Area Under a Curve (Edexcel A-Level Mathematics): Revision Notes

The area under a curve $y = f(x)$ between $x = a$ and $x = b$ is represented by the definite integral:

$\text{Area} = \int_a^b f(x) \, dx$

This integral computes the net area between the curve and the $x$-axis over the interval $[a, b]$.

1. Set Up the Integral: Identify the function $f(x)$ and the interval $[a, b]$ over which you want to calculate the area.
2. Integrate: Find the antiderivative $F(x)$ of $f(x)$ .
3. Evaluate the Definite Integral: Use the Fundamental Theorem of Calculus to evaluate the integral:

$\text{Area} = \int_a^b f(x) \, dx = F(b) - F(a)$

Example 1: Area Under $y = x^2$ from $x = 0$ to $x = 2$

• Step 1: Set up the integral:

$\text{Area} = \int_0^2 x^2 \, dx$

• Step 2: Find the antiderivative:

$\int x^2 \, dx = \frac{x^3}{3} + C$ So, $F(x) = \frac{x^3}{3} .$

• Step 3: Evaluate the integral from $0$ to $2$:

$\text{Area} = \left[\frac{x^3}{3}\right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} - 0 = \frac{8}{3} \text{ square units}$

Example 2: Area Under $y = \sin(x)\ from\ x = 0\ to\ x = \pi$

• Step 1: Set up the integral:

$\text{Area} = \int_0^\pi \sin(x) \, dx$

• Step 2: Find the antiderivative:

$\int \sin(x) \, dx = -\cos(x) + C$

So, $F(x) = -\cos(x) .$

• Step 3: Evaluate the integral from $0$ to $\pi:$

$\text{Area} = \left[-\cos(x)\right]_0^\pi = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 \text{ square units}$

• Area Between the Curve and the x-Axis: If the curve lies below the x-axis (i.e., $f(x) \leq 0$ over the interval), the definite integral gives a negative value. To find the actual area, take the absolute value of the integral:

$\text{Area} = \left|\int_a^b f(x) \, dx\right|$

• Area Between Two Curves: To find the area between two curves $y = f(x)$ and $y = g(x)$ over [a, b], where $f(x) \geq g(x)$ for all $x$ in $[a, b]$, subtract the integral of the lower curve from the integral of the upper curve:

$\text{Area} = \int_a^b \left[f(x) - g(x)\right] \, dx$

Example: Area Between $y =$ $x^2$ and $y = x + 2$ from $x = 0$ to $x = 1$

• Step 1: Set up the integral:

$\text{Area} = \int_0^1 \left[(x + 2) - x^2\right] \, dx = \int_0^1 \left(x + 2 - x^2\right) \, dx$

• Step 2: Integrate:

$\int \left(x + 2 - x^2\right) \, dx = \frac{x^2}{2} + 2x - \frac{x^3}{3} + C$

• Step 3: Evaluate from $0$ to $1$:

$\text{Area} = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_0^1 = \left(\frac{1}{2} + 2 - \frac{1}{3}\right) - \left(0\right) = \frac{1}{2} + 2 - \frac{1}{3} = \frac{3}{6} + \frac{12}{6} - \frac{2}{6} = \frac{13}{6} \text{ square units}$

• The area under a curve between two points is found using definite integration.
• The area represents the accumulated quantity, such as distance, revenue, or probability, depending on the context.
• The definite integral $\int_a^b f(x) \, dx$ gives the net area between the curve and the $x$-axis from $x = a$ to $x = b$ .
• Special cases like the area between curves or regions below the x-axis are handled by modifying the basic integration process.

Example: Find the area of the shaded region in the following diagram.

1. $y = x^2$

2. $\int_{2}^{4} x^2 \, dx = \left[ \frac{1}{3} x^3 \right]_{2}^{4}\\ = \left( \frac{1}{3} \times 4^3 \right) - \left( \frac{1}{3} \times 2^3 \right)\\ = \frac{64}{3} - \frac{8}{3}\\ = \frac{56}{3}$

3. Area is $\frac{56}{3}$

However, such integrals will not always give the area between a curve and the $x$-axis.

Example:

$\therefore$ If we want to know the area and we know part of the curve lies under the $x$-axis, we should integrate this separately from any positive areas.

Thus,

$|x|$ is the modulus/magnitude/length of a quantity (i.e., distance from $0$).

Examples: 4. $|-7| = 7$ 5. $|2| = 2$