Reading the Sign of a Graph
A definite integral can summarize contributions from both sides of the horizontal axis. The sign of a graph’s height determines the sign of its contribution: a region above the axis contributes positively, while a region below it contributes negatively. If the graph crosses the axis, contributions from its different sides must be considered separately.
This gives a practical way to interpret a graph. Identify where the function is above, below, or on the horizontal axis; calculate the contribution from each region; then combine those signed values. A positive and a negative contribution can partially or completely cancel.
Signed Contributions on an Interval
Suppose \(f\) is bounded and Riemann integrable on an interval \([a,b]\), where \(a<b\). Its definite integral is the limit of Riemann sums as the widths of the subintervals shrink. Each term in such a sum is a function value multiplied by a subinterval width.
Here \(x_i^*\) is a sample point in a subinterval, and \(\Delta x_i\) is its positive width. Thus the sign of a term comes from \(f(x_i^*)\), the graph’s height at the sample point. If the graph is above the axis, its heights are nonnegative; if it is below, they are nonpositive. The integral preserves this sign information.
“Nonnegative” includes zero, and “nonpositive” includes zero. The sign condition alone does not guarantee a strictly positive or strictly negative contribution. For instance, if \(f(x)=0\) throughout an interval, every Riemann-sum term is zero, so the contribution is zero. More generally, strict positivity or negativity requires the corresponding region to have positive area.
Why the Integral Respects the Sign
The sign rule follows directly from the definition by Riemann sums. Consider an interval on which \(f(x)\geq 0\) everywhere. In every tagged partition, each width \(\Delta x_i\) is positive and each sampled height \(f(x_i^*)\) is nonnegative. Consequently every term, and therefore the whole Riemann sum, is nonnegative. As the mesh shrinks, these sums converge to the integral. A limit of nonnegative real numbers is nonnegative, so the integral cannot be negative. The corresponding argument for \(f(x)\leq 0\) shows that its integral cannot be positive.
If \(f(x)\geq 0\) on the interval, each sampled height is nonnegative.
Every subinterval width \(\Delta x_i\) is positive, so each product \(f(x_i^*)\Delta x_i\) is nonnegative.
The Riemann sums converge to the integral, which must therefore be nonnegative. Reversing all the inequalities proves the nonpositive case.
This argument explains why the sign convention is built into the integral rather than added afterward. The same principle applies to a graph below the axis: its negative heights produce nonpositive Riemann-sum terms and hence a nonpositive integral.
When a Graph Crosses the Axis
If a graph changes sign, split the interval at its crossings and any other points needed to keep each piece on one side of the axis. Then calculate the signed contribution on each piece. For simple straight-line graph regions, the magnitude can be found using familiar triangle or rectangle formulas; attach a positive sign above the axis and a negative sign below it.
A zero at an endpoint of a region does not by itself give that region a zero contribution. A triangle can meet the axis at its endpoints and still have positive area. What matters is the size of the region, not merely whether the graph touches or crosses the axis.
Worked Example: A Triangle Above Followed by One Below
Suppose a piecewise-linear graph forms a triangle above the axis from \(x=0\) to \(x=4\), with height \(3\), and then a triangle below the axis from \(x=4\) to \(x=8\), with depth \(2\). Find the net signed area over \([0,8]\).
The first triangle has base \(4-0=4\) and height \(3\). Since it lies above the axis, its signed contribution is positive:
The second triangle has base \(8-4=4\) and depth \(2\). Its area magnitude is
but it lies below the axis, so its signed contribution is \(-4\). Adding the two contributions gives
The net signed area is \(2\) square units. As a check, the positive contribution exceeds the negative contribution by \(6-4=2\), agreeing with the result.
Worked Example: Splitting a Crossing Triangle at Its Zero
A straight segment joins the graph points \((2,3)\) and \((4,-1)\). Find its signed contribution from \(x=2\) to \(x=4\). Because its endpoint heights have opposite signs, the segment crosses the horizontal axis and the interval must be split at the crossing.
The slope of the segment is
Starting at \((2,3)\), the height falls by \(2\) for each unit increase in \(x\). To fall from \(3\) to \(0\) takes \(3/2\) units of horizontal distance, so the zero occurs at
The above-axis triangle runs from \(x=2\) to \(x=7/2\), so its base is \(7/2-2=3/2\), and its height is \(3\). Its positive contribution is
The below-axis triangle runs from \(x=7/2\) to \(x=4\), so its base is \(4-7/2=1/2\), and its depth is \(1\). Its signed contribution is
Therefore the contribution over the whole segment is
As a check, the net contribution is positive because the larger triangle above the axis contributes \(9/4\), while the smaller triangle below contributes only \(-1/4\). Their difference is \(2\), as calculated.
Combining Several Graph Regions
A graph may move above and below the axis several times. Make a list of the regions in their left-to-right order, record each contribution with its sign, and add the signed values. A zero contribution stays zero; it must not be changed into a positive or negative value just because the graph is on a particular side of the axis elsewhere.
Worked Example: A Step Graph with Four Contributions
Define \(f\) on \([0,5]\) by setting \(f(x)=2\) for \(0\leq x<1\), \(f(x)=-1\) for \(1\leq x<3\), \(f(x)=3\) for \(3\leq x<4\), and \(f(x)=-2\) for \(4\leq x\leq 5\). The graph consists of four constant-height blocks. Find the net signed area.
On \([0,1)\), the graph is above the axis with height \(2\) and width \(1\). The contribution is
On \([1,3)\), the height is \(-1\) and the width is \(3-1=2\), giving
On \([3,4)\), the height is \(3\) and the width is \(4-3=1\), giving
Finally, on \([4,5]\), the height is \(-2\) and the width is \(5-4=1\), so the contribution is
Add all four signed contributions:
The net signed area is \(1\) square unit. Checking by grouping the positive terms and negative terms gives \(2+3=5\) above the axis and \(-2-2=-4\) below it; their sum is \(5+(-4)=1\), confirming the calculation.
A Common Sign Error
A frequent mistake is to calculate the area magnitude of every region and then add all the results as positive numbers. That discards the information carried by the graph’s position. For example, in the first worked example, adding the triangle magnitudes gives \(6+4=10\), but that is not the net signed area: the second triangle is below the axis and contributes \(-4\), so the net is \(6-4=2\).
The opposite error is also possible: assigning a strictly positive or negative contribution just because the function is nonnegative or nonpositive. The correct conclusions are nonnegative and nonpositive, respectively. For example, if a graph lies on the horizontal axis throughout an interval, its function values are both nonnegative and nonpositive, but its contribution is exactly zero.
Check Your Understanding
For each question, keep the sign of every graph contribution explicit.
- A triangle above the axis has base \(6\) and height \(2\). What is its signed contribution?
- A rectangle below the axis has width \(3\) and height \(4\). What is its signed contribution?
- A graph is nonnegative throughout an interval. Must its integral be strictly positive? Explain.
- A graph has contributions \(5\), \(-2\), and \(-1\) on three successive regions. What is the net signed area?
- Why should an interval containing a sign change usually be split before calculating its graph contribution?
