From a Plotted Curve to an Integral Estimate
A graph gives a picture of a function, but a definite integral measures accumulated signed change over an interval. To estimate the integral from a graph, we must translate the picture into statements about how high or low the function is across each part of the interval.
The key distinction is between a point reading and a bound across an interval. A labeled point may tell us an exact value such as \(f(1)=2\). But the integral depends on all values between the endpoints, not just the values at a few selected points. A reliable estimate therefore uses graph information that encloses the curve throughout each subinterval.
Read the Axes Before the Curve
First identify the horizontal and vertical scales. If adjacent vertical grid lines represent \(0.5\) units of \(x\), then the width of a grid cell is \(0.5\), not \(1\). Likewise, a vertical grid interval might represent \(2\) units of \(f(x)\). The product of a horizontal width and a function height has the units of the integral.
Next decide what the graph actually tells you. A point marked with coordinates, or a function specified exactly, can provide exact data. A curve traced on a screen or printed page usually gives only approximate visual readings. A thick line, coarse grid, or low-resolution image makes it impossible to claim more precision than the plot supports.
For a continuous function on a closed interval, the definite integral exists. If the graph shows that throughout a subinterval the function stays between two horizontal levels, those levels can be used to bound that subinterval’s integral. This is the graph-based use of pointwise bounds and integral comparison.
The Basic Enclosure Principle
Suppose a partition divides \([a,b]\) into subintervals \([x_{i-1},x_i]\), each with width \(\Delta x_i=x_i-x_{i-1}\). Suppose the graph is known to satisfy \(L_i\leq f(x)\leq U_i\) at every point of the \(i\)-th subinterval. Then the integral over that subinterval is bounded by the height bounds multiplied by its width.
Adding the bounds across all subintervals gives a bound for the full integral. The bounds \(L_i\) and \(U_i\) may be negative: the integral is signed, so a graph below the horizontal axis contributes negatively.
Why the Enclosure Gives Valid Bounds
Here is the reason the method works. On each subinterval, the assumed graph enclosure says \(L_i\leq f(x)\leq U_i\) throughout. The integral comparison principle therefore gives the corresponding inequality between the integrals. The integral of a constant \(c\) over an interval of width \(\Delta x_i\) is \(c\Delta x_i\). Combining these facts gives the subinterval bound, and splitting the integral across the partition gives the total bound.
Summing these inequalities is valid because the integral over \([a,b]\) is the sum of the integrals over the subintervals. This argument does not require the function to be positive. It uses signed integral comparison, so it applies when some or all of the graph lies below the axis.
Worked Example: Endpoint Readings on an Increasing Graph
Worked Example: Endpoint Readings on an Increasing Graph
A velocity graph \(v(t)\) is known to increase throughout \(0\leq t\leq2\) seconds. The labeled values are \(v(0)=1\), \(v(1)=2\), and \(v(2)=4\) meters per second. Bound the displacement \(\int_0^2 v(t)\,dt\) using the two one-second subintervals.
Because the graph increases on \([0,1]\), its values there lie between \(1\) and \(2\) meters per second. That interval contributes between \(1\cdot1=1\) meter and \(2\cdot1=2\) meters.
On \([1,2]\), the values lie between \(2\) and \(4\) meters per second. That interval contributes between \(2\cdot1=2\) meters and \(4\cdot1=4\) meters.
Check: both subinterval widths are \(1\) second, so multiplying a velocity in meters per second by a width in seconds gives meters. The lower and upper totals are \(3\) and \(6\) meters, respectively. The bounds are valid because the stated increase guarantees that the endpoint heights bound the graph on each subinterval.
Worked Example: A Graph That Crosses the Axis
Worked Example: A Graph That Crosses the Axis
A velocity graph is known to stay between \(-2\) and \(-1\) meters per second on \([0,1]\), between \(-1\) and \(1\) meters per second on \([1,2]\), and between \(1\) and \(3\) meters per second on \([2,3]\). Bound the displacement over the three seconds.
Each interval has width \(1\) second. The first contribution is between \((-2)(1)=-2\) meters and \((-1)(1)=-1\) meter. The second is between \((-1)(1)=-1\) meter and \(1(1)=1\) meter. The third is between \(1(1)=1\) meter and \(3(1)=3\) meters.
Check: the lower contributions add to \(-2-1+1=-2\), and the upper contributions add to \(-1+1+3=3\). The negative bounds on the first interval correctly represent motion in the negative direction; treating all three regions as positive area would give the wrong quantity.
Worked Example: Allowing for Visual Resolution
Worked Example: Allowing for Visual Resolution
A plotted rate \(r(t)\), measured in kilograms per second, is shown for \(0\leq t\leq2\) seconds. The horizontal scale gives four intervals of width \(0.5\) second. Because the curve is thick and the grid is coarse, careful reading only establishes these height enclosures:
| Time interval (s) | Graph height throughout the interval (kg/s) |
|---|---|
| \([0,0.5]\) | from \(1.0\) to \(1.4\) |
| \([0.5,1]\) | from \(1.3\) to \(1.8\) |
| \([1,1.5]\) | from \(1.7\) to \(2.2\) |
| \([1.5,2]\) | from \(2.0\) to \(2.5\) |
Each lower contribution is its lower height times \(0.5\) second. Their total is \(0.5(1.0+1.3+1.7+2.0)=0.5(6.0)=3.0\) kilograms. The upper contributions total \(0.5(1.4+1.8+2.2+2.5)=0.5(7.9)=3.95\) kilograms.
Check: the lower heights sum to \(6.0\), and the upper heights sum to \(7.9\); multiplying each sum by the interval width \(0.5\) second gives the stated bounds. The units are \((\text{kg/s})(\text{s})=\text{kg}\). These are justified bounds only if the stated height enclosures really contain the curve across each interval. If they come from visual readings, report them as approximate graph-based bounds rather than exact coordinate facts.
The midpoint of the displayed bounds is \((3.0+3.95)/2=3.475\) kilograms. It can serve as a single rough estimate, but it is not an exact value or a guarantee that the true integral is especially close to it. The enclosure, not the midpoint, is the rigorous information provided by the graph.
When Endpoint Readings Are Enough—and When They Are Not
Endpoint values can provide interval bounds when the graph is known to be monotone on the interval. For an increasing function, the left endpoint is a lower height and the right endpoint is an upper height. For a decreasing function, their roles reverse. In either case, monotonicity is the reason the endpoint readings bound every intermediate value.
Without such information, the curve could rise above both endpoint values or fall below them between the endpoints. A small number of plotted points does not reveal every possible bend in the graph. In particular, do not infer that a curve stays between two heights merely because the visible samples do.
Common Pitfall: Treating Sampled Heights as Bounds
Consider \(f(x)=4x(1-x)\) on \([0,1]\). Its endpoint readings are \(f(0)=0\) and \(f(1)=0\). A mistaken calculation might use those two values as both the lower and upper height on the interval and conclude that \(\int_0^1 f(x)\,dx=0\).
The error is that the endpoint values do not bound the graph throughout \([0,1]\). In fact, the graph rises between the endpoints: \(f(1/2)=4(1/2)(1/2)=1\). The corrected graph information is \(0\leq f(x)\leq1\) on the interval, so the valid enclosure is
Check: the graph is nonnegative and reaches height \(1\), so the corrected height bounds include its interior values. The endpoint-only calculation falsely treats the entire curve as if it had height zero.
What a Graph Can—and Cannot—Establish
Graph-based estimates are useful when an exact antiderivative is unavailable, when only a plotted curve is supplied, or when a quick bound is more important than an exact value. The strength of the conclusion depends on the strength of the graph information.
- Exact labeled coordinates can give exact point values, but do not automatically bound neighboring values.
- Known monotonicity lets endpoint heights bound the function over an interval.
- A reliable horizontal enclosure gives a direct interval-wide height bound.
- A visually read curve supports only the precision allowed by its scale, thickness, and resolution.
- A graph below the axis contributes negatively to the definite integral.
Use more, narrower subintervals when the graph supports them: narrower enclosures can produce tighter integral bounds. But extra arithmetic cannot overcome missing information. A fine-looking grid is not proof that a thick, low-resolution curve has been read exactly.
Check Your Understanding
Use the graph-enclosure method and explain what information justifies each bound.
- A function increases from \(2\) to \(5\) on an interval of width \(3\). What lower and upper integral bounds follow from that information?
- On an interval of width \(0.5\), a graph stays between \(-4\) and \(-2\). What bounds does this give for the integral over that interval?
- Why are two exact endpoint values not enough, in general, to bound the integral between them?
- A graph’s horizontal grid spacing is \(0.25\) units and its vertical spacing is \(2\) units. What product of scale values corresponds to one grid rectangle?
- What should you report if the curve’s plotted thickness makes its height uncertain by \(0.1\) units?
