Area Between Curves and Changing Roles

Overview

Central question: Why must top minus bottom be rebuilt when curves trade places? You will construct area from thin rectangles, decide whether vertical or horizontal slices make the region simplest, and split an integral at every point where the description of a slice changes. Area is nonnegative, so the integral must represent a nonnegative slice length throughout each interval.

Consider the curves $y=x$ and $y=x^2$ on $0\le x\le1$. Before writing an integral, sketch their values at $x=0$, $x=\frac12$, and $x=1$. Between the intersections, the line lies above the parabola. A vertical slice therefore has height $x-x^2$ and width $dx$.

Build area from one slice

The area of one narrow vertical rectangle is approximately

$$\left(\text{top function}-\text{bottom function}\right)\Delta x.$$

Adding increasingly narrow rectangles produces the exact integral

$$A=\int_0^1\left(x-x^2\right)\,dx =\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 =\frac16.$$

The integrand is a vertical distance, not the difference of two arbitrary formulas. Its order comes from the geometry of the slice. If the functions were reversed, the result would be negative even though geometric area cannot be negative.

Find intersections before choosing bounds

The visible boundary can change only at intersections or at stated endpoints. Solve $x=x^2$ to obtain $x(x-1)=0$, so the intersection inputs are $0$ and $1$. These values are not merely where curves touch on a graph; they determine the interval on which one top-minus-bottom description remains true.

For a more revealing contrast, consider $y=x^3$ and $y=x$. They intersect at $x=-1$, $0$, and $1$. On $-1<x<0$, $x^3$ lies above $x$. On $0<x<1$, $x$ lies above $x^3$. One unsplit integral would contain a sign change and measure signed cancellation rather than total area. The correct model is

$$A=\int_{-1}^{0}\left(x^3-x\right)\,dx +\int_{0}^{1}\left(x-x^3\right)\,dx.$$

Choose the direction that keeps slices simple

Vertical slices use $dx$ and require top minus bottom. Horizontal slices use $dy$ and require right minus left. Either direction can describe the same region, but one may avoid splitting a function into branches or solving for an inverse.

Suppose a region is bounded by $x=y^2$ and $x=y+2$. A horizontal slice has length $(y+2)-y^2$, so it is natural to use $dy$. Before integrating, solve $y^2=y+2$ to find the $y$-bounds. The choice is not about a personal preference for $dx$ or $dy$; it is about whether the slice crosses the region in one uninterrupted segment with a clear pair of boundaries.

Validate a geometric model

An area setup should pass three quick tests before it is evaluated. First, choose a sample point in each interval and verify which boundary is above, below, left, or right. A single test point is enough only after intersections have divided the domain into intervals where the order cannot change without another intersection. Second, check the dimensions: a slice height has units of length, and multiplying by a slice width gives square units. Third, compare the answer with a rough picture. A small region between nearby curves should not produce an enormous area.

These checks matter when equations are more complicated than the examples above. A graphing tool can help locate likely intersections, but it does not replace solving or checking them. A low-resolution graph can hide a tangency, a narrow lobe, or an intersection outside the visible window. Use the graph to form a geometric hypothesis, then establish the exact boundaries symbolically when the problem supplies formulas.

The same framework applies to areas bounded by more than two curves. Trace the outer boundary first, identify the slice direction that crosses the region in one segment, and list every place where that segment changes an endpoint. If one vertical slice crosses two disjoint pieces of the region, horizontal slices may be simpler. The best representation is the one that makes “one small area” unambiguous.

Go deeper: signed difference versus total area

The integral $\int_a^b[f(x)-g(x)]\,dx$ is a signed net vertical difference. It is equal to geometric area only when $f(x)\ge g(x)$ everywhere on the interval. When order changes, total area is $\int_a^b|f(x)-g(x)|\,dx$, evaluated by splitting at every zero of $f-g$. This absolute-value statement packages the same geometry as the split integrals, but it does not eliminate the need to find the intersections.

Stop and check

A tempting setup for the region between $y=x$ and $y=x^3$ is $\int_{-1}^{1}(x-x^3)\,dx$. The misconception is treating a signed difference as geometric area. Over the negative half of the interval, this integrand is negative because the functions have exchanged vertical order. The repair is to split at $x=0$ and write the positive slice height separately on each side.

Retrieval check

Set up the area between $y=2x$ and $y=x^2$ from their intersections. First solve $x^2=2x$ to obtain $x=0$ and $x=2$. On $0<x<2$, test $x=1$: the line has value $2$ and the parabola has value $1$, so the setup is

$$\int_0^2\left(2x-x^2\right)\,dx.$$

Explain why the integrand is nonnegative on the entire interval. The next lesson turns a planar slice into a cross-sectional area and then accumulates those areas to make a volume.