Why intersections determine the limits
When a region is described as lying between two curves, its leftmost and rightmost points often occur where the curves meet. Those points give the horizontal limits for vertical slices, and therefore the integration bounds for an area integral.
Finding the bounds is not just a matter of solving an equation. An algebraic equation may have several solutions, while the region described in a problem may use only some of them. A solution may also fail to satisfy the original equation or the stated domain. The task is to find candidate intersections, check them, and identify which points actually form the edges of the specified region.
From curve intersections to horizontal coordinates
Suppose the boundary curves are graphs \(y=f(x)\) and \(y=g(x)\). At an intersection, they have the same \(x\)-coordinate and the same \(y\)-coordinate. Thus their intersection coordinates satisfy \(f(x)=g(x)\). Solving this equation gives candidate horizontal coordinates; substituting a candidate into either function gives the corresponding \(y\)-coordinate.
For a region whose left and right edges are intersections of two graphs, the smaller relevant \(x\)-coordinate is the left bound and the larger is the right bound. If the description includes a vertical boundary, such as \(x=0\), that boundary can supply one of the horizontal limits instead. In either case, the bounds must come from the described region, not merely from a list of algebraic answers.
A short justification for using the intersection coordinates
Here is why intersection coordinates serve as limits when a region is enclosed between two graphs. Suppose the region extends horizontally from \(x=a\) to \(x=b\), and its upper and lower boundaries are the graphs \(y=f(x)\) and \(y=g(x)\). A vertical slice at a coordinate strictly between the endpoints runs from one graph to the other. At an endpoint where those two graphs close the region, the slice has zero height: the upper and lower boundary values coincide. That coincidence means \(f(a)=g(a)\) or \(f(b)=g(b)\).
Therefore, the horizontal edges of this kind of enclosed region must be found among the solutions of \(f(x)=g(x)\). Solving the equation identifies candidates; the region description tells us which candidates are its left and right edges. If a vertical boundary is specified instead, its \(x\)-coordinate supplies the corresponding limit directly. This reasoning explains the role of intersections without assuming that every solution of the equality belongs to the region.
Worked Example: A parabola and a horizontal boundary
Worked Example: A parabola and a horizontal boundary
Find the horizontal bounds of the finite region enclosed by \(y=x^2\) and \(y=4\).
Step 1: Set the boundary functions equal. At an intersection, the values of \(y\) must agree:
Step 2: Solve for the candidate coordinates. Subtracting \(4\) gives \(x^2-4=0\), which factors as \((x-2)(x+2)=0\). Thus the candidates are \(x=2\) and \(x=-2\).
Step 3: Check both candidates in the original equation. For \(x=2\), \(x^2=2^2=4\), so the intersection is \((2,4)\). For \(x=-2\), \(x^2=(-2)^2=4\), so the intersection is \((-2,4)\). Both candidates satisfy the original equation.
Step 4: Select the bounds. These are the two intersections closing the described region. The smaller horizontal coordinate, \(-2\), is the left bound; the larger, \(2\), is the right bound. Thus the horizontal interval is \([-2,2]\). If setting up its area with vertical slices, the limits are \(-2\) and \(2\).
Worked Example: Two curves with different intersection heights
Worked Example: Two curves with different intersection heights
Find the horizontal bounds of the finite region enclosed by \(y=x^2\) and \(y=2x+3\).
Step 1: Equate the boundary values.
Step 2: Rearrange and factor. Subtract \(2x+3\) from both sides:
The candidate horizontal coordinates are \(x=3\) and \(x=-1\).
Step 3: Substitute each value into both curve equations. At \(x=3\), the parabola gives \(y=3^2=9\), and the line gives \(y=2(3)+3=9\). The point is \((3,9)\). At \(x=-1\), the parabola gives \(y=(-1)^2=1\), and the line gives \(y=2(-1)+3=1\). The point is \((-1,1)\). Both values are actual intersections.
Step 4: Identify the bounds from the region. The finite region enclosed by these two curves runs between their two intersections. The smaller \(x\)-coordinate is \(-1\), and the larger is \(3\), so the horizontal limits are \([-1,3]\). These are limits on \(x\), not the \(y\)-coordinates \(1\) and \(9\).
Checking candidates against the region description
Solving the equality of boundary functions is necessary, but not always sufficient. For example, if a curve contains a square root, its domain constrains the equation. Squaring both sides can also produce a value that solves the squared equation but not the original one. The following example shows why checking against both the equation and the described region matters.
Worked Example: Rejecting a candidate that fails the original equation
Find the horizontal bounds of the region enclosed by \(y=\sqrt{x}\), \(y=2-x\), and the vertical boundary \(x=0\).
Step 1: Find candidate intersections of the two curves. Set their \(y\)-values equal:
The square root requires \(x\geq 0\). Also, because \(\sqrt{x}\geq 0\), equality requires \(2-x\geq 0\), so \(x\leq 2\). Together, these conditions give \(0\leq x\leq 2\) for a valid intersection.
Step 2: Solve, keeping track of the restrictions. Squaring both sides gives \(x=(2-x)^2\). Expanding and collecting terms gives \(x^2-5x+4=0\), which factors as \((x-1)(x-4)=0\). The algebraic candidates are \(x=1\) and \(x=4\).
Step 3: Check each candidate in the original equation. For \(x=1\), \(\sqrt{1}=1\) and \(2-1=1\), so \(x=1\) is valid and the intersection is \((1,1)\). For \(x=4\), \(\sqrt{4}=2\), while \(2-4=-2\). Since \(2\neq -2\), \(x=4\) is not an intersection. It also violates the required condition \(x\leq 2\).
Step 4: Use the other boundary to complete the bounds. The region has a vertical edge at \(x=0\), so its left limit is \(0\). Its other end is the valid intersection at \(x=1\). The horizontal bounds are therefore \([0,1]\). As a consistency check, at \(x=0\) the two curve values are \(0\) and \(2\), and at \(x=1\) they both equal \(1\). Between these endpoints, for \(0\leq x<1\), \(\sqrt{x}<1\) while \(2-x>1\); the curves meet at \(x=1\), as required by the described region.
A practical method for choosing the bounds
When the region is described in words or shown in a diagram, use the description to decide which boundary pieces form its left and right edges. Then apply this sequence:
Identify the curves that meet at the edges of the region. For graphs \(y=f(x)\) and \(y=g(x)\), their intersection candidates satisfy \(f(x)=g(x)\).
Show the algebra that produces the candidates, and note any restrictions imposed by the functions or the problem.
Substitute into the original equations, not just a transformed equation. Reject candidates that fail an original equation or a domain restriction.
Use the stated boundaries to decide which valid points are the left and right edges. A specified vertical boundary can provide an endpoint without being an intersection of the two curves.
For vertical slices, the left limit is the smaller horizontal coordinate and the right limit is the larger one.
Common mistakes and how to avoid them
A frequent mistake is to use every root of an equation as an integration limit without checking whether it is an actual intersection. In the square-root example, the equation obtained after squaring has \(x=4\) as a root. The incorrect calculation would be:
The last equality is false, so \(x=4\) cannot be an intersection. The correction is to substitute each candidate into the original equation and retain only candidates that satisfy it, while respecting the function domains.
Another mistake is to confuse a point with one of its coordinates. An intersection such as \((3,9)\) is a point in the plane, but the vertical-slice bound is \(x=3\), not \(y=9\). Finally, do not assume an endpoint must be an intersection of the two curves: a stated vertical boundary can supply an endpoint, as \(x=0\) did above.
Why careful bounds matter
The limits of integration describe the horizontal extent of the region being measured. If a valid intersection is missed, an extraneous candidate is included, or a vertical boundary is ignored, the integral will describe a different interval from the one in the problem. Checking the original equations and the region description before writing the limits prevents these errors. It also makes the setup auditable: each bound can be traced to a specific point or boundary in the plane.
Check Your Understanding
For each question, explain how you would verify the bounds, not only how you would find candidates.
- For the region enclosed by \(y=x^2\) and \(y=9\), what equation gives the candidate horizontal coordinates, and which two coordinates form the bounds?
- If solving an equation gives two candidates, what should you check before treating both as intersections?
- For \(y=\sqrt{x}\) and \(y=3-x\), what domain or sign restriction should be considered before squaring the intersection equation?
- A region has a vertical edge at \(x=-2\) and its other edge is where two curves meet at \(x=5\). What are its horizontal bounds?
- Why is an intersection point \((a,b)\) associated with the horizontal bound \(a\), rather than \(b\), when using vertical slices?
