To find the area between a curve and the x-axis, integrate the function and subtract the value at the lower limit from the value at the upper limit. This lesson covers a curve that stays above the axis. The next lesson handles curves that cross it.
It is the first lesson in areas and motion, and it relies on integrating a power with the correct constant and evaluating a definite integral with correct limits.
What does the integral actually measure?
Imagine slicing the region into very thin vertical strips. Each strip has a height y and a tiny width, so its area is about y × width. Adding all the strips from x = a to x = b gives the definite integral of y from a to b.
If F is an antiderivative of y, then the area is F(b) − F(a). The constant of integration cancels, so leave it out.
How to set out an area question
- Identify the limits. Use the given x-values, or solve y = 0 if the region is bounded by the curve and the axis.
- Sketch quickly. Check the curve is above the x-axis between the limits. Skip this and you will miss a sign change.
- Integrate each term: raise the power by one and divide by the new power.
- Substitute the upper limit, then the lower limit, in brackets.
- Subtract and simplify. Write the answer with units squared if the question gives units.
Worked example
Find the area between the curve y = 3x² + 2, the x-axis and the lines x = 1 and x = 3.
Step 1, limits: x = 1 and x = 3. The curve 3x² + 2 is always positive, so it stays above the axis.
Step 2, integrate: 3x² becomes x³ and 2 becomes 2x, so F(x) = x³ + 2x.
Step 3, upper limit: F(3) = 27 + 6 = 33.
Step 4, lower limit: F(1) = 1 + 2 = 3.
Step 5, subtract: 33 − 3 = 30.
Area = 30 square units.
Check: at x = 1 the height is 5 and at x = 3 it is 29. The width is 2, so the area must lie between 10 and 58. Thirty fits.
The mistake to watch for
A common slip is to substitute only the upper limit and stop.
Mistaken answer: F(3) = 33, so the area is 33.
The student treated the integral as “the area up to x = 3”, forgetting that the region starts at x = 1.
The correction is to write both brackets every time: F(3) − F(1). The lower limit removes the area to the left of x = 1. Here it is worth 3, so the answer drops from 33 to 30.
Check yourself
Try these on paper, then open each answer.
1. Find the area between y = 2x + 3, the x-axis, x = 0 and x = 4.
Show answer
F(x) = x² + 3x. F(4) = 16 + 12 = 28 and F(0) = 0. Area = 28 − 0 = 28.
Check with a trapezium: heights 3 and 11, width 4, so ½ × (3 + 11) × 4 = 28.
28 square units
2. Find the area under y = √x between x = 0 and x = 9.
Show answer
Write √x as x1/2. F(x) = (2/3) x3/2. F(9) = (2/3) × 27 = 18 and F(0) = 0.
18 square units
3. The curve y = x(4 − x) meets the x-axis at two points. Find the area of the region enclosed by the curve and the axis.
Show answer
The roots are x = 0 and x = 4, and the curve is above the axis between them. Expand: y = 4x − x². F(x) = 2x² − x³/3.
F(4) = 32 − 64/3 = 32/3 and F(0) = 0.
32/3 square units (10⅔)
Where this leads next
Next, see what to do when the signed area changes sign, then use the non-calculator working trainer to practise the fraction arithmetic in limits. The quadratic structure explorer helps you find roots and sketch a parabola before you integrate.
If your integration is fine but you cannot tell which area the question wants, a teacher in online one-to-one Additional Mathematics tuition can go through your own past papers with you.