Why Split an Integral at Sign Changes?
A definite integral adds signed contributions across an interval. Where the integrand is positive, its contribution is positive; where the integrand is negative, its contribution is negative. As a result, contributions from different parts of an interval can partly or completely cancel.
That cancellation is useful when finding net change, but it can hide how much total area lies between the graph and the horizontal axis. To keep track of both, split the interval at zeros of the integrand, determine its sign on each part, and calculate each contribution separately.
Zeros and Sign Intervals
Let \(f\) be continuous on a closed interval \([a,b]\). A zero of \(f\) is a value \(c\) for which \(f(c)=0\). The zeros between \(a\) and \(b\), together with the endpoints, divide the interval into smaller intervals. If \(f\) has no zero inside one of these smaller intervals, continuity ensures that \(f\) has the same sign throughout that interval.
A zero may or may not be a sign change. For example, \(f(x)=x\) changes from negative to positive at \(0\), while \(g(x)=x^2\) is positive on both sides of \(0\). Both have a zero there, but only the first changes sign. Checking the sign on each subinterval, rather than assuming every zero changes sign, prevents errors.
On a subinterval where \(f\) is positive, its integral contributes positively to net change and contributes the same amount to total area. On a subinterval where \(f\) is negative, its integral contributes negatively to net change, while the corresponding area is the positive magnitude of that integral.
Why Splitting Works
The result follows from additivity of definite integrals: if \(a=c_0<c_1<\cdots<c_n=b\), then the integral over the full interval is the sum of the integrals over the consecutive subintervals. This lets us compute each sign interval separately. On each one, the absolute value in the area integral can be removed by using the known sign of \(f\).
List the zeros of \(f\) that lie inside \([a,b]\), and include the endpoints as interval boundaries.
Establish whether \(f\) is positive or negative on each resulting subinterval. Do not assume every zero changes the sign.
Use an antiderivative to calculate the signed integral on every subinterval.
Add signed values for net change; add their absolute values for total geometric area.
To justify the area rule, let \(I_i=\int_{c_{i-1}}^{c_i} f(x)\,dx\) for each subinterval. If \(f\) is positive there, \(\int_{c_{i-1}}^{c_i}|f(x)|\,dx=I_i\). If \(f\) is negative there, \(|f(x)|=-f(x)\), so \(\int_{c_{i-1}}^{c_i}|f(x)|\,dx=-I_i=|I_i|\). Thus the total area is the sum of the magnitudes of the signed subinterval integrals, while the net change is their signed sum.
These equalities require the intervals to cover \([a,b]\) without gaps or overlap in their interiors, and the sign of \(f\) to be constant on each subinterval. For continuous \(f\), splitting at all interior zeros provides such intervals.
Worked Examples
Worked Example: A Quadratic with Two Sign Changes
Find the net change and total geometric area for \(f(x)=(x+1)(x-2)\) on \([-2,3]\).
Step 1: Find the zeros and signs. Solving \((x+1)(x-2)=0\) gives \(x=-1\) and \(x=2\). These divide the interval into \([-2,-1]\), \([-1,2]\), and \([2,3]\). The factors show that \(f\) is positive on the first interval, negative on the second, and positive on the third.
Step 2: Find an antiderivative. Expand \(f(x)=x^2-x-2\). An antiderivative is \(F(x)=\frac{x^3}{3}-\frac{x^2}{2}-2x\).
Step 3: Calculate each signed contribution.
For the middle interval, \(F(2)=\frac{8}{3}-2-4=-\frac{10}{3}\), so
At the right endpoint, \(F(3)=9-\frac{9}{2}-6=-\frac{3}{2}\). Therefore,
Step 4: Combine and check. The net change is the signed sum:
Directly, \(F(3)-F(-2)=-\frac{3}{2}-(-\frac{2}{3})=-\frac{5}{6}\), confirming the net change. The total area is
The negative middle contribution reduces net change but adds positive area.
Worked Example: A Trigonometric Integrand
Find the net change and total geometric area for \(f(x)=\sin x\) on \([0,2\pi]\).
Step 1: Find the zeros and signs. The zeros in the interval are \(0\), \(\pi\), and \(2\pi\). The sine function is positive on \((0,\pi)\) and negative on \((\pi,2\pi)\).
Step 2: Integrate on each interval. An antiderivative is \(F(x)=-\cos x\). Thus
Step 3: Combine and check. The net change is \(2+(-2)=0\). The endpoint calculation confirms this: \(F(2\pi)-F(0)=-1-(-1)=0\). The total area is \(|2|+|-2|=4\). The equal positive and negative contributions cancel in the definite integral, but not in the area.
Worked Example: A Zero That Does Not Change Sign
Find the net change and total geometric area for \(f(x)=(x-1)^2(x+1)\) on \([-2,2]\).
Step 1: Find the zeros and signs. The zeros are \(-1\) and \(1\). Since \((x-1)^2\) is nonnegative and does not change sign at \(1\), the sign is determined by \(x+1\), except that \(f(1)=0\). The function is negative on \((-2,-1)\) and positive on \((-1,1)\) and \((1,2)\). The zero at \(1\) does not reverse the sign, but it can still be included as a split point.
Step 2: Find an antiderivative. Expanding gives \(f(x)=x^3-x^2-x+1\), so an antiderivative is \(F(x)=\frac{x^4}{4}-\frac{x^3}{3}-\frac{x^2}{2}+x\).
Step 3: Evaluate at the split points.
The three signed contributions are
Step 4: Combine and check. The net change is
Directly, \(F(2)-F(-2)=\frac{4}{3}-\frac{8}{3}=-\frac{4}{3}\), as required. The total area is
The split at \(1\) leaves the sign unchanged, so its two positive contributions add to the same amount as the integral over \([-1,2]\). The split is valid; it simply is not a sign change.
A Common Mistake
A frequent error is to report the definite integral as the total area even when the integrand is negative on part of the interval. In the first example, the incorrect calculation would be to call \(-\frac{5}{6}\) the total area because it is the value of the full definite integral. That value is the net change, not the total geometric area.
The correction is to split at the zeros and take the magnitude of each signed contribution for area. In that example, the pieces are \(\frac{11}{6}\), \(-\frac{9}{2}\), and \(\frac{11}{6}\). Their signed sum is \(-\frac{5}{6}\), while their magnitudes sum to \(\frac{49}{6}\). Taking the absolute value only after computing the full integral would give \(\frac{5}{6}\), which is still not the total area: cancellation has already occurred.
Using the Result Carefully
Before calculating, identify which quantity the question asks for. “Net change” or “the definite integral” calls for a signed sum. “Total area” calls for a sum of magnitudes. If a problem asks for both, keep the signed contributions in a list until the final step rather than combining them too early.
For a continuous integrand, finding every zero inside the interval and checking signs between consecutive zeros gives a systematic method. A zero where the sign does not change is harmless as a split point, but it does not justify changing a positive contribution into a negative one. The sign, not merely the presence of a zero, determines how each piece contributes.
Check Your Understanding
Use zeros and signs to decide how each integral should be split and combined.
- For \(f(x)=x-3\) on \([1,5]\), identify the sign on each subinterval and state which contributions add for total area.
- If the signed contributions on three intervals are \(4\), \(-7\), and \(2\), find the net change and total area.
- Why is \(\left|\int_a^b f(x)\,dx\right|\) not generally equal to the total geometric area?
- For \(f(x)=(x-2)^2\), does the zero at \(x=2\) change the sign? Explain.
- What condition on the sign of \(f\) within a subinterval lets you replace the integral of \(|f|\) there by either the integral of \(f\) or its negative?
