# The definite integral and areas · Year 12

> The definite integral and areas, Year 12: The definite integral and more. Explanation, worked examples and free practice for the national curriculum in England.

In The definite integral and areas (Year 12) you practise 5 skills: The definite integral · The area between a graph and the axis · An area on both sides of the axis · The area between two graphs · Finding a parameter.

The definite integral is worked out with an antiderivative: find an antiderivative, put in the upper limit and the lower limit, and take the second value away from the first. A number over the square of a linear expression has an antiderivative with the expression itself in the denominator, and a minus sign.

When the graph is above the horizontal axis, the integral is equal to the area between the graph and the axis. When the graph is below the axis, the integral is negative, and the area is its negative. When the graph crosses the axis inside the interval, split the interval where it crosses and add the absolute values of the parts.

The area between two graphs is the integral of the upper graph minus the lower graph, between the points where they meet. When a bound or a coefficient is not known, write the integral with it and set it equal to the given value.

**Tip:** Before working out an area, check where the graph is above the axis and where it is below it.

## What you learn

### The definite integral

- Example: Work out the integral ∫₁⁴ 7 dx .
- Answer: 21
- Solution: An antiderivative is F(x) = 7x . The integral: F(4) − F(1) , that is 28 − 7 = 21 .

### The area between a graph and the axis

- Example: Find the area between the line y = x , the horizontal axis and the line x = 6 . (A coordinate grid with the graph of the function, and the region between it and the horizontal axis shaded)
- Answer: 18
- Solution: The integral: ∫₀⁶ x dx = 18 . The area is 18 .

### An area on both sides of the axis

- Example: Find the total area between y = x and the horizontal axis on the interval −1 ≤ x ≤ 6 . (A coordinate grid with the graph of the function, and the region between it and the horizontal axis shaded)
- Answer: 37/2
- Solution: The graph crosses the axis at x = 0 . The first part: ∫₋₁⁰ x dx = −1/2 . The second part: ∫₀⁶ x dx = 18 . The total area: 1/2 + 18 = 37/2 .

### The area between two graphs

- Example: Find the area enclosed by the graphs y = 1 and y = x² . (A coordinate grid with the two graphs, and the closed region between them shaded)
- Answer: 4/3
- Solution: The graphs meet at x = −1 and at x = 1 . Between them the graph y = 1 is the upper one, so the area is ∫₋₁¹ (−x² + 1) dx , that is 4/3 .

### Finding a parameter

- Example: The definite integral of 2x from x = 0 to x = a is 9 . Find a , where a > 0 .
- Answer: 3
- Solution: An antiderivative is x² , so the integral is a² − 0 = 9 . Since a > 0 , this gives a = 3 .

The exercises are made by rules and the answers are worked out in code. The topic follows selected AS and A level mathematics practice in England (the GCE subject content).

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