A definite integral is evaluated by integrating, then computing F(upper) − F(lower). The result is a number, and the constant c is not needed.
This follows integrating a power and brackets and exponentials. It is the step that turns an antiderivative into a signed area or a total change.
How do you set it out clearly?
Use square brackets with the limits on the right. The layout matters because it stops you substituting into the wrong place.
[F(x)] from a to b = F(b) − F(a)
Write each substitution in its own brackets, such as (−1)³ rather than −1³, because (−1)³ = −1 while −1³ is ambiguous in a quick read.
How do you do it, step by step?
- Integrate the expression and write it in square brackets with the limits.
- Substitute the upper limit into every x, using brackets.
- Substitute the lower limit in the same way.
- Subtract: upper result minus lower result.
- Simplify and state the number, with units if the question has any.
Worked example
Evaluate ∫ from −1 to 2 of (3x² + 2x) dx.
Step 1, integrate: [x³ + x²] with limits −1 to 2.
Step 2, upper limit x = 2: 2³ + 2² = 8 + 4 = 12.
Step 3, lower limit x = −1: (−1)³ + (−1)² = −1 + 1 = 0.
Step 4, subtract: 12 − 0 = 12.
12
Check: an independent method is to integrate from −1 to 0 and from 0 to 2 and add. The first part is [x³ + x²] from −1 to 0 = 0 − 0 = 0, and the second is 12 − 0 = 12. The total is 12.
The mistake to watch for
A common slip is to mishandle the negative lower limit.
Mistaken working: lower limit: (−1)³ + (−1)² = 1 + 1 = 2, so 12 − 2 = 10
The student treated (−1)³ as +1. An odd power of a negative number stays negative, so the answer 10 is wrong.
The correction is to put every substituted value in brackets and work out the powers first. (−1)³ = −1 and (−1)² = +1, so the lower value is 0, and the answer is 12.
What if the answer is negative?
A negative result is not an error. For example, ∫ from 0 to 2 of (x² − 2x) dx = [x³/3 − x²] = (8/3 − 4) − 0 = −4/3.
The curve lies below the x-axis between 0 and 2, so the signed area is negative. The geometric area would be 4/3. Sketch the curve if the question asks for an area rather than an integral.
Check yourself
Try these on paper, then open each answer.
1. Evaluate ∫ from 0 to 2 of 4x³ dx.
Show answer
[x⁴] from 0 to 2 = 2⁴ − 0⁴ = 16. 16
2. Evaluate ∫ from 1 to 4 of √x dx.
Show answer
The integral of x1/2 is (2/3)x3/2. At x = 4: (2/3) × 8 = 16/3. At x = 1: 2/3. The difference is 14/3. 14/3
3. Evaluate ∫ from −2 to 1 of (x² + 1) dx.
Show answer
[x³/3 + x]. At x = 1: 1/3 + 1 = 4/3. At x = −2: −8/3 − 2 = −14/3. Subtract: 4/3 − (−14/3) = 18/3 = 6. 6
Where this leads next
Next, build the habit of checking an antiderivative by differentiation, then try the integration methods practice set. The non-calculator working trainer helps with fraction arithmetic, and the quadratic structure explorer helps you see where a curve crosses the axis.
If negative limits and signs keep costing you marks, a teacher in online one-to-one Additional Mathematics tuition can fix your layout with you.