Why Three Curves Require Careful Setup
For a region between two curves, the area setup depends on which curve is above the other. With three curves, that boundary may change as \(x\) varies: one curve can form the top of the region on one interval, while another takes over later. The lower boundary can change as well.
The key is not to choose one pair of curves and use it across the entire interval automatically. Instead, locate the relevant intersections, divide the interval there, and compare the curves on each resulting piece. Each integral then uses the correct vertical distance: top minus bottom.
A Reliable Setup Procedure
Pairwise intersections are potential change points, not automatic change points. A crossing matters when it changes which curve forms the boundary of the particular region being considered. After finding and sorting the relevant intersections, compare the curves between consecutive intersections. One test value in each interval is often enough to establish their order, since continuous curves cannot reverse order without meeting.
State the left and right \(x\)-bounds and determine which of the three curves can form its boundary.
Solve the equations for pairs of curves, then retain the intersection \(x\)-values that lie on or within the region’s extent and could affect its boundary.
Test the curves on each open interval to identify the top and bottom boundaries.
Use upper curve minus lower curve on every piece, then add the integrals.
Why the Piecewise Integral Gives the Area
On a short vertical slice at \(x\), the region’s height is the difference between its upper and lower boundaries. If those boundaries are \(U(x)\) and \(L(x)\), the height is \(U(x)-L(x)\). The area is built by adding these heights across the interval, which is the definite integral of the height.
If the boundary changes at \(c\), split the interval there. The definite integral’s additivity gives \(\int_a^b h(x)\,dx=\int_a^c h(x)\,dx+\int_c^b h(x)\,dx\). Applying the correct height function on each side therefore gives the total area. For multiple changes, repeat this partition at every relevant change point.
where \(x_0<x_1<\cdots<x_n\) are the partition points, and \(U_i\) and \(L_i\) are the upper and lower boundaries on the \(i\)th interval. The comparison must be checked on each interval; the formula does not mean that the same two curves are used everywhere.
Worked Example: A Top Boundary Switching Between Three Curves
Consider the region between \(y=0\) and the upper boundary formed by the lowest of \(y=x+1\), \(y=2\), and \(y=4-x\), for \(0\le x\le3\). Find its area.
First locate where the candidate upper curves meet. The equations \(x+1=2\) and \(2=4-x\) both give \(x=1\) and \(x=2\), respectively. The remaining pair meets when \(x+1=4-x\), so \(2x=3\) and \(x=\frac32\). This last intersection does not change which curve is the lowest: the constant curve \(y=2\) is lower than both candidates around that crossing.
Compare the curves on the intervals determined by the actual boundary changes. On \(0\le x\le1\), \(x+1\le2\), and \(x+1\le4-x\), so the upper boundary is \(x+1\). On \(1\le x\le2\), \(2\le x+1\) and \(2\le4-x\), so it is \(2\). On \(2\le x\le3\), \(4-x\le2\) and \(4-x\le x+1\), so it is \(4-x\). The lower boundary throughout is \(0\).
Compute each piece:
Thus \(A=\frac32+2+\frac32=5\) square units. As a check, each integrand is nonnegative on its interval, as required for top minus bottom. Differentiating the antiderivatives gives \(x+1\), \(2\), and \(4-x\), respectively; evaluating them at the stated bounds gives the three areas above.
When the Bottom Boundary Changes
The same method applies when the upper curve stays fixed but the lower boundary switches. In that case, identify which candidate lower curve is highest on each interval. The highest of the lower curves forms the edge closest to the fixed upper boundary.
Worked Example: A Bottom Boundary Switching Between Three Curves
Find the area between \(y=4\) and the highest of the curves \(y=0\), \(y=x-1\), and \(y=2x-3\), for \(0\le x\le3\).
The candidates \(0\) and \(x-1\) meet at \(x=1\). The candidates \(x-1\) and \(2x-3\) meet when \(x-1=2x-3\), giving \(x=2\). The curves \(0\) and \(2x-3\) meet at \(x=\frac32\); this intersection does not change the highest curve because \(x-1\) is higher there.
On \(0\le x\le1\), \(0\) is at least as high as both \(x-1\) and \(2x-3\), so it is the lower boundary of the region. On \(1\le x\le2\), \(x-1\ge0\), and \((x-1)-(2x-3)=2-x\ge0\), so \(x-1\) is highest. On \(2\le x\le3\), \(2x-3\ge x-1\) and \(2x-3\ge0\), so \(2x-3\) is highest. The fixed upper boundary \(y=4\) is above all three lower curves on this interval.
Evaluate the pieces, keeping the order upper minus lower:
The area is \(4+\frac72+2=\frac{19}{2}\) square units. Each integrand is nonnegative on its interval, and differentiating the antiderivatives \(4x\), \(5x-\frac{x^2}{2}\), and \(7x-x^2\) recovers \(4\), \(5-x\), and \(7-2x\). This confirms both the integration and the chosen order.
One Region Using All Three Curves
A useful way to check the method is to examine an enclosed region whose boundary consists of three different curves. In the next example, the upper curve stays the same, but the lower boundary changes from a horizontal line to a parabola at an irrational intersection.
Worked Example: Switching Boundaries at an Irrational Intersection
Consider the region enclosed by \(y=2x\), \(y=2\), and \(y=x^2\), between the points where the line meets the horizontal line and where the line meets the parabola. Find its area.
Find the pairwise intersections. The line and horizontal line meet when \(2x=2\), so \(x=1\). The horizontal line and parabola meet when \(x^2=2\), giving \(x=\pm\sqrt2\); the intersection relevant to this region is \(x=\sqrt2\). The line and parabola meet when \(2x=x^2\), or \(x(x-2)=0\); the relevant right-hand endpoint is \(x=2\). Therefore the interval is \(1\le x\le2\), with a possible boundary switch at \(\sqrt2\).
To check the curve order, at \(x=\frac65\), the line has value \(\frac{12}{5}\), the horizontal curve has value \(2\), and the parabola has value \(\frac{36}{25}\). Thus the line is uppermost and the horizontal curve lies above the parabola. At \(x=\frac32\), the values are \(3\), \(2\), and \(\frac94\), respectively; the line is still uppermost, but the parabola now lies above the horizontal curve. The bottom boundary changes at \(x=\sqrt2\).
For the first part:
For the second part:
Adding the two pieces gives
square units, rounded to three decimal places. Both integrands are nonnegative on their intervals: \(2x-2\ge0\) for \(x\ge1\), and \(2x-x^2=x(2-x)\ge0\) for \(1\le x\le2\). Differentiating the antiderivatives \(x^2-2x\) and \(x^2-\frac{x^3}{3}\) returns the corresponding integrands, and substituting the bounds gives the displayed exact area.
Common Mistake: Using One Lower Curve Across the Whole Interval
In the three-curve example, a tempting but incorrect setup is to use the horizontal curve \(y=2\) as the lower boundary all the way from \(x=1\) to \(x=2\):
This calculation is correct for the area between the line and the horizontal curve, but it is not the area of the enclosed region for the entire interval. For \(x>\sqrt2\), the parabola lies above \(y=2\), so the region’s lower boundary is \(y=x^2\), not \(y=2\). The correction is to split at \(x=\sqrt2\): use \(2x-2\) on \([1,\sqrt2]\) and \(2x-x^2\) on \([\sqrt2,2]\). That gives \(\frac{7-4\sqrt2}{3}\), not \(1\).
Check Your Understanding
For each question, identify the relevant intervals and boundary curves before evaluating any integrals.
- For \(0\le x\le2\), the region lies above \(y=0\) and below the lowest of \(y=x\), \(y=1\), and \(y=2-x\). At what \(x\)-values does its upper boundary change?
- For \(0\le x\le3\), the upper boundary is \(y=5\) and the lower boundary is the highest of \(y=0\), \(y=x-1\), and \(y=2x-4\). Which curve is lowest on each interval?
- Why might a pairwise intersection of two curves fail to be a boundary-change point for a particular region?
- For the region between \(y=2x\), \(y=2\), and \(y=x^2\) in the worked example, which curve is the lower boundary when \(x=\frac32\)?
- What sign check should be true of each integrand in a correct area setup using vertical slices?
