From a Representation to an Integral
A graph can show where two curves meet and which one lies above the other. A table can give the same information through function values and equality points. Turning either representation into an area integral requires careful reading: the bounds must cover the intended region, and the integrand must measure the distance between its boundaries.
This tutorial focuses on the setup. The goal is not to calculate an area from a complicated integral, but to translate the region shown or described into a definite integral that represents its area.
What to Read from a Graph or Table
Before writing an integral, identify four pieces of information:
- The intersection locations. These usually provide the endpoints of the interval or intervals being integrated over.
- The slice direction. A vertical slice measures a height; a horizontal slice measures a width.
- The boundary curves. Decide which graph gives the upper and lower boundary for vertical slices, or the right and left boundary for horizontal slices.
- Any change in boundary order. If the curves cross inside the region, the top-minus-bottom expression may change and the integral must be split there.
A graph may give exact intersection coordinates, such as points marked at grid locations, or only estimates. If a crossing is read approximately, use an approximate bound and identify it as such. A table can establish that the functions are equal at listed entries and show their order at sampled points. But a few table entries alone do not prove that no additional crossing occurs between those points. That conclusion needs more information, such as a graph or a stated fact about the curves.
Why the Difference of Boundaries Gives Area
Consider a region between continuous curves \(y=f(x)\) and \(y=g(x)\) from \(x=a\) to \(x=b\), with \(f(x)\ge g(x)\) throughout that interval. A thin vertical slice at \(x\) has height \(f(x)-g(x)\). If the interval is divided into small widths \(\Delta x\), each narrow strip has approximate area \((f(x)-g(x))\Delta x\). Adding the strip areas and taking the limit gives the definite integral.
This argument explains why a graph is more than a picture: it tells us which function supplies each boundary and where the bounds belong. The same reasoning for horizontal strips uses their width, \(x_{\text{right}}(y)-x_{\text{left}}(y)\), and accumulates with respect to \(y\).
These formulas require that the stated boundaries and order apply throughout the interval of integration. When the order changes, use separate intervals so that each strip length is written as a nonnegative boundary distance.
Worked Example: Reading Bounds and Order from a Graph
Worked Example: Reading Bounds and Order from a Graph
A graph shows the curves \(y=4-x^2\) and \(y=x^2\), enclosing one region. Set up an integral for its area using vertical slices.
Step 1: Find the intersections. At an intersection the \(y\)-values are equal:
Add \(x^2\) to both sides, then divide by \(2\):
These are the left and right bounds of the enclosed region.
Step 2: Determine the upper curve. Test an \(x\)-value between the intersections, such as \(x=0\):
Thus \(4-x^2\) is above \(x^2\) between the intersections. The graph shows no additional crossing within these bounds, so the order holds across the region.
Step 3: Write the setup. Use top minus bottom, with respect to \(x\):
Check: At \(x=-\sqrt{2}\) and \(x=\sqrt{2}\), both functions equal \(2\), so the vertical distance is zero at each endpoint. At \(x=0\), the integrand is \(4-0=4\), the positive distance between the curves there. The bounds and integrand therefore match the graph.
Using Table Values to Detect a Change in Order
A table is especially useful when it lists equal function values at the boundary points and provides interior values that reveal which curve is higher. If the order changes at an intersection inside the region, the setup must use separate integrals on either side.
Worked Example: Splitting a Table-Described Region
A table describes two curves \(y=f(x)\) and \(y=g(x)\). The graph confirms that their only intersections on the shown region occur at \(x=0\), \(x=2\), and \(x=4\).
| \(x\) | \(f(x)\) | \(g(x)\) |
|---|---|---|
| 0 | 1 | 1 |
| 1 | 3 | 2 |
| 2 | 3 | 3 |
| 3 | 2 | 3 |
| 4 | 1 | 1 |
Set up the total area of the two enclosed parts using vertical slices.
Step 1: Read the intersections. The functions are equal at \(x=0\), \(x=2\), and \(x=4\). These divide the region into two intervals.
Step 2: Compare values in each interval. At \(x=1\), which lies between \(0\) and \(2\), the table gives \(f(1)=3\) and \(g(1)=2\). Therefore \(f\) is above \(g\) on the first part, as the graph confirms. At \(x=3\), between \(2\) and \(4\), \(f(3)=2\) and \(g(3)=3\), so \(g\) is above \(f\) on the second part.
Step 3: Set up each area using upper minus lower. The order switches at \(x=2\), so write:
Check: At every endpoint listed for the two pieces, the table shows equal function values: at \(x=0,2,4\), the vertical distance is zero. At the interior test value \(x=1\), the first integrand is \(3-2=1>0\); at \(x=3\), the second is \(3-2=1>0\). Each integral therefore represents a nonnegative area on its interval.
The graph’s confirmation matters here. Without it or another guarantee, the table’s samples would not rule out an extra crossing between listed \(x\)-values. The correct conclusion is supported by the stated graph information, not by assuming that sparse samples describe every point between them.
Worked Example: Setting Up with Horizontal Slices
A region is shown on a graph with \(y\) as the convenient direction for reading its boundaries. Its left boundary is \(x=y^2\), and its right boundary is \(x=2-y\). Set up its area using horizontal slices.
Worked Example: Setting Up with Horizontal Slices
Step 1: Find the \(y\)-coordinates of the intersections. Equate the \(x\)-values of the boundaries:
Move all terms to one side and factor:
Thus the intersection levels are \(y=-2\) and \(y=1\).
Step 2: Check left-right order. Use an intermediate value, \(y=0\). The boundary values are \(x=0^2=0\) and \(x=2-0=2\). Since \(0<2\), \(y^2\) is the left boundary and \(2-y\) is the right boundary. The graph confirms that this order holds between the intersections.
Step 3: Write the setup. Horizontal slice width is right minus left:
Check: At \(y=-2\), both boundary values are \(4\): \((-2)^2=4\) and \(2-(-2)=4\). At \(y=1\), both equal \(1\). At \(y=0\), the integrand is \(2-0-0=2>0\), matching the graph’s positive horizontal width.
A Reliable Setup Procedure
When the region is provided visually or in a table, use a deliberate sequence rather than guessing an integral from the shape alone.
Identify the specific bounded component whose area is requested. Do not assume every visible portion belongs to the same region.
For vertical slices, use the relevant \(x\)-coordinates; for horizontal slices, use the relevant \(y\)-coordinates. Label bounds as approximate if they come from a visual estimate.
Read upper and lower curves for vertical slices, or right and left curves for horizontal slices. Use an interior point in each interval to check the order.
If an intersection lies inside the region and reverses the order, write one integral for each interval.
At a shared endpoint the slice length should be zero. Within each interval, the chosen boundary difference should be nonnegative.
Common Mistake: Using One Order Across a Crossing
In the table example, a tempting but incorrect setup is:
This treats \(f\) as the upper curve all the way from \(0\) to \(4\). But the table gives \(f(3)=2\) and \(g(3)=3\), so near that sample \(f\) is below \(g\). The expression \(f-g\) is negative there and represents signed accumulation, not the area of that part of the region.
The correction is to split at the crossing \(x=2\), then use the upper curve minus the lower curve on each part:
Another frequent error is treating a table’s sampled values as proof that no crossings occur between them. A table only reports values at its listed inputs. If a complete setup depends on knowing there are no unlisted intersections, that fact must come from the graph, a stated condition, or additional information about the functions.
Why This Skill Matters
An integral setup encodes a geometric claim. Its bounds say which part of the region is included; its integrand says which two boundaries determine each slice. A correct-looking calculation cannot repair a setup that omits a component or reverses the boundaries. Reading the representation carefully before integrating is therefore an essential part of solving area problems, not merely a preliminary step.
Check Your Understanding
For each question, identify the information needed to write a valid area setup.
- A graph shows two curves meeting at \(x=-1\) and \(x=3\), with one curve above the other throughout. What bounds and boundary order belong in a vertical-slice setup?
- A table gives \(f(1)>g(1)\) and \(f(4)<g(4)\), and a graph shows one crossing between those inputs. Why should the area setup be split at that crossing?
- A graph estimates an intersection at \(x\approx 2.4\). How should that bound be reported in an integral setup?
- For horizontal slices, which boundary difference gives the slice width?
- Why can a sparse table fail to establish that two curves have no other intersections between listed inputs?
