From Intersections to an Area Integral
When two graphs meet, their intersection points can provide the left and right boundaries of a region. To express that region’s area with vertical slices, identify the curve above and the curve below between those boundaries. The area is then accumulated from the vertical distance between the curves.
The order of these steps matters. Correct intersection points do not by themselves tell you which function belongs first in the integrand. That ordering must be established on the interval, not guessed from the appearance of a sketch.
Intersections Set the Candidate Limits
An intersection of the graphs \(y=f(x)\) and \(y=g(x)\) occurs at an \(x\)-value where the functions have equal outputs. To find intersection coordinates, solve
Each solution gives an \(x\)-coordinate; substituting it into either function gives the corresponding \(y\)-coordinate. If two such intersections bound the region being considered, their \(x\)-coordinates are the candidate limits for an integral using vertical slices. Always identify which pair of intersections bounds the particular region in the question.
This formula assumes the functions are continuous on \([a,b]\), the region is bounded by the two graph segments and the vertical lines at \(x=a\) and \(x=b\), and \(f(x)\ge g(x)\) throughout the interval. If the lower curve is named \(f\) instead, the subtraction order must be reversed.
Determining Which Curve Is Above
A graph can suggest which curve is higher, but a calculation provides a reliable check. Define the difference \(h(x)=f(x)-g(x)\). A positive value means \(f\) is above \(g\) at that input; a negative value means \(g\) is above \(f\).
For example, if \(f\) and \(g\) are continuous on an interval and have no intersections in its interior, then \(h\) is continuous and has no zeros there. Its sign cannot change within the interval: a change from positive to negative would force a zero by the Intermediate Value Theorem. Thus one test value strictly between the endpoints determines the ordering throughout the interior. At intersection endpoints, the difference is zero.
The neighborhood claim follows directly from continuity. Let \(h=f-g\), with \(h(c)>0\). By continuity, there is a neighborhood of \(c\) in which
Consequently, \(h(x)>h(c)-h(c)/2=h(c)/2>0\) there. Without continuity, a positive value at just one point would not establish the ordering nearby.
Why Vertical Distance Gives the Area
Divide \([a,b]\) into narrow subintervals. In a subinterval of width \(\Delta x_i\), choose a sample input \(x_i^*\). The vertical distance between the graphs at that input is \(f(x_i^*)-g(x_i^*)\), provided \(f\) is the upper curve. A thin rectangle with this height and width has area
Adding these approximate strip areas gives a Riemann sum. As the widths approach zero, continuity ensures that the sums approach the area of the region, and the sums of these heights times widths approach the definite integral. Therefore,
This also explains why the subtraction order is upper minus lower: each slice has a nonnegative height. When the coordinates have units of length, the integrand has units of length and \(dx\) contributes another length, so the result has units of square length.
Worked Examples
Worked Example: A Line and a Parabola
Find the area of the region bounded by \(y=x\) and \(y=x^2\).
Step 1: Find the intersections. Set the functions equal:
Thus \(x=0\) or \(x=1\). The intersection points are \((0,0)\) and \((1,1)\), so the candidate limits are 0 and 1.
Step 2: Determine the ordering. The functions are continuous, and their difference \(x-x^2=x(1-x)\) is positive for \(0<x<1\). For a test value, at \(x=\frac12\), the line has value \(\frac12\), while the parabola has value \(\frac14\). The line is above the parabola.
Step 3: Set up and evaluate the area.
Check: The integrand is zero at \(x=0\) and \(x=1\), as expected where the curves meet, and is positive between them. Differentiating the antiderivative gives \(x-x^2\), confirming the integration. The area is \(\frac16\) square units.
Worked Example: A Downward-Opening Parabola and a Line
Set up and evaluate the area bounded by \(y=4-x^2\) and \(y=x+2\).
Step 1: Find the intersections.
The intersection \(x\)-coordinates are \(-2\) and \(1\). Substituting into \(y=x+2\) gives \(y=0\) at \(x=-2\) and \(y=3\) at \(x=1\). The points are \((-2,0)\) and \((1,3)\).
Step 2: Identify the upper curve. The difference, parabola minus line, is \(2-x-x^2\). At the test value \(x=0\), this difference is \(2>0\). Since the difference is continuous and the curves have no other intersections between \(-2\) and \(1\), the parabola is above the line throughout that interval.
Step 3: Set up and evaluate the area.
Check: At each intersection, \(2-x-x^2=0\). The antiderivative differentiates to \(2-x-x^2\), and the area is positive. The bounded area is \(\frac92\) square units.
Worked Example: A Constant Function and a Parabola
Find the area of the region bounded by \(y=1\) and \(y=x^2-1\).
Step 1: Find the intersections.
The corresponding \(y\)-coordinate is 1, so the intersection points are \((-\sqrt{2},1)\) and \((\sqrt{2},1)\).
Step 2: Determine the ordering. At \(x=0\), the constant function has value 1 and the parabola has value \(-1\). The difference \(1-(x^2-1)=2-x^2\) is positive for \(-\sqrt{2}<x<\sqrt{2}\). Thus \(y=1\) is above \(y=x^2-1\) on the interval.
Step 3: Set up and evaluate the area.
Check: The integrand \(2-x^2\) is zero at both limits and positive at \(x=0\). Its antiderivative differentiates to \(2-x^2\). The resulting area, \(\frac{8\sqrt{2}}{3}\), is positive.
A Common Mistake: Reversing the Subtraction
In the second example, at \(x=0\) the parabola has value 4 and the line has value 2. Therefore the parabola is above the line. Reversing the subtraction gives the incorrect calculation
This is a signed integral of lower minus upper, not the area. A negative answer signals that the order has been reversed. The correction is to subtract the lower function from the upper function:
Before integrating, check the ordering with a test input and confirm that the proposed integrand is nonnegative across the interval. At the intersection limits, it should be zero.
A Reliable Setup Process
Set the functions equal, solve for the relevant \(x\)-coordinates, and identify the intersection points.
Use the pair of intersection points that bound the region in the question, and put their \(x\)-coordinates in increasing order.
Use a test value between the limits. For a conclusion throughout the interval, rely on continuity and verify there are no other intersections inside it.
Integrate the vertical distance from the left limit to the right limit.
Confirm the integrand is nonnegative on the interval and is zero at the intersection endpoints.
Why the Setup Matters
In an exam response, showing only an integral may hide an incorrect choice of limits or curve order. A clear setup makes the reasoning visible: the equality equation gives the intersections, the test value establishes which function is above, and the integral records the vertical distance over the bounded interval.
Keep the claims tied to their conditions. One test value does not establish an ordering everywhere if the functions could cross again; continuity and the absence of interior intersections are what allow the ordering to persist. Likewise, a negative integral obtained from lower minus upper is not a negative geometric area—it indicates that the subtraction order needs attention.
Check Your Understanding
For each question, identify the relevant intersections and explain how the ordering determines the integral.
- Find the intersection \(x\)-coordinates of \(y=2x\) and \(y=x^2\). Which graph is above between the two intersections?
- For \(y=3\) and \(y=x^2-1\), find the limits and write (but do not evaluate) the area integral for the bounded region.
- Why is a test value sufficient to establish the ordering when the difference of the functions is continuous and has no interior zeros?
- What does a negative value from integrating lower minus upper indicate in an area setup?
- For two continuous curves intersecting at the limits of an interval, what additional fact must be checked before using one test value to determine which curve is above throughout?
