x^2-1\) and \(y=0\) on \([-2,2]\).
Set the functions equal to find the crossings:
Test \(x=-\tfrac32\), \(x=0\), and \(x=\tfrac32\) in the three resulting pieces. The difference \(x^2-1\) is \(\tfrac54>0\), \(-1<0\), and \(\tfrac54>0\), respectively. Therefore, the parabola is above the \(x\)-axis on the two outer pieces, and the \(x\)-axis is above it on the middle piece.
An antiderivative of \(x^2-1\) is \(F(x)=\tfrac{x^3}{3}-x\). The left area is:
On the middle piece, the upper-minus-lower integrand is \(0-(x^2-1)=1-x^2\):
The right area is:
Hence the total area is:
Check: \(F'(x)=x^2-1\), while the middle integrand \(1-x^2\) has antiderivative \(x-\tfrac{x^3}{3}\). The sign tests agree with the selected upper curves, and all three computed areas are positive.
Worked Example: Two Crossings Inside the Interval
Worked Example: Two Crossings Inside the Interval
Find the area between \(y=f(x)=x^2\) and \(y=g(x)=4x-3\) on \([0,4]\).
The intersections satisfy:
Thus the crossings are \(x=1\) and \(x=3\). The difference is \(f(x)-g(x)=x^2-4x+3\). At \(x=\tfrac12\), it is \(\tfrac14-2+3=\tfrac54>0\), so \(f\) is above \(g\) on \([0,1]\). At \(x=2\), it is \(4-8+3=-1<0\), so \(g\) is above \(f\) on \([1,3]\). At \(x=\tfrac72\), it is \(\tfrac{49}{4}-14+3=\tfrac54>0\), so \(f\) is above \(g\) on \([3,4]\).
An antiderivative of \(x^2-4x+3\) is \(F(x)=\tfrac{x^3}{3}-2x^2+3x\). The three pieces are:
Adding gives \(A=\tfrac43+\tfrac43+\tfrac43=4\). Notice that integrating \(f-g\) just once from \(0\) to \(4\) would give \(F(4)-F(0)=\tfrac43\), not \(4\): the negative contribution between the crossings cancels part of the positive contributions. The split calculation reverses the subtraction on that middle piece.
Check: Differentiating \(F\) gives \(F'(x)=x^2-4x+3=f(x)-g(x)\). The middle integrand is \(g-f\), so its integral equals \(F(1)-F(3)\), as used. All three values are positive and their sum is the total area.
A Common Mistake: Using One Subtraction Everywhere
A frequent error is to calculate \(\int_a^b(f(x)-g(x))\,dx\) over the whole interval and call the result the area, even though the curves cross. That integral is a signed accumulation: portions where \(f\) is below \(g\) contribute negatively. In the previous example, the one-piece calculation is \(\tfrac43\), while the actual area is \(4\).
Another way this mistake appears is to keep the same integrand but change the limits carelessly. Reversing the limits of an integral changes its sign; it does not turn a signed integral into geometric area. The integrand on each piece must represent a nonnegative vertical height.
A Reliable Procedure
Set the two functions equal and solve for the \(x\)-values belonging to the interval of interest.
List the relevant crossing locations from left to right and use them, with the endpoints, as the bounds of consecutive subintervals.
Compare the functions at a test point in each open subinterval. Use the result to select upper minus lower.
Compute each definite integral, check that its value is nonnegative, and sum the piece areas.
This approach is useful whenever an area problem has more than one crossing that changes the curves’ order. The essential bookkeeping is local: each subinterval has its own upper curve and its own correct subtraction order. A final sign and arithmetic check can reveal a missed switch or an incorrectly chosen integrand.
Check Your Understanding
For each problem, identify the crossing partition and decide which function is upper on every piece before evaluating any integrals.
- For \(y=x^2-4\) and \(y=0\) on \([-3,3]\), where should the interval be split?
- Why can integrating \(f-g\) over an interval containing a crossing fail to give the total geometric area?
- For \(y=x^2-x\) and \(y=0\) on \([-1,2]\), which function is upper on \([0,1]\)?
- What should you verify about each piece’s integral before adding the pieces as areas?
- If a crossing is at an endpoint of the interval, does it create another interior subinterval boundary?
