1. Idea of Area Between Curves
These notes explain how to measure the area of a region trapped between two curves. When two functions form the upper and lower boundaries of a region, the area is found by subtracting one curve from the other and integrating. The idea is simple: if one curve lies above the other, then at each position the small vertical slice of area has height equal to the vertical difference between the two curves.
The full area is obtained by adding all such thin slices from the left boundary to the right boundary. This is why definite integrals naturally appear in this topic.
1.1. The basic picture
Suppose two curves \(y_1(x)\) and \(y_2(x)\) enclose a region. If \(y_1(x) \ge y_2(x)\) throughout the interval \([a, b]\), then the region between them can be viewed as made of vertical slices whose height is \(y_1(x) - y_2(x)\). Each slice has width \(dx\), so its area is \( [y_1(x) - y_2(x)]\, dx \). Adding all slices gives a definite integral.
This leads directly to the standard formula that will be used repeatedly.
2. Definition of Area Between Curves
Let \(y_1(x)\) and \(y_2(x)\) be continuous on a closed interval \([a, b]\), and suppose that \(y_1(x) \ge y_2(x)\) for all \(x\) in the interval. Then the area of the region between the two curves is defined as:
\(\text{Area} = \int_a^b [y_1(x) - y_2(x)]\, dx.\)
Here, \(y_1\) forms the upper boundary, \(y_2\) forms the lower boundary, and the vertical lines \(x = a\) and \(x = b\) serve as the left and right boundaries.
2.1. When curves intersect inside the interval
If the curves intersect at some point within the interval, the upper and lower functions may switch roles. In that case, the interval has to be split at the intersection point(s), and the difference is taken with correct ordering on each sub-interval.
The intersection points therefore play a central role in calculating area between curves.
3. Finding Points of Intersection
The left and right boundaries of the region are often determined by the intersection of the two curves. To find these points, set
\(y_1(x) = y_2(x).\)
Solving this equation gives the required intersection points. These points become the limits of integration if the region is enclosed entirely between them.
Sometimes the boundaries are explicitly given instead of being intersection points, but the method remains the same: identify left and right boundaries, then subtract the lower function from the upper one.
3.1. How many intersection points to expect
Sometimes two curves intersect once, sometimes twice, and sometimes not at all. When there are two intersection points, the area between curves is typically the region between them on that entire interval. If there is only one intersection point, the other boundary may come from a vertical line given in the problem. If the curves never intersect, limits are taken from the described boundaries.
4. Standard Formula for Area Between Curves
Once the upper and lower curves are identified, the formula for area between them is:
\(\boxed{\text{Area} = \int_a^b \big( y_1(x) - y_2(x) \big)\, dx}\)
This formula applies when vertical slices are used. The process can also be reversed by using horizontal slices, but this is only needed when the curves cannot be expressed easily as functions of \(x\).
4.1. Horizontal slicing (less common)
If the curves are easier to express as \(x = g(y)\) and \(x = h(y)\), the region can be sliced horizontally. The left boundary is \(x = g(y)\) and the right boundary is \(x = h(y)\). The height of each slice is \(dy\), and its width is \(h(y) - g(y)\). The area formula becomes:
\(\text{Area} = \int_c^d [h(y) - g(y)]\, dy.\)
This method is helpful when vertical slicing becomes impossible or leads to multiple integrals.
5. Graph Picture and Intuition
To understand the region between curves, it helps to sketch a rough diagram with the following elements:
- The two curves clearly drawn.
- The left and right boundaries marked, usually the intersection points.
- The curve that stays above shown distinctly from the one below.
- One or two sample vertical slices drawn to visualize how the height is computed.
A correct sketch makes it immediately clear whether one needs to split the region into parts or whether a single integral is enough.
5.1. Understanding non-overlapping regions
Some problems involve holes or parts of the graph where the two curves do not meet. These cases require careful sketching. The main rule is simple: take slices only from the region that lies between the curves. If the curves cross multiple times, the problem splits naturally into separate integrals.
6. Worked Examples
These examples illustrate how to apply the idea of taking the upper curve minus the lower curve and integrating between the correct limits. The steps remain the same in every problem: sketch, find intersections (if needed), identify upper and lower curves, then integrate their difference.
6.1. Example 1: Area between \(y = x\) and \(y = x^2\)
The curves \(y = x\) and \(y = x^2\) intersect when:
\(x = x^2 \Rightarrow x^2 - x = 0 \Rightarrow x(x - 1) = 0.\)
So the intersection points are \(x = 0\) and \(x = 1\). On this interval, \(y = x\) lies above \(y = x^2\).
Thus the area is:
\(\text{Area} = \int_0^1 (x - x^2)\, dx.\)
Compute the antiderivative:
\(\int (x - x^2)\, dx = \frac{x^2}{2} - \frac{x^3}{3}.\)
Now evaluate from \(0\) to \(1\):
\(\left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \left( \frac{1}{2} - \frac{1}{3} \right) - 0 = \frac{1}{6}.\)
So the area between the two curves is \(\frac{1}{6}\) square units.
6.2. Example 2: Area between curves with no intersection inside the interval
Take the curves \(y = 4\) and \(y = x^2\) on the interval \([-1, 1]\). Here, \(y = 4\) is always the upper curve and \(y = x^2\) is the lower curve.
The area is:
\(\text{Area} = \int_{-1}^1 (4 - x^2)\, dx.\)
Find the antiderivative:
\(\int (4 - x^2)\, dx = 4x - \frac{x^3}{3}.\)
Evaluate from \(-1\) to \(1\):
\(\left[4x - \frac{x^3}{3}\right]_{-1}^1 = \left(4 - \frac{1}{3}\right) - \left(-4 + \frac{1}{3}\right).\)
Simplify:
\(= \left(\frac{12}{3} - \frac{1}{3}\right) - \left(-\frac{12}{3} + \frac{1}{3}\right) = \frac{11}{3} - (-\frac{11}{3}) = \frac{22}{3}.\)
So the area between the curves \(y = 4\) and \(y = x^2\) on \([-1, 1]\) is \(\frac{22}{3}\) square units.
7. Observations and Notes
Some useful ideas to keep in mind while working with area between curves:
- The upper curve is always the one with the larger \(y\)-value at each \(x\).
- Intersection points give natural limits for the integral.
- If the curves cross, the integral must be split at the crossing point.
- Horizontal slicing is rarely needed but becomes useful when the region cannot be captured by a single vertical slice.
- A quick sketch helps avoid sign errors and shows whether the computed area is reasonable.