From a Region to Thin Strips
A definite integral can represent geometric area, but for a region bounded by two curves we first need to describe what each small piece of that region looks like. Vertical slices provide one systematic way to do this. Each slice has a small horizontal width and extends vertically from the lower curve to the upper curve.
The key is to keep the two dimensions of a slice distinct. Its height comes from the vertical distance between the curves; its width comes from a small change in \(x\). Adding the areas of many such thin strips leads to an integral with respect to \(x\).
Describing the Region
Suppose \(f\) and \(g\) are continuous on the closed interval \([a,b]\), and suppose \(f(x)\ge g(x)\) for every \(x\) in that interval. Consider the region consisting of points whose \(x\)-coordinates lie between \(a\) and \(b\), and whose \(y\)-coordinates lie between the two curves:
At a fixed value of \(x\), a vertical line meets the region from \(y=g(x)\) at the bottom to \(y=f(x)\) at the top. The vertical distance between those two values is
This is why the slice’s height is upper function minus lower function. The subtraction measures a distance in the vertical direction: the top \(y\)-coordinate minus the bottom \(y\)-coordinate.
To approximate the area, divide \([a,b]\) into small subintervals. A subinterval from \(x_{i-1}\) to \(x_i\) has width \(\Delta x_i=x_i-x_{i-1}\). Choose a sample value \(\xi_i\) in that subinterval. The vertical distance at the sample value is \(f(\xi_i)-g(\xi_i)\), so a thin rectangle with that height and width has area
Adding these approximate areas gives an approximation for the area of the whole region. The widths are horizontal changes in \(x\), so the sums use \(\Delta x_i\). As the subintervals become arbitrarily narrow, the sums approach the definite integral with respect to \(x\).
Deriving the Formula
The derivation makes precise how the strip areas become an integral. Let \(P\) be a partition of \([a,b]\), written \(a=x_0<x_1<\cdots<x_n=b\). For each subinterval, let \(\Delta x_i=x_i-x_{i-1}\) and choose \(\xi_i\in[x_{i-1},x_i]\). The sum of the sample-rectangle areas is
As the greatest subinterval width approaches zero, these sums approach the area. This limit is the definite integral of the continuous height function \(f-g\). The continuity condition ensures that the height varies without jumps and that the rectangle sums converge to the integral.
The formula has three parts that each come from the geometry. The limits \(a\) and \(b\) mark the region’s horizontal extent. The difference \(f(x)-g(x)\) is the vertical height of a slice. The \(dx\) indicates that the slices have horizontal widths, so the sum is taken across \(x\).
At a particular \(x\), identify the top curve \(f(x)\) and the bottom curve \(g(x)\).
Subtract the bottom \(y\)-coordinate from the top \(y\)-coordinate: \(f(x)-g(x)\).
A strip near \(x\) has approximate area \((f(x)-g(x))\Delta x\).
As the widths shrink, the sum becomes \(\int_a^b(f(x)-g(x))\,dx\).
Worked Examples
Worked Example: A Line Below a Horizontal Curve
Find the area between \(y=3\) and \(y=x\) from \(x=0\) to \(x=2\).
First determine which function is upper. On this interval, \(x\le2<3\), so \(y=3\) is above \(y=x\). The vertical slice height is \(3-x\), and the width direction is \(x\). Therefore,
An antiderivative of \(3-x\) is \(3x-\frac{x^2}{2}\). Evaluating at the endpoints gives
Check: The vertical heights at the left and right endpoints are \(3\) and \(1\), and the horizontal width is \(2\). The trapezoid area is \(\frac{3+1}{2}\cdot2=4\), agreeing with the integral.
Worked Example: A Parabola Above a Horizontal Curve
Find the area between \(y=x^2+2\) and \(y=2\) from \(x=0\) to \(x=1\).
Since \(x^2\ge0\), the function \(x^2+2\) is at least \(2\) throughout the interval. Thus the slice height is \((x^2+2)-2=x^2\). The area is
An antiderivative of \(x^2\) is \(\frac{x^3}{3}\), so
Check: Differentiating \(\frac{x^3}{3}\) gives \(x^2\), the slice height. The height is nonnegative on \([0,1]\), and the computed area \(\frac13\) is positive, as an area should be.
Worked Example: A Constant Upper Curve
Find the area between \(y=5\) and \(y=x^2+1\) from \(x=-2\) to \(x=2\).
On this interval, \(x^2\le4\), so \(x^2+1\le5\). The constant curve is upper, and each vertical slice has height \(5-(x^2+1)=4-x^2\). Hence
An antiderivative is \(4x-\frac{x^3}{3}\). Substituting both endpoints yields
Check: The region fits inside a rectangle of width \(4\) and height \(4\), whose area is \(16\). The area between \(y=1\) and \(y=x^2+1\) is \(\int_{-2}^{2}x^2\,dx=\frac{16}{3}\). Removing that portion from the rectangle gives \(16-\frac{16}{3}=\frac{32}{3}\), matching the result.
Why the Difference Must Be in the Right Order
The formula depends on identifying the upper and lower functions correctly. Reversing them changes the sign of every slice height. For the first example, an incorrect setup would be
The calculation is consistent with its integrand, but \(x-3\) is negative on the interval: it measures bottom minus top, not a positive geometric height. The correction is to use upper minus lower, \(3-x\), which gives area \(4\). A negative result here signals that the subtraction order does not match the geometry.
The requirement that one function stay above the other across the interval matters. If the ordering does not hold throughout, \(f-g\) is not a nonnegative slice height everywhere, so this single formula does not directly describe the total geometric area. The formula derived here applies when the upper and lower roles are consistent on the interval.
Why Vertical Slices Lead to \(dx\)
A vertical slice is narrow in the horizontal direction and tall in the vertical direction. Its small width is a change in \(x\), denoted \(\Delta x\); when the widths shrink in the limit, that becomes \(dx\). Its height is measured by subtracting \(y\)-values. This geometric pairing explains the structure of the integral rather than treating \(dx\) as an arbitrary symbol.
For an area measured in square units, the dimensions also make sense: the slice height has vertical units, while its width has horizontal units. Multiplying them gives square units. For example, if the functions represent heights in meters and \(x\) is measured in meters, then \((f(x)-g(x))\,dx\) is measured in square meters.
The vertical-slice formula is useful because it translates a two-dimensional region into a one-variable accumulation. Once the boundaries are described by functions of \(x\) and their order is known, each slice contributes its height times its width, and the integral adds those contributions across the interval.
Check Your Understanding
Use the slice-height reasoning in each question before evaluating any integral.
- For a region between \(y=f(x)\) and \(y=g(x)\), with \(f(x)\ge g(x)\) on \([a,b]\), what is the height of a vertical slice?
- Why does adding the areas of vertical strips lead to an integral with respect to \(x\)?
- Set up, but do not evaluate, the area between \(y=4\) and \(y=x+1\) on \([0,2]\).
- If the lower function is mistakenly subtracted from the upper function in the wrong order, what does a negative integral indicate in this setting?
- State the continuity and ordering conditions used in the vertical-slice area formula.
