When a curve crosses the x-axis inside your limits, split the interval at each crossing, integrate each part on its own, and add the sizes. Areas below the axis give negative integrals, so you take the positive value of each part before adding.
This is the natural next step after finding area between a curve and an axis. It also prepares you for distance travelled versus displacement, which uses exactly the same idea.
Why does the integral cancel?
A definite integral adds up y × width for every thin strip. Above the axis y is positive, so the strip adds area. Below the axis y is negative, so the strip subtracts.
The integral therefore gives signed area: area above minus area below. If the two are equal, the answer is zero, even though the region is not empty.
How to find the true area
- Find the roots. Solve y = 0 to see where the curve meets the axis.
- Keep the roots inside the interval. Each one is a split point.
- Sketch. Mark which sections are above and which are below the axis.
- Integrate each section separately, using F(upper) − F(lower).
- Make each section positive. A negative answer just means the section is below the axis, so drop the minus sign.
- Add the positive areas.
Worked example
Find the total area between y = x² − 4x + 3 and the x-axis from x = 0 to x = 3.
Step 1, roots: x² − 4x + 3 = (x − 1)(x − 3) = 0, so x = 1 and x = 3.
Step 2, split point: x = 1 lies inside the interval. The other root, x = 3, is the upper limit already.
Step 3, sketch: the curve opens upward. It is above the axis for x < 1 and below it between x = 1 and x = 3.
Step 4, integrate: F(x) = x³/3 − 2x² + 3x.
F(0) = 0, F(1) = 1/3 − 2 + 3 = 4/3, F(3) = 9 − 18 + 9 = 0.
Step 5, each section:
- From 0 to 1: F(1) − F(0) = 4/3. Above the axis, so the area is 4/3.
- From 1 to 3: F(3) − F(1) = 0 − 4/3 = −4/3. Below the axis, so the area is 4/3.
Step 6, add: 4/3 + 4/3 = 8/3.
Total area = 8/3 square units.
Check: the dip between x = 1 and x = 3 is a parabola segment of width 2 and depth 1, and two-thirds of the bounding rectangle is (2/3) × 2 × 1 = 4/3. That matches.
The mistake to watch for
A common slip is to integrate from 0 to 3 in one go.
Mistaken answer: F(3) − F(0) = 0 − 0 = 0, so the area is 0.
The positive 4/3 and the negative 4/3 cancelled, so the student concluded there was no area.
The correction is to check for roots inside the interval before integrating. Whenever the question says “area” and the curve crosses the axis, split first. Zero from a whole-interval integral is a warning, not an answer.
Check yourself
Try these on paper, then open each answer.
1. Find the area enclosed between y = x² − 9 and the x-axis.
Show answer
Roots: x = −3 and x = 3. The curve is below the axis between them. F(x) = x³/3 − 9x.
F(3) = 9 − 27 = −18 and F(−3) = −9 + 27 = 18. The integral is −18 − 18 = −36.
The area is the positive value: 36 square units.
Check: a parabola segment of width 6 and depth 9 has area (2/3) × 6 × 9 = 36.
2. Find the total area between y = x² − 2x and the x-axis from x = 1 to x = 4.
Show answer
Roots: x = 0 and x = 2. Only x = 2 is inside the interval. F(x) = x³/3 − x².
F(1) = −2/3, F(2) = −4/3, F(4) = 64/3 − 16 = 16/3.
From 1 to 2: −4/3 − (−2/3) = −2/3, so the area is 2/3. From 2 to 4: 16/3 + 4/3 = 20/3.
Total = 2/3 + 20/3 = 22/3 square units (7⅓)
3. Show that the integral of y = x³ from −1 to 1 is zero, then find the true area between the curve and the x-axis over that interval.
Show answer
F(x) = x⁴/4. F(1) = 1/4 and F(−1) = 1/4, so the integral is 0.
The curve crosses the axis at x = 0. From 0 to 1 the area is 1/4. From −1 to 0 the value is F(0) − F(−1) = −1/4, so its area is 1/4.
Total area = 1/2 square unit
Where this leads next
The same splitting idea drives calculating displacement from velocity, where “below the axis” means moving backwards. Try the quadratic structure explorer to see roots and a sketch before you integrate, and use the non-calculator working trainer to check your fractions.
Students who know the method but forget the roots step often need a routine written into their own working. That is something our teachers build in online one-to-one Additional Mathematics tuition.