Area under a curve: the idea and the method
The area between a curve, the x-axis, x = a and x = b is ∫ y dx from a to b. Take absolute values below the x-axis.
Where the formula comes from
Cut the region under a curve into very thin vertical strips. A strip at x has height y and width dx, so its area is about dA = y dx. Adding all the strips from x = a to x = b is exactly what a definite integral does: A = ∫ y dx from a to b = ∫ f(x) dx from a to b. If the curve is easier to write as x = g(y), use thin horizontal strips of length x and width dy: A = ∫ x dy from c to d.
Vertical strip vs horizontal strip
Each strip must run from one boundary of the region to the other. Choose the strip that gives one simple integral.
Rules that protect your marks
- Area is never negative. If the curve is below the x-axis on [a, b], A = |∫ f(x) dx from a to b|.
- If the curve is partly above and partly below, split at the crossing point and add: A = |A₁| + A₂. Integrating straight across can give a wrong answer, even 0.
- Use symmetry: find the area of one part and multiply, e.g. 4 × the first-quadrant area for a circle or an ellipse.
- Always end with sq units.
How to draw the figure for an area question
- Draw the x-axis and the y-axis and mark the origin O.
- Sketch the curve from its standard shape (line, circle, parabola or ellipse) and write its equation beside it.
- Draw the boundary lines, such as x = a and x = b, and mark where they meet the curve.
- Shade the required region.
- Draw one thin strip inside the shaded region and label its width dx or dy.
- Label every point you use as a limit of integration.
Example 1: a vertical strip
Find the area of the region bounded by the curve y = x², the x-axis and the lines x = 2 and x = 3.
- The region lies above the x-axis, between x = 2 and x = 3. Use vertical strips.
- A = ∫ y dx from 2 to 3 = ∫ x² dx from 2 to 3
- = [x³/3] from 2 to 3
- = 27/3 − 8/3 = 19/3
Area = 19/3 sq units.
Example 2: a horizontal strip
Find the area of the region in the first quadrant bounded by the curve y = x², the y-axis and the lines y = 1 and y = 4.
- The boundaries are the y-axis and the lines y = 1 and y = 4, so use horizontal strips.
- In the first quadrant, y = x² gives x = √y.
- A = ∫ x dy from 1 to 4 = ∫ √y dy from 1 to 4
- = [(2/3) y√y] from 1 to 4
- = (2/3)(8 − 1) = 14/3
Area = 14/3 sq units.
Surveyors find the area of a plot with a curved boundary, such as land beside a river bend in a village map, by adding thin strips. That is exactly what ∫ y dx does.
1 marks: “Find the area bounded by the curve y = 3x², the x-axis and the lines x = 0 and x = 2.” A = ∫ 3x² dx from 0 to 2 = [x³] from 0 to 2 = 8 sq units.
2 marks: “Write the integral, with limits, for the area of the region bounded by y = √x, the x-axis and the line x = 4, and evaluate it.” A = ∫ √x dx from 0 to 4 = (2/3)[x√x] from 0 to 4 = (2/3)(8) = 16/3 sq units.
3 marks: “Find the area of the region bounded by the curve x = 2y², the y-axis and the lines y = 1 and y = 3.” Use horizontal strips: A = ∫ x dy from 1 to 3 = ∫ 2y² dy from 1 to 3 = (2/3)[y³] from 1 to 3 = (2/3)(27 − 1) = 52/3 sq units.
Strip, sum, sign, sq units. Draw a strip, integrate, take the absolute value, write the unit.