Integration and Areas for ESAT Mathematics 2

Definite Integration and the Area Between a Curve and an Axis

In the context of the ESAT, you must understand the distinction between a definite integral and the geometric area between a curve and an axis. While area is typically a positive value, a definite integral is a calculation that treats regions below the $x$-axis as negative. Specifically, a definite integral sums all areas above the $x$-axis and subtracts all areas below the $x$-axis in a single operation.

This occurs because integration can be viewed as the sum of infinitely many thin rectangles. The contribution of each rectangle is its height, given by the $y$-value, multiplied by its width, $dx$. When a curve is below the $x$-axis, the $y$-values are negative, leading to a negative contribution to the total integral.

Example 1

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

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

Example 2

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

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

Example 3

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

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

The result is zero because $x^5$ is an odd function: the negative area between $-3$ and $0$ exactly cancels out the positive area between $0$ and $3$.

Finding Total Area Between a Curve and the x-Axis

To find the total geometric area (where all parts are positive), you cannot simply compute the integral across the whole range if the curve crosses the $x$-axis. Instead, you must:

1. Identify where the curve crosses the $x$-axis.
2. Calculate the definite integral for each section separately.
3. Take the absolute (positive) value of each result and add them together.

Worked Example

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

The curve crosses the $x$-axis at $x = 1$. We split the calculation:

For the region 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 the region from $1$ to $2$: $\int_1^2 (x^2 - 1) dx = [\frac{x^3}{3} - x]_1^2 = (\frac{8}{3} - 2) - (\frac{1}{3} - 1) = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$

The total area is $|-\frac{2}{3}| + |\frac{4}{3}| = \frac{2}{3} + \frac{4}{3} = 2$. Note that the single definite integral $\int_0^2 (x^2 - 1) dx$ would only give $\frac{2}{3}$.

Integrating with respect to y

While less common, you may encounter integrals with respect to $y$, written as $\int f(y) dy$. These are calculated using the same rules as $x$. For example:

$\int_{-2}^{3} y^3 dy = [\frac{y^4}{4}]_{-2}^{3} = \frac{3^4}{4} - \frac{(-2)^4}{4} = \frac{81}{4} - \frac{16}{4} = \frac{65}{4}$

Using Symmetry to Simplify Integrals

Graphs with specific symmetries allow for faster integration:

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

1. Antisymmetric (Odd functions): If $f(-x) = -f(x)$, then $\int_{-b}^{-a} f(x) dx = -\int_{a}^{b} f(x) dx$. This implies $\int_{-a}^{a} f(x) dx = 0$.

Even though trigonometric integration is not explicitly in the ESAT specification, you should be able to deduce that $\int_0^{2\pi} \cos x dx = 0$ or $\int_{-\pi}^{\pi} \sin x dx = 0$ based on these symmetry properties.

Rules for Integrating Powers of x

In the ESAT, you are expected to integrate sums of terms in powers of $x$ using the rule:

$\int k x^n dx = \frac{k x^{n+1}}{n+1} + c$ for $n \neq -1$

If an expression does not look like a simple power of $x$, you must simplify it first. For example, to integrate $\int (x-2)^2 dx$, expand the brackets first: $\int (x^2 - 4x + 4) dx = \frac{x^3}{3} - 2x^2 + 4x + c$. Integration can be performed term by term.

Caution: You cannot integrate a fraction by integrating the numerator and denominator separately. You must simplify the expression into a sum of powers first, such as $\int \frac{(3x-5)^2}{x^{1/2}} dx$, which requires expanding and dividing each term by $x^{1/2}$.

The Fundamental Theorem of Calculus

The theorem links differentiation and integration in two main ways:

1. Evaluation: $\int_a^b f(x) dx = F(b) - F(a)$, where $F'(x) = f(x)$. This justifies why swapping limits introduces a negative sign: $\int_a^b f(x) dx = -\int_b^a f(x) dx$.

2. Differentiation of an Integral: $\frac{d}{dx} \int_a^x f(t) dt = f(x)$. This shows that differentiating an integral returns the original function.

You can also split integrals over contiguous ranges: $\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx$. This holds even if $c$ is not between $a$ and $b$, provided the function is defined across the entire relevant interval.

Approximating Areas with the Trapezium Rule

The Trapezium Rule estimates the area under a curve using $n$ trapezia of equal width $h$. The width $h$ is calculated as $h = \frac{b-a}{n}$.

The formula for a single trapezium is $Area = \frac{h}{2}(a+b)$. For multiple strips, the heights (y-values) at each interval $x_0, x_1, \dots, x_n$ are used:

$Area \approx \frac{h}{2}(y_0 + 2y_1 + 2y_2 + \dots + 2y_{n-1} + y_n)$

Overestimates and Underestimates

You can determine if an estimate is an overestimate or underestimate by looking at the curve's shape:

• If the curve bends away from the $x$-axis (convex/concave up), the trapezium rule gives an overestimate.
• If the curve bends toward the $x$-axis (concave/concave down), it gives an underestimate.

Solving Differential Equations of the Form dy/dx = f(x)

To solve $\frac{dy}{dx} = f(x)$, you must find the function $y$ that differentiates to $f(x)$ by integrating both sides.

Method 1: Using the Constant of Integration

Find the general solution $y = \int f(x) dx + c$. If a boundary condition is given (e.g., $y=5$ when $x=1$), substitute these values to solve for $c$.

Method 2: Definite Integration

Set up a definite integral with corresponding limits: $\int_5^y dy = \int_1^x f(t) dt$ This leads directly to the specific solution without needing to solve for $c$ separately.