A definite integral gives signed area, and geometric area is always positive. When a curve crosses the x-axis between your limits, the two answers differ, and the question decides which one it wants.
The habit that fixes this is simple: find where the curve meets the axis before you integrate.
What does a definite integral really measure?
Think of the integral as adding small strips of height y. A strip above the axis has positive height and adds. A strip below has negative height and subtracts.
So the integral is “area above minus area below”. That can be positive, negative or zero. Area in the everyday sense is “area above plus area below”, each taken as a positive size.
When do the two agree?
They agree only when the curve stays on one side of the axis for the whole interval. If it is all above, the integral is the area. If it is all below, the area is the integral with the sign reversed.
The moment the curve crosses the axis inside your limits, you have to split.
Worked example
Find (a) the value of the integral of x² − 4 from x = 0 to x = 3, and (b) the total area between the curve y = x² − 4, the x-axis and the lines x = 0 and x = 3.
Step 1: the antiderivative. ∫(x² − 4) dx = x³/3 − 4x.
(a) Signed value. At x = 3: 9 − 12 = −3. At x = 0: 0. So the integral is −3 − 0 = −3.
(b) Area. First find the crossing: x² − 4 = 0 gives x = 2 (x = −2 is outside the interval). The curve is below the axis from 0 to 2 and above it from 2 to 3.
- From 0 to 2: [x³/3 − 4x] = (8/3 − 8) − 0 = −16/3. As an area, 16/3.
- From 2 to 3: (−3) − (−16/3) = 7/3. This is above the axis, so the area is 7/3.
Total area = 16/3 + 7/3 = 23/3, which is 7 2/3.
Check: adding the signed pieces, −16/3 + 7/3 = −9/3 = −3, which matches part (a). The two answers are different numbers for different questions.
What mistake do students make?
A student finds −3 and either writes “area = −3” or “area = 3” after dropping the sign. Both are wrong, for different reasons.
“Area = −3” fails because area cannot be negative.
“Area = 3” fails because the negative part (16/3) and the positive part (7/3) partly cancelled inside the integral. Removing the sign afterwards cannot restore what cancelled. The only repair is to split at x = 2 before adding.
A routine that prevents it
- Read the wording: does it say “evaluate the integral” or “find the area”?
- Solve y = 0 and note the crossing points inside the limits.
- Integrate each section separately.
- For area, take each section as a positive value and add.
- Compare the total with the signed answer as a check.
Self-check
- Evaluate the integral of (x − 2) from x = 1 to x = 3, then find the area between the curve and the x-axis over the same interval.
- Evaluate the integral of x³ from x = −1 to x = 1, then find the area.
- A particle has velocity v = t − 4 m/s for 0 ≤ t ≤ 6. Find its displacement and the distance it travels.
Show answer
- Antiderivative x²/2 − 2x. At 3: 4.5 − 6 = −1.5. At 1: 0.5 − 2 = −1.5. Integral = 0. The crossing is at x = 2. From 1 to 2: (2 − 4) − (−1.5) = −0.5, so area 0.5. From 2 to 3: (−1.5) − (−2) = 0.5. Total area = 1.
- Antiderivative x⁴/4. At 1: 1/4. At −1: 1/4. Integral = 0. By symmetry the area is 2 × 1/4 = 1/2.
- Antiderivative t²/2 − 4t. At 6: 18 − 24 = −6. So displacement is −6 m. The velocity is zero at t = 4. From 0 to 4: 8 − 16 = −8. From 4 to 6: −6 − (−8) = 2. Distance = 8 + 2 = 10 m.
Where do I practise this?
The same splitting idea is taught in areas and motion: see splitting an interval when signed areas change and distance travelled versus displacement. To check the integrating step itself, revisit integration methods.
You can test any quadratic’s crossing points in the quadratic structure explorer, and practise the arithmetic of fractions like 16/3 in the non-calculator working trainer. Words such as signed area and displacement appear in the Additional Mathematics terminology guide.
If the picture still does not decide your working, online one-to-one Additional Mathematics tuition gives you an assigned teacher to sketch each question with you, one at a time.