Definite Integration and Area for the TMUA

Introduction to Integration

There are two primary ways to conceptualise integration. These methods are deeply related, though the specific mechanics of that relationship are beyond the scope of the TMUA.

First, integration is the reverse of differentiation. It identifies what function must have been differentiated to produce a given expression. This is written as an indefinite integral:

$\int x^2 dx = \frac{x^3}{3} + c$

We include the constant of integration, $c$, because the derivative of any constant is zero. Thus, multiple functions can share the same derivative.

Second, integration can be viewed as finding the area between a curve and the $x$ axis. When we use numbers at the top and bottom of the integral sign, called limits, we are performing definite integration:

$\int_{1}^{3} x^2 dx$

Definite Integration as Signed Area

In definite integration, the term area must be used with caution. Geometric area is traditionally a positive value, but a definite integral calculates a signed sum. It sums the areas above the $x$ axis and subtracts the areas below it in a single calculation.

This occurs because integration can be modeled as the sum of infinitely thin rectangles of width $dx$ and height $y$. For regions below the $x$ axis, the $y$ values are negative, so the product $y \times dx$ contributes a negative value to the total integral.

Worked Example 1

Calculate $\int_{0}^{3} x^2 dx$ and illustrate the result.

$\int_{0}^{3} x^2 dx = [\frac{x^3}{3}]_{0}^{3} = \frac{27}{3} - \frac{0}{3} = 9$

Worked Example 2

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

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

Worked Example 3

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

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

The answer is zero because the function $y = x^5$ is antisymmetric. the negative area between $-3$ and $0$ perfectly cancels the positive area between $0$ and $3$.

The Difference Between Integrals and Geometric Area

If a TMUA question asks for the area between a curve and the $x$ axis, a single definite integral across the whole range will likely give the wrong answer if the curve crosses the axis. To find the total geometric area, you must:

1. Identify the roots of the function to see where it crosses the $x$ axis.
2. Calculate the definite integral for each region separately.
3. Take the absolute (positive) value of each result.
4. Add these positive values together.

Worked Example: Finding Total Area

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

We must split the calculation at the root $x = 1$.

For Region A (from $0$ to $1$): $\int_{0}^{1} (x^2 - 1) dx = [\frac{x^3}{3} - x]_{0}^{1} = (\frac{1}{3} - 1) - 0 = -\frac{2}{3}$

For Region B (from $1$ to $2$): $\int_{1}^{2} (x^2 - 1) dx = [\frac{x^3}{3} - x]_{1}^{2} = (\frac{8}{3} - 2) - (\frac{1}{3} - 1) = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$

Area A is $\frac{2}{3}$ (positive magnitude) and Area B is $\frac{4}{3}$. The total area is $\frac{2}{3} + \frac{4}{3} = 2$. Note that the single definite integral $\int_{0}^{2} (x^2 - 1) dx$ would have given $\frac{2}{3}$, which is the signed sum, not the total area.

Integration with respect to y

You may also encounter integrals with respect to $y$, such as $\int y^3 dy$. These are calculated using the same power rules as $x$, but the regions are measured between the curve and the $y$ axis.

Integration Symmetry Tricks

Symmetry can significantly simplify definite integrals.

Symmetric (Even) Functions: If $f(x)$ is symmetric about the $y$ axis, then $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$. This applies to functions like $y = x^2$ or $y = \cos x$.

Antisymmetric (Odd) Functions: If $f(x)$ is antisymmetric (reflecting in $y$ and then $x$ returns the same curve), then $\int_{-a}^{a} f(x) dx = 0$. This applies to $y = x^3$, $y = \sin x$, and $y = \tan x$.