Two Different Quantities
A definite integral can be negative, zero, or positive. Geometric area, by contrast, is never negative. This distinction matters when an integrand represents the vertical difference between two curves: its sign tells us which curve is higher, while its integral records signed contributions that may cancel.
Suppose two curves are given by \(y=f(x)\) and \(y=g(x)\) over an interval. At a particular input \(x\), the difference \(f(x)-g(x)\) compares their heights. If this difference is positive, \(f\) is above \(g\) there; if it is negative, \(f\) is below \(g\). The sign of the difference therefore gives a local description of the region between the curves.
The Sign of a Difference Tells Which Curve Is Higher
For curves \(y=f(x)\) and \(y=g(x)\), define their vertical difference at \(x\) by \(f(x)-g(x)\). The sign gives the ordering at that input:
| Sign of \(f(x)-g(x)\) | Curve ordering at \(x\) |
|---|---|
| \(f(x)-g(x)>0\) | \(f(x)>g(x)\): \(f\) is above \(g\). |
| \(f(x)-g(x)=0\) | The curves have the same height. |
| \(f(x)-g(x)<0\) | \(f(x)<g(x)\): \(f\) is below \(g\). |
Thus \(\int_a^b (f(x)-g(x))\,dx\) is a signed total of the differences in height. It is not automatically the geometric area between the curves. If \(f\) stays above \(g\), those signed contributions are nonnegative. If \(f\) stays below \(g\), they are nonpositive. If the curves switch order, contributions with opposite signs can cancel.
The integral measures net signed contribution, not the total amount of separation regardless of direction. Geometric area counts the region on both sides of a crossing as positive area. That difference is the central idea of this lesson.
Why the Integral Records the Sign
The definite integral can be understood as the limit of sums of contributions over small intervals. Divide \([a,b]\) into pieces of width \(\Delta x_i\), and choose a sample input \(x_i^*\) in each piece. For \(h(x)=f(x)-g(x)\), one contribution to the sum is approximately \(h(x_i^*)\Delta x_i\).
Each width \(\Delta x_i\) is positive. Therefore, the sign of that contribution is the sign of \(h(x_i^*)\). A sample where \(f\) is above \(g\) contributes positively; a sample where \(f\) is below \(g\) contributes negatively. For continuous functions, the limit of these sums is the definite integral.
This also explains the fixed-ordering case. If \(f(x)-g(x)\geq 0\) everywhere on the interval, every term in each sum is nonnegative, so the limit cannot be negative. If \(f(x)-g(x)\leq 0\) everywhere, every term is nonpositive, so the limit cannot be positive. If the difference takes both positive and negative values, the sum includes contributions with opposite signs; their net may be smaller than either total magnitude.
Worked Example: One Curve Stays Above the Other
Worked Example: One Curve Stays Above the Other
Compare the signed integral of \(f(x)-g(x)\) on \([0,2]\) with the geometric area between \(f(x)=x+2\) and \(g(x)=1\).
Step 1: Determine the ordering. Subtract the functions:
For every \(x\) from \(0\) to \(2\), \(x+1>0\). Therefore \(f\) is above \(g\) throughout the interval, and the signed contributions are all positive.
Step 2: Compute the signed integral.
Step 3: Compare with geometric area. The two boundary lines are \(y=x+2\) and \(y=1\). The region between them is a trapezoid: its parallel vertical sides have lengths \(1\) and \(3\), and the horizontal distance between them is \(2\). Its geometric area is
The geometric area and signed integral are both \(4\). This agreement occurs because \(f\) stays above \(g\), so none of the signed contributions are negative.
Check: An antiderivative of \(x+1\) is \(\frac{x^2}{2}+x\), since \(\frac{d}{dx}\left(\frac{x^2}{2}+x\right)=x+1\). Evaluating it at the endpoints gives \(4-0=4\), as calculated.
Worked Example: Reversing the Curves Reverses the Sign
Worked Example: Reversing the Curves Reverses the Sign
Now take \(f(x)=1\) and \(g(x)=x+2\) on \([0,2]\). Find the signed integral of \(f(x)-g(x)\) and compare it with the geometric area between the curves.
Step 1: Determine the ordering.
For every \(x\in[0,2]\), \(-x-1<0\). Thus \(f\) is below \(g\) everywhere on the interval.
Step 2: Compute the signed integral.
Step 3: Compare with geometric area. These are the same two lines as in the previous example, so they enclose the same trapezoid, with parallel sides of lengths \(1\) and \(3\) and horizontal distance \(2\). Its geometric area is \(\frac{1+3}{2}\cdot2=4\), not \(-4\).
The negative sign does not mean that the region has negative area. It records that the first function, \(f\), lies below the second, \(g\). The geometric area is \(4\) square units.
Check: Differentiating the antiderivative gives \(\frac{d}{dx}\left(-\frac{x^2}{2}-x\right)=-x-1\). Its endpoint difference is \(-4-0=-4\), confirming the signed integral.
Worked Example: Contributions Cancel at a Crossing
Worked Example: Contributions Cancel at a Crossing
Let \(f(x)=x\) and \(g(x)=1\) on \([0,2]\). Find the signed integral of \(f(x)-g(x)\), then determine the geometric area between the curves.
Step 1: Locate where the ordering changes. The difference is
This difference is negative for \(0\leq x<1\), zero at \(x=1\), and positive for \(1<x\leq2\). Thus \(f\) is below \(g\) to the left of \(1\) and above \(g\) to the right.
Step 2: Compute the signed integral over the full interval.
Step 3: Find the geometric area on each side of the crossing. From \(x=0\) to \(x=1\), the region is a triangle with base \(1\) and height \(1\), so its area is \(\frac12(1)(1)=\frac12\). From \(x=1\) to \(x=2\), the region is another triangle with base \(1\) and height \(1\), so its area is also \(\frac12\). The total geometric area is
The signed integral is \(0\) because the negative contribution on the first part cancels the positive contribution on the second part. The geometric area is \(1\) square unit because both triangles count as positive area.
Check: The antiderivative \(\frac{x^2}{2}-x\) differentiates to \(x-1\). Its value at \(2\) is \(0\), and its value at \(0\) is \(0\), so the full signed integral is \(0-0=0\). The two nonnegative triangle areas add to \(1\), independently confirming that zero net contribution does not imply zero area.
A Common Mistake: Calling a Zero Integral Zero Area
A tempting but incorrect conclusion is that the region in the previous example has area zero because its signed integral is zero. The calculation may look persuasive:
The error is in the last inference. The integral is zero, but the integrand changes sign at \(x=1\). On \([0,1]\), the curve \(y=x\) lies below \(y=1\); on \([1,2]\), it lies above. The negative and positive signed contributions cancel, while the two triangular regions each have area \(\frac12\). The actual geometric area is \(1\), not \(0\).
What to Keep in Mind
When comparing two curves, keep three questions separate:
- Which function is higher at a given input? Compare \(f(x)\) and \(g(x)\), or inspect the sign of \(f(x)-g(x)\).
- What does the definite integral report? It gives a signed total: positive where \(f\) is higher, negative where \(f\) is lower.
- What does geometric area report? It is the nonnegative size of the region, regardless of which curve is higher.
If the curves do not change order, the signed integral has the sign determined by their ordering, and its magnitude agrees with the area between them. If the curves change order, the signed integral can include cancellation. Consequently, neither a negative nor a zero signed integral should be read as negative or zero geometric area.
Check Your Understanding
For each question, distinguish what the sign of the integral tells you from what geometric area tells you.
- If \(f(x)-g(x)>0\) throughout an interval, which curve is above the other, and what sign must \(\int (f(x)-g(x))\,dx\) have?
- Can the geometric area between two curves be negative? Explain briefly.
- If \(\int_a^b(f(x)-g(x))\,dx=0\), must the curves coincide on the interval? Why or why not?
- For \(f(x)=x\) and \(g(x)=1\) on \([0,2]\), where does the ordering change, and why does the signed integral fail to give the geometric area?
- What does a negative value of \(\int_a^b(f(x)-g(x))\,dx\) tell you if \(f(x)-g(x)\) is negative throughout \([a,b]\)?
