Integration (AQA A-Level Mathematics): Revision Notes

1. Find the Points of Intersection: Determine the points $\ x = a \ and \ x = b$ where the curve $\ y = f(x)$ and the line $\ y = g(x)$ intersect. These points are found by solving the equation $f(x) = g(x) .$
2. Set Up the Integral: The area between the curve and the line is found by integrating the difference between the two functions over the interval $[a, b]$: $\\ \text{Area} = \int_a^b \left| f(x) - g(x) \right| \, dx$ If $\ f(x) \geq g(x)$ over the interval, the absolute value is unnecessary, and the formula simplifies to:

$\text{Area} = \int_a^b \left( f(x) - g(x) \right) \, dx\\$ If the functions cross each other within the interval, you need to split the integral at the crossing point(s).

3. Evaluate the Integral: Compute the definite integral to find the area.

Example Let's find the area between the curve $\ y = x^2$ and the line $\ y = x + 2$ over the interval where they intersect.

Step 1: Find the Points of Intersection

Set the equations equal to each other to find the points of intersection: $\\ x^2 = x + 2$

Rearrange to form a quadratic equation: $\\ x^2 - x - 2 = 0$

Factor the quadratic equation: $\\ (x - 2)(x + 1) = 0$

The solutions are $\ x = 2 \ and \ x = -1 .$

So, the curve and the line intersect at $\ x = -1 \ and \ x = 2$.

Step 2: Set Up the Integral

Since $\ x^2 \$is generally less than $\ x + 2$ over the interval $\ [-1, 2]$ (we know this from checking or plotting the functions), the area is given by: $\\ \text{Area} = \int_{-1}^2 \left( (x + 2) - x^2 \right) \, dx$

Step 3: Evaluate the Integral

First, set up the integrand: $\\ \text{Area} = \int_{-1}^2 \left( x + 2 - x^2 \right) \, dx$

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

Evaluate this antiderivative from $\ -1 \ to \ 2$: $\\ \text{Area} = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^2$

Substitute the limits: $\\ \text{Area} = \left( \frac{2^2}{2} + 2(2) - \frac{2^3}{3} \right) - \left( \frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^3}{3} \right)$

$\\ \text{Area} = \left( \frac{4}{2} + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right)$

$\\ \text{Area} = \left( 2 + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right)$

Simplify each part: $\\ \text{Area} = \left( 6 - \frac{8}{3} \right) - \left( -\frac{3}{2} + \frac{1}{3} \right)$

Let's simplify further:

• For the first part: $\ 6 = \frac{18}{3} \ , so \ \frac{18}{3} - \frac{8}{3} = \frac{10}{3} .$
• For the second part: Convert to a common denominator: $\\ -\frac{3}{2} = -\frac{9}{6}, \quad \frac{1}{3} = \frac{2}{6}, \quad \Rightarrow \quad -\frac{9}{6} + \frac{2}{6} = -\frac{7}{6}$

Finally, compute the total area: $\\ \text{Area} = \frac{10}{3} + \frac{7}{6} = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} = :highlight[4.5 square units]$

• Intersection Points: Find where the curve and line intersect to determine the limits of integration.
• Integrate: Set up the integral of the difference between the line and the curve, and integrate over the interval.
• Compute the Area: Evaluate the integral to find the area between the curve and the line.