A definite integral is the change in the integrated function between two limits. Integrate first, then substitute the upper limit, then the lower limit, and subtract.
This lesson is part of SPM Additional Mathematics integration. It builds on finding an indefinite integral and its constant.
How do I evaluate one?
Write the integral in square brackets with the limits, then subtract. The notation is [F(x)] from a to b = F(b) − F(a).
Evaluate ∫ (2x + 3) dx from 1 to 3.
[x² + 3x] from 1 to 3 = (3² + 3×3) − (1² + 3×1) = (9 + 9) − (1 + 3) = 18 − 4 = 14.
Evaluate ∫ (3x² − 2x) dx from 0 to 2: [x³ − x²] from 0 to 2 = (8 − 4) − (0 − 0) = 4.
Worked example: an unknown limit
Given ∫ 2x dx from 1 to k equals 15, find k, where k > 1.
[x²] from 1 to k = k² − 1.
So k² − 1 = 15, which gives k² = 16 and k = 4 or k = −4.
The condition k > 1 rejects −4, so k = 4. Check: 4² − 1 = 15.
The mistake that costs marks
The common slip is to substitute only the upper limit, or to subtract in the wrong direction. Both give a wrong number, and the working still looks tidy.
Take ∫ (2x + 3) dx from 1 to 3 again.
| Step | Wrong | Right |
|---|---|---|
| Substitute | Upper only: 18 | Upper 18, lower 4 |
| Subtract | (no subtraction) | 18 − 4 |
| Reversed | 4 − 18 = −14 | Upper minus lower |
| Answer | 18 or −14 | 14 |
A simple habit removes the error. Write each substitution in its own bracket, then subtract the brackets.
Properties you can use
Two properties save working in exam questions:
- Adjacent limits join: the integral from a to b plus the integral from b to c equals the integral from a to c.
- A constant multiplier moves outside: the integral of k f(x) equals k times the integral of f(x).
If ∫ f(x) dx from 1 to 4 equals 7, then ∫ (2f(x) + 3) dx from 1 to 4 = 2(7) + 3(4 − 1) = 14 + 9 = 23. The integral of the constant 3 over an interval of length 3 is 3 × 3 = 9.
Check yourself
Evaluate ∫ (x² + 1) dx from −1 to 2, then find k if ∫ 3x² dx from 0 to k equals 27.
Answer
[x³/3 + x] from −1 to 2 = (8/3 + 2) − (−1/3 − 1).
The first bracket is 14/3, and the second is −4/3. So the result is 14/3 + 4/3 = 18/3 = 6.
For the second part, [x³] from 0 to k = k³ = 27, so k = 3.
What to study next
The next lesson turns the integral into an area. Continue with finding area between a curve and an axis, where the sign of the integral needs care. Then test the chapter with the integration practice set.
For a teacher to go through limits and signs with you, see online one-to-one Additional Mathematics tuition. The mistake log records which step slips.