A definite integral has limits and evaluates to a number: integrate, write the result in square brackets, then substitute the upper limit and subtract the value at the lower limit, F(b) − F(a). No + c is needed, because it cancels. Definite integrals measure area: under a curve, between two graphs, and between limits found from points of intersection.
Definite integrals
- ∫ f(x) dx from a to b = [F(x)] from a to b = F(b) − F(a)
- Swapping the limits changes the sign
- Equal limits give zero
- Area under a curve above the x-axis = ∫ y dx from a to b
- Area between two graphs = ∫ (upper − lower) dx
How do you evaluate a definite integral?
Integrate as usual, but instead of adding + c, write the result in square brackets with the limits. Then substitute the upper limit, substitute the lower limit, and subtract. The constant of integration would appear in both and cancel, which is why it is left out.
Take particular care substituting negative limits: use brackets, such as (−2)³ = −8.
Surds and useful properties
Definite integrals involving roots work the same way once the roots are written as powers. Three properties save time: swapping the limits changes the sign of the answer, an integral with equal limits is zero, and an integral from a to c can be split at any point b in between.
Area under a curve
For a curve above the x-axis between x = a and x = b, the area of the region beneath it is the definite integral of y from a to b. This holds while the curve stays above the axis. A region below the axis gives a negative integral, so its area is the negative of that value.
Area between two graphs
The area enclosed between two graphs is the integral of the upper graph minus the lower graph. Always subtract lower from upper; checking a point in the middle of the interval tells you which is which. If the graphs cross inside the interval, the order swaps, so the region must be split.
Finding the limits from points of intersection
Often the limits are not given. Solve the two equations simultaneously to find where the graphs meet, and use those x-values as the limits. Writing the coordinates of the intersection points is often a separate mark.
Worked examples
Example 1. Evaluate ∫ 3x² dx from x = 1 to x = 2.
- Integrate: [x³] from 1 to 2.
- Substitute: 2³ − 1³ = 8 − 1.
Answer: 7.
Marks: M1 for integrating to x³, A1 for 7.
Example 2. Evaluate ∫ √x dx from x = 1 to x = 9.
- Integrate: [⅔x^(3/2)] from 1 to 9.
- Substitute: ⅔ × 27 − ⅔ × 1 = 18 − ⅔.
Answer: 52/3.
Marks: M1 for ⅔x^(3/2), A1 for 52/3.
Example 3. Find the area of the region under y = 6x − x² between x = 0 and x = 3.
- Area = ∫ (6x − x²) dx from 0 to 3 = [3x² − x³/3] from 0 to 3.
- Substitute: (27 − 9) − 0 = 18.
Answer: 18 square units.
Marks: M1 for setting up the integral, A1 for the integration, A1 for 18.
Example 4. Find the area enclosed between y = 4x and y = x².
- Intersections: x² = 4x, so x = 0 or x = 4.
- Between them, 4x is above x², so the area is ∫ (4x − x²) dx from 0 to 4.
- [2x² − x³/3] from 0 to 4 = 32 − 64/3.
Answer: 32/3 square units.
Marks: M1 for the limits, M1 for the integral, A1 for 32/3.
Example 5. The line y = x + 6 meets the curve y = x² at two points. Find them, and the area enclosed.
- x² = x + 6 gives x² − x − 6 = 0, so x = −2 or x = 3. The points are (−2, 4) and (3, 9).
- Area = ∫ ((x + 6) − x²) dx from −2 to 3 = [x²/2 + 6x − x³/3] from −2 to 3.
- = (9/2 + 18 − 9) − (2 − 12 + 8/3) = 27/2 + 22/3.
Answer: 125/6 square units.
Marks: M1 for the intersections, M1 for the integral, A1 for 125/6.
Common mistakes
- Subtracting the wrong way round. Evaluate F(upper) − F(lower).
- Sign errors with negative limits. Use brackets when substituting a negative value.
- Adding + c to a definite integral. It cancels, so leave it out.
- Integrating lower minus upper. For the area between graphs, subtract the lower graph from the upper one.
- Treating a negative integral as an area. A region below the x-axis gives a negative value; its area is the magnitude.
How definite integrals are examined
WMA12, the Edexcel International A-Level Pure 2 paper, almost always includes a definite integral or an area; in Edexcel A-Level (9MA0) the topic belongs to Year 1. Questions commonly:
- evaluate a definite integral, often with surds or fractional powers
- find the area of a shaded region under a curve
- find intersection points, then the area between a curve and a line
- show that an area equals a given exact value
For diagram-based area questions with full marking, go to the Pure 2 integration exam questions. Use the Pure 2 integration worksheet and the integration revision engine for more practice.
For regions that cross the axis or lie next to the y-axis, see area under a curve. For harder regions, see area between two curves.
Frequently asked questions
Why is there no + c in a definite integral?
Because it appears in both F(b) and F(a) and cancels when you subtract.
What does a negative definite integral mean?
That more of the region lies below the x-axis than above. The area is always positive, so take the magnitude of each part.
How do you find the area between a curve and a line?
Find where they meet, then integrate (upper − lower) between those x-values.