Introduction
When two curves meet inside an interval, they can exchange which one lies above the other. The vertical difference between them then changes sign. A definite integral of that difference records this sign: contributions from one side of the crossing can cancel contributions from the other.
Geometric area, by contrast, is nonnegative. If the curves enclose a region on both sides of their intersection, the total area includes both parts; it does not subtract one from the other. This distinction explains why integrating the difference just once across the whole interval can give a value that is not the total area.
The Difference of the Curves
Let \(f\) and \(g\) be continuous functions on \([a,b]\), and define their difference by \(D(x)=f(x)-g(x)\). At a point where the curves intersect, their values agree, so \(D(x)=0\). An interior crossing at \(c\), where \(a<c<b\), means that the sign of \(D\) changes there: one curve is higher immediately to the left, and the other is higher immediately to the right.
For a single crossing at \(c\), suppose, for definiteness, that \(f\) is below \(g\) on the left and above \(g\) on the right. The sign conditions, including the intersection itself, are
For a genuine crossing, the inequalities are strict away from \(c\): \(D(x)<0\) on \([a,c)\) and \(D(x)>0\) on \((c,b]\). The weak inequalities on the closed intervals are useful because \(D(c)=0\); strict inequalities on those closed intervals would incorrectly exclude the crossing point.
Signed Integral and Geometric Area
The definite integral \(\int_a^b D(x)\,dx\) is signed. Where \(D\) is positive, it contributes positively; where \(D\) is negative, it contributes negatively. By contrast, area between curves is found from the vertical distance between them, which is always nonnegative. At a crossing, the curve on top changes, so the relevant vertical distance is \(g-f\) on the left in this example and \(f-g\) on the right.
More precisely, let \(A_L\) denote the area between the curves from \(a\) to \(c\), and \(A_R\) the area from \(c\) to \(b\). Under the stated sign conditions, \(A_L=\int_a^c(g-f)\,dx\) and \(A_R=\int_c^b(f-g)\,dx\). The signed integral and total area therefore satisfy
These equations apply when the curves are continuous and have the stated sign pattern. If the sign pattern is reversed, the signed integral is \(A_L-A_R\) instead. In either case, opposite-sign contributions cancel in the signed integral rather than adding as they do in geometric area.
Why the Signed Integral Has This Form
The reason is the additivity of definite integrals and the fact that a function’s sign determines the sign of its integral on an interval where that sign is fixed. Splitting at the crossing gives a direct justification.
For continuous \(D=f-g\), additivity gives \(\int_a^b D(x)\,dx=\int_a^c D(x)\,dx+\int_c^b D(x)\,dx\).
Since \(D\leq0\) on \([a,c]\), the area there is \(A_L=\int_a^c(-D(x))\,dx\). Thus \(\int_a^cD(x)\,dx=-A_L\).
Since \(D\geq0\) on \([c,b]\), the area there is \(A_R=\int_c^bD(x)\,dx\).
Adding the two integrals gives \(\int_a^bD(x)\,dx=-A_L+A_R\), whereas the geometric area is \(A_L+A_R\).
This argument uses the usual interpretation of the definite integral as signed accumulation and the area between vertically sliced curves. It does not require the two pieces to have equal areas. If they happen to be equal, the signed integral is zero even though the total geometric area is positive.
Worked Examples
Worked Example: Two Lines with Equal Areas on Either Side
Let \(f(x)=x\) and \(g(x)=2-x\) on \([0,2]\). Find their crossing, the signed integral of \(f-g\) over the full interval, and the total area between the curves.
At an intersection, \(x=2-x\), so \(2x=2\) and \(x=1\). The common height is \(f(1)=g(1)=1\). The difference is
For \(0\leq x<1\), \(2x-2<0\); for \(1<x\leq2\), \(2x-2>0\). Thus \(g\) is above \(f\) on the left, and \(f\) is above \(g\) on the right.
The signed integral over the full interval is
The antiderivative checks because \(\frac{d}{dx}(x^2-2x)=2x-2\). The zero results from cancellation, not from an absence of area. On the left, the vertical distance \(g-f=2-2x\) forms a triangle with base \(1\) and height \(2\), so its area is \(\frac12(1)(2)=1\). On the right, \(f-g=2x-2\) forms another triangle with base \(1\) and height \(2\), giving area \(\frac12(1)(2)=1\). The total geometric area is \(1+1=2\), while the signed integral is \(0\).
As an integral check, the left area is \(\int_0^1(2-2x)\,dx=[2x-x^2]_0^1=1\), and the right area is \(\int_1^2(2x-2)\,dx=[x^2-2x]_1^2=0-(-1)=1\). Both agree with the triangle calculations.
Worked Example: A Crossing with Unequal Areas
Let \(f(x)=x\) and \(g(x)=1\) on \([0,3]\). Find the crossing, the signed integral, and the total geometric area.
The curves meet where \(x=1\), so the crossing is at \(c=1\). Their difference is \(D(x)=x-1\), which is negative for \(0\leq x<1\) and positive for \(1<x\leq3\).
The signed integral is
The antiderivative checks because \(\frac{d}{dx}\left(\frac{x^2}{2}-x\right)=x-1\). The left area is the triangle between \(y=1\) and \(y=x\) from \(0\) to \(1\): its base and height are both \(1\), so \(A_L=\frac12\). Equivalently, \(\int_0^1(1-x)\,dx=[x-\frac{x^2}{2}]_0^1=\frac12\).
On the right, \(f\) is above \(g\). The area is
This is also a triangle with base \(2\) and height \(2\), giving area \(\frac12(2)(2)=2\). Therefore the total geometric area is \(\frac12+2=\frac52\). The signed result, \(\frac32\), agrees with \(A_R-A_L=2-\frac12=\frac32\), and is not the total area.
Worked Example: Polynomial Curves with One Crossing on the Interval
Let \(f(x)=x^2\) and \(g(x)=1\) on \([0,2]\). Find the signed integral and total area between the curves.
Solving \(x^2=1\) gives \(x=1\) or \(x=-1\); only \(x=1\) lies in \([0,2]\). The difference is \(D(x)=x^2-1\). It is negative for \(0\leq x<1\) and positive for \(1<x\leq2\), so the curves cross at \(x=1\).
Over the full interval,
The antiderivative checks since \(\frac{d}{dx}\left(\frac{x^3}{3}-x\right)=x^2-1\). On the left, \(g\) is above \(f\), so
On the right, \(f\) is above \(g\), and
Thus the total area is \(\frac23+\frac43=2\), while the signed integral is \(-\frac23+\frac43=\frac23\). The sign and magnitude agree with the predicted cancellation.
A Common Mistake
A common error is to integrate \(f-g\) across the entire interval and report the result as the total area, without checking whether the curves cross. For the lines in the first example, that incorrect calculation would be:
This cannot be the total area: the curves are distinct except at one point and bound two visible triangular regions. The calculation is a correct signed integral, but its interpretation as total area is wrong. The correction is to account for the sign change: the left region has area \(1\), the right region has area \(1\), and the total geometric area is \(1+1=2\). The signed integral is \(1-1=0\).
Why the Distinction Matters
A definite integral of a difference answers a signed question: which side contributes more after positive and negative contributions are combined? Geometric area answers a different question: how much region lies between the curves, without cancellation? When there is an interior crossing, these questions need not have the same answer.
Before interpreting an integral as area, check the sign of \(f-g\) throughout the interval. If it changes sign at an interior crossing, the single signed integral over the whole interval is not, in general, the total geometric area. The direction of the sign change also matters: it determines which portion contributes positively and which negatively. Equal portions can cancel completely, while unequal portions leave a signed remainder.
Check Your Understanding
Use the sign of the difference to decide what a full-interval signed integral tells you.
- If \(D(x)=f(x)-g(x)\) is negative to the left of an interior crossing and positive to the right, which curve is above on each side?
- Why can the signed integral over an interval containing a crossing be zero even when the geometric area is positive?
- For the curves \(f(x)=x\) and \(g(x)=1\) on \([0,3]\), what are the signed integral and total area?
- What is wrong with interpreting \(\int_a^b(f-g)\,dx\) as total geometric area without checking the sign of \(f-g\)?
- If the areas on the two sides of a crossing are unequal, what determines whether the signed integral is positive or negative?
