1. Idea of Area Using Definite Integrals
These notes focus on using definite integrals to compute the exact area of regions bounded by curves, lines, and axes. The definite integral acts like an addition machine that totals infinitely many thin slices of area. Once the boundaries of the region are understood, setting up the integral becomes straightforward.
The key idea: area is computed by taking the height of a thin slice of the region and multiplying by an extremely small width. When these contributions are added from one boundary to another, the definite integral gives the final area.
1.1. Where definite integrals fit in
Definite integrals provide the most general method of calculating area. Instead of relying on geometric formulas for triangles or circles, integrals allow the calculation of curved or irregular regions. Whenever a region is described using curves and lines, a definite integral helps convert the picture into an exact numerical answer.
2. Definition of Area Using a Definite Integral
Suppose a region is bounded above by a function \(y = f(x)\), below by the \(x\)-axis or another curve, and lies between two vertical boundaries \(x = a\) and \(x = b\). If the function is non-negative over this interval, the area is defined as:
\(\text{Area} = \int_a^b f(x)\, dx.\)
In cases where the bounds depend on multiple curves, the region may require subtracting two functions or even switching between \(x\)-integration and \(y\)-integration, depending on the shape.
2.1. Signed area and actual area
The definite integral \(\int_a^b f(x)\, dx\) gives the signed area. If \(f(x)\) dips below the \(x\)-axis, the integral becomes negative on that portion. For pure area, the region is split wherever necessary, and absolute values are taken.
These notes stick to regions where the sign is clear, or the boundaries are easy to handle.
3. Area Bounded by a Curve and the Coordinate Axes
Many regions are enclosed by a single curve and one or both coordinate axes. These are often the simplest to evaluate using integrals. The boundaries give the limits directly.
3.1. Region bounded by a curve and the x-axis
If the region lies between the graph of \(y = f(x)\), the \(x\)-axis, and the vertical lines at \(x = a\) and \(x = b\), the area is:
\(\text{Area} = \int_a^b f(x)\, dx.\)
This is the most common use of definite integrals for area.
3.2. Region bounded by a curve and the y-axis
If the region is enclosed by the curve \(y = f(x)\) and the y-axis (i.e., the vertical line \(x = 0\)), the area extends from \(x = 0\) to another boundary \(x = a\). The setup remains the same:
\(\text{Area} = \int_0^a f(x)\, dx.\)
The y-axis simply acts as a left boundary.
4. Area Bounded by Curves and Vertical or Horizontal Lines
Regions may also be bounded by lines of the form \(x = a\), \(x = b\), or \(y = c\). These lines usually serve as clean boundaries for the integral, making the region easy to visualize.
4.1. Vertical boundaries (x = a and x = b)
These lines give immediate limits for the integral. The region stretches from \(x = a\) to \(x = b\), and the height is determined by the upper minus lower function.
\(\text{Area} = \int_a^b [f_{\text{upper}}(x) - f_{\text{lower}}(x)]\, dx.\)
4.2. Horizontal boundaries (y = c and y = d)
When the region is easier to describe by horizontal slices, the integral is taken with respect to \(y\). If the left boundary is \(x = g(y)\) and the right boundary is \(x = h(y)\), the area is:
\(\text{Area} = \int_c^d [h(y) - g(y)]\, dy.\)
This method is especially useful if the curves cannot be written easily as \(y = f(x)\).
5. Using Integration with Respect to y
Sometimes a region cannot be captured using vertical slices because one x-value corresponds to multiple y-values. In such cases, switching to horizontal slicing simplifies the region dramatically.
The idea remains the same: width times height. But now width is the horizontal thickness \(dx = h(y) - g(y)\), and height is the small change \(dy\).
5.1. Choosing between x-integration and y-integration
To pick the right method, sketch the region first. If vertical slices intersect the region in one continuous segment, use \(dx\). If vertical slices break the region into multiple parts, switch to horizontal slices and integrate with respect to \(dy\).
6. Mixed-Boundary Regions
Some regions are enclosed by a combination of a curve, the x-axis, the y-axis, and a vertical or horizontal line. The essential task is to identify left, right, top, and bottom boundaries and decide whether slicing should be vertical or horizontal.
Sketching helps to detect whether the region needs to be split into multiple simpler parts, each handled with a separate integral.
6.1. When the region has multiple curves
In such cases, different sections of the region may have different upper or lower boundaries. Splitting the region ensures the formula always subtracts the correct lower function from the upper one.
7. Worked Examples
These examples show how definite integrals are used to compute area once the boundaries are clear. A sketch is always recommended before setting up the integrals.
7.1. Example 1: Area under a simple curve
Find the area bounded by the curve \(y = 3x - x^2\), the x-axis, and the vertical lines \(x = 0\) and \(x = 3\).
The region lies above the x-axis on this interval. Set up the integral:
\(\text{Area} = \int_0^3 (3x - x^2)\, dx.\)
Find the antiderivative:
\(\int (3x - x^2)\, dx = \frac{3x^2}{2} - \frac{x^3}{3}.\)
Evaluate from 0 to 3:
\(\left[\frac{3x^2}{2} - \frac{x^3}{3}\right]_0^3 = \left(\frac{27}{2} - 9\right) - 0 = \frac{27 - 18}{2} = \frac{9}{2}.\)
So the area of the region is \(\frac{9}{2}\) square units.
7.2. Example 2: Area using horizontal slicing
Find the area enclosed by the curves \(x = y^2\) and \(x = 4 - y^2\).
A sketch shows the region is symmetric about the x-axis. The curves intersect when:
\(y^2 = 4 - y^2 \Rightarrow 2y^2 = 4 \Rightarrow y^2 = 2 \Rightarrow y = \pm \sqrt{2}.\)
Using horizontal slices, the left boundary is \(x = y^2\) and the right boundary is \(x = 4 - y^2\). Thus the width of a slice is:
\((4 - y^2) - (y^2) = 4 - 2y^2.\)
The area is:
\(\text{Area} = \int_{-\sqrt{2}}^{\sqrt{2}} (4 - 2y^2)\, dy.\)
Find the antiderivative:
\(\int (4 - 2y^2)\, dy = 4y - \frac{2y^3}{3}.\)
Evaluate from \(-\sqrt{2}\) to \(\sqrt{2}\):
\(\left[4y - \frac{2y^3}{3}\right]_{-\sqrt{2}}^{\sqrt{2}} = \left(4\sqrt{2} - \frac{2(\sqrt{2})^3}{3}\right) - \left(-4\sqrt{2} + \frac{2(-\sqrt{2})^3}{3}\right).\)
Simplify step by step:
- \((\sqrt{2})^3 = 2\sqrt{2}\)
- \(- ( -4\sqrt{2} ) = +4\sqrt{2}\)
The result becomes:
\(= \left(4\sqrt{2} - \frac{4\sqrt{2}}{3}\right) - \left(-4\sqrt{2} + \left(-\frac{4\sqrt{2}}{3}\right)\right) = \frac{8\sqrt{2}}{3} + \frac{8\sqrt{2}}{3} = \frac{16\sqrt{2}}{3}.\)
The area of the region is \(\frac{16\sqrt{2}}{3}\) square units.
8. Observations and Notes
Some useful points that help when setting up area problems:
- The boundaries dictate the limits of integration.
- A sketch often reveals whether the region requires one integral or multiple pieces.
- Pick vertical or horizontal slices depending on which gives a simpler description of the region.
- Subtract lower from upper for vertical slices, and left from right for horizontal slices.
- Always check whether the integrand stays positive on the chosen interval.