Definite Integration and Area Under Curves

The Relationship Between Integrals and Area

When calculating the region between a curve and an axis, we must distinguish between the numerical value of a definite integral and the physical area. In definite integration, the calculation behaves like a sum of areas where regions below the $x$ axis are subtracted rather than added. Specifically, it sums the areas of all regions above the axis and subtracts all regions that lie below it. This calculation is performed as a single operation.

This occurs because of how integration is constructed. You may have seen diagrams where integration is represented as a sum of many extremely thin rectangles between a curve and the $x$ axis. Each rectangle contributes a value equal to its height ($y$) multiplied by its infinitesimal width ($dx$). Since the width $dx$ is always positive, the sign of the contribution depends entirely on $y$. If a rectangle is beneath the $x$ axis, its $y$ value is negative, making its contribution to the integral negative. This is why definite integration treats these sections as negative areas.

Examples of Signed Integration

To understand this mechanism, consider the following examples where we calculate integrals and observe their geometric interpretations.

Example 1

Calculate $\int_0^3 x^2 \, dx$ and illustrate the result.

$\int_0^3 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^3 = \frac{27}{3} - \frac{0}{3} = 9$

In this case, the entire region is above the $x$ axis, so the definite integral and the geometric area are identical.

Example 2

Calculate $\int_{-3}^{0} x^3 \, dx$ and illustrate the result.

$\int_{-3}^{0} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{-3}^{0} = \frac{0}{4} - \frac{(-3)^4}{4} = -\frac{81}{4}$

Here, the region is entirely below the $x$ axis, resulting in a negative definite integral. However, the geometric area of this region would be reported as the positive value $\frac{81}{4}$.

Example 3

Calculate $\int_{-3}^{3} x^5 \, dx$ and illustrate the result.

$\int_{-3}^{3} x^5 \, dx = \left[ \frac{x^6}{6} \right]_{-3}^{3} = \frac{3^6}{6} - \frac{(-3)^6}{6} = 0$

The answer is zero because the function is antisymmetric about the origin. The negative area from $-3$ to $0$ exactly cancels out the positive area from $0$ to $3$.

Finding Geometric Area by Splitting Integrals

If an ESAT question asks for the total area between a curve and an axis between two $x$ values, simply calculating the definite integral over the whole range might produce an incorrect answer if the curve crosses the axis. To find the true geometric area, you must identify where the curve crosses the $x$ axis, calculate the area of each section separately using definite integrals, and then sum the absolute values (the positive versions) of these results.

Worked Example: Splitting the Region

Find the area between the $x$ axis, the lines $x = 0$ and $x = 2$, and the curve $y = x^2 - 1$.

First, we sketch the curve to identify where it lies relative to the axis:

The diagram shows the curve crosses the $x$ axis at $x = 1$. This splits our region into two parts: Area A (from 0 to 1) and Area B (from 1 to 2). Area A is below the axis and Area B is above it.

1. Calculate the integral for Area A: $\int_0^1 x^2 - 1 \, dx = \left[ \frac{x^3}{3} - x \right]_0^1 = \frac{1}{3} - 1 = -\frac{2}{3}$
2. Calculate the integral for Area B: $\int_1^2 x^2 - 1 \, dx = \left[ \frac{x^3}{3} - x \right]_1^2 = \left[ \frac{2^3}{3} - 2 \right] - \left[ \frac{1^3}{3} - 1 \right] = \frac{4}{3}$

Since Area A is $-\frac{2}{3}$, its geometric area is $\frac{2}{3}$. Adding this to the geometric area of B ($\frac{4}{3}$), we get a total area of $\frac{6}{3} = 2$. Note that the single definite integral from 0 to 2 would have given $\frac{4}{3} - \frac{2}{3} = \frac{2}{3}$, which is incorrect for the total area.

Integration with Respect to y

While not explicitly required by the specification, it is helpful to understand integrals with respect to $y$, such as $\int y^3 \, dy$. These follow the same principles as $x$ integrals, but they represent areas between the curve and the $y$ axis. Ensure the function is expressed in terms of $y$ before integrating.

Example: $\int_{-2}^{3} y^3 \, dy = \left[ \frac{y^4}{4} \right]_{-2}^{3} = \frac{3^4}{4} - \frac{(-2)^4}{4}$.

In the diagram above, the integral for section A would be negative while the integral for section B would be positive.

Using Symmetry and Antisymmetry

Symmetry can significantly simplify definite integrals.

1. Symmetry about the y axis (Even Functions): If $f(x)$ is symmetric about the $y$ axis, then $\int_{-a}^{-b} f(x) \, dx = \int_{b}^{a} f(x) \, dx$.

1. Antisymmetry (Odd Functions): If $f(x)$ is antisymmetric (reflecting across the $y$ axis and then the $x$ axis returns the same shape), then $\int_{-b}^{-a} f(x) \, dx = - \int_{a}^{b} f(x) \, dx$.

Using these properties, you can deduce that many integrals evaluate to zero or can be doubled. For instance, $\int_{-10}^{10} x^3 \, dx = 0$ because $x^3$ is antisymmetric. Similarly, $\int_{-\pi}^{\pi} x^2 \, dx = 2 \int_{0}^{\pi} x^2 \, dx$ because $x^2$ is symmetric. Although trigonometric integration is not on the syllabus, you may be expected to deduce through symmetry that $\int_{0}^{2\pi} \cos x \, dx = 0$ or that $\int_{-\pi}^{\pi} \sin x \, dx = 0$.