Areas Under Curves and Between Curves

Using definite integration to calculate areas bounded by curves, lines, and axes

SPEC 8.2 • INTEGRATION

Area Under a Curve

Definite integration can be used to find the area between a curve and the x-axis.

Key Formula

For a curve

y = f (x)

where

f (x) \geq 0

:

Area = \int_{a}^{b} y d x

Remember

This formula only gives the correct area when the curve is above the x-axis (y ≥ 0).

Worked Example01

Find the area enclosed by the curve y = x², the x-axis, and the lines x = 0 and x = 3.

1

Since y = x² ≥ 0 for all x, we can integrate directly.

2

Area =

\int_{0}^{3} x^{2} d x

3

=

[\frac{x ^{3}}{3}]_{0}^{3}

4

=

\frac{27}{3} - \frac{0}{3} = 9 - 0 = 9

Answer: Area = 9 square units

Areas Below the x-axis

When a curve lies below the x-axis, the integral gives a negative value.

Important

The definite integral calculates signed area:

Above x-axis → positive

Below x-axis → negative

Actual Area

For y < 0:

Area = \int_{a}^{b} y d x

Worked Example02

Find the area bounded by the curve y = x² − 4, the x-axis, and the lines x = 0 and x = 2.

1

Check: When x = 0, y = −4. When x = 2, y = 0. The curve is below the x-axis.

2

Integral =

\int_{0}^{2} (x^{2} - 4) d x = [\frac{x ^{3}}{3} - 4 x]_{0}^{2}

3

=

(\frac{8}{3} - 8) - (0) = \frac{8}{3} - \frac{24}{3} = - \frac{16}{3}

4

Since the integral is negative (area below x-axis), actual area =

\frac{16}{3}

Answer: Area = 16/3 square units ≈ 5.33 square units

Curves Crossing the x-axis

When a curve crosses the x-axis within your interval, you must split the integral at each crossing point.

Method

Find roots: Solve y = 0

Split: Separate integrals

Calculate: Take |negative areas|

Sum: Add all positive areas

Common Error

Without splitting, positive and negative areas cancel out!

Worked Example03

Find the total area enclosed by the curve y = x³ − x and the x-axis.

1

Find roots: x³ − x = 0 → x(x² − 1) = 0 → x(x−1)(x+1) = 0 → x = −1, 0, 1

2

For x ∈ [−1, 0]: y > 0 (above axis). For x ∈ [0, 1]: y < 0 (below axis)

3

Area₁ =

\int_{- 1}^{0} (x^{3} - x) d x = [\frac{x ^{4}}{4} - \frac{x ^{2}}{2}]_{- 1}^{0} = 0 - (\frac{1}{4} - \frac{1}{2}) = \frac{1}{4}

4

Area₂ =

\int_{0}^{1} (x^{3} - x) d x = [\frac{x ^{4}}{4} - \frac{x ^{2}}{2}]_{0}^{1} = \frac{1}{4} - \frac{1}{2} = \frac{1}{4}

5

Total area = 1/4 + 1/4 = 1/2

Answer: Total area = 1/2 square unit

Area Between Two Curves

To find the area between two curves, subtract the lower curve from the upper curve.

Key Formula

For f(x) ≥ g(x) between a and b:

Area = \int_{a}^{b} [f (x) - g (x)] d x

(top minus bottom)

Key Steps

Identify which curve is on top

Find limits (intersection points)

Integrate (top − bottom)

Worked Example04

Find the area enclosed between the curves y = x² and y = x.

1

Find intersections: x² = x → x² − x = 0 → x(x − 1) = 0 → x = 0 or x = 1

2

Between x = 0 and x = 1: y = x is above y = x² (check: at x = 0.5, line gives 0.5, parabola gives 0.25)

3

Area =

\int_{0}^{1} (x - x^{2}) d x = [\frac{x ^{2}}{2} - \frac{x ^{3}}{3}]_{0}^{1}

4

=

(\frac{1}{2} - \frac{1}{3}) - (0) = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}

Answer: Area = 1/6 square unit

Finding Intersection Points

When limits aren't given, you must find where the curves intersect to determine your bounds.

Method

To find where y = f(x) and y = g(x) intersect:

f (x) = g (x)

Solve this equation for x to find the limits of integration

Exam Tip

Always check your intersection points are correct by substituting back into both equations. They should give the same y-value.

Worked Example05

Find the area enclosed between y = 6x − x² and y = x.

1

Find intersections: 6x − x² = x → 5x − x² = 0 → x(5 − x) = 0 → x = 0 or x = 5

2

Check which is on top: At x = 2, curve gives 6(2) − 4 = 8, line gives 2. Curve is above.

3

Area =

\int_{0}^{5} [(6 x - x^{2}) - x] d x = \int_{0}^{5} (5 x - x^{2}) d x

4

=

[\frac{5 x ^{2}}{2} - \frac{x ^{3}}{3}]_{0}^{5} = \frac{5 ( 25 )}{2} - \frac{125}{3} = \frac{125}{2} - \frac{125}{3}

5

=

\frac{375}{6} - \frac{250}{6} = \frac{125}{6}

Answer: Area = 125/6 square units ≈ 20.83 square units

When Curves Cross Each Other

Sometimes two curves swap positions - one is on top in part of the region, and the other is on top in another part.

Method

Find ALL intersection points

Split at each intersection

For each section, determine which curve is on top

Integrate (top − bottom) for each section

Add all positive areas

Watch Out

If you don't split where curves cross, areas will cancel and give the wrong answer!

Example

For y = x and y = x³ between x = −1 and x = 1: the curves cross at x = −1, 0, and 1. You must calculate the area from −1 to 0 separately from the area from 0 to 1, then add them.

Worked Example06

Find the total area enclosed between y = x³ and y = x for −1 ≤ x ≤ 1.

1

Find intersections: x³ = x → x³ − x = 0 → x(x² − 1) = 0 → x = −1, 0, 1

2

For x ∈ [−1, 0]: At x = −0.5: cubic = −0.125, line = −0.5. Cubic is above.

3

For x ∈ [0, 1]: At x = 0.5: cubic = 0.125, line = 0.5. Line is above.

4

Area₁ =

\int_{- 1}^{0} (x^{3} - x) d x = [\frac{x ^{4}}{4} - \frac{x ^{2}}{2}]_{- 1}^{0} = 0 - (\frac{1}{4} - \frac{1}{2}) = \frac{1}{4}

5

Area₂ =

\int_{0}^{1} (x - x^{3}) d x = [\frac{x ^{2}}{2} - \frac{x ^{4}}{4}]_{0}^{1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}

6

Total = 1/4 + 1/4 = 1/2

Answer: Total area = 1/2 square unit

Using Geometric Shapes

When finding area between a curve and a straight line, use geometric shapes instead of integrating the line. Here are some examples:

∫curveTrapAnswer

Green = Blue − Yellow

Trap∫curveAnswer

Green = Yellow − Blue

Triangle∫curveAnswer

Green = Yellow − Blue

Rect∫curveAnswer

Green = Yellow − Blue

Key Formulas

Triangle: ½ × base × height Trapezoid: ½(a + b) × h Rectangle: base × height

Worked Example07

Find the area enclosed between the curve y = x² and the straight line joining the origin to the point (3, 9).

1

The line from (0, 0) to (3, 9) has equation y = 3x (gradient = 9/3 = 3)

2

Method: Area under line − Area under curve

3

Area under line = Triangle with base 3 and height 9 = ½ × 3 × 9 = 27/2

4

Area under curve =

\int_{0}^{3} x^{2} d x = [\frac{x ^{3}}{3}]_{0}^{3} = \frac{27}{3} = 9

5

Enclosed area = 27/2 − 9 = 27/2 − 18/2 = 9/2

Answer: Area = 9/2 = 4.5 square units

Worked Example08

Find the area between the curve y = √x and the chord joining (1, 1) to (4, 2).

1

The curve y = √x is above the chord (check: at x = 2, √2 ≈ 1.41, chord gives ≈ 1.33)

2

Method: Area under curve − Area of trapezoid

3

Area under curve =

\int_{1}^{4} x d x = [\frac{2 x ^{3/2}}{3}]_{1}^{4} = \frac{2 ( 8 )}{3} - \frac{2 ( 1 )}{3} = \frac{16}{3} - \frac{2}{3} = \frac{14}{3}

4

Trapezoid: parallel sides y₁ = 1 and y₂ = 2, width = 4 − 1 = 3

5

Area of trapezoid = ½(1 + 2) × 3 = ½ × 3 × 3 = 9/2

6

Enclosed area = 14/3 − 9/2 = 28/6 − 27/6 = 1/6

Answer: Area = 1/6 square units

Practice

Question 1: Find the area enclosed by the curve y = x² − 2x, the x-axis, and the lines x = 0 and x = 3.

Step 1: Find where curve crosses x-axis: x² − 2x = 0 → x(x − 2) = 0 → x = 0 or x = 2

Step 2: For x ∈ [0, 2]: y < 0. For x ∈ [2, 3]: y > 0

Step 3: Area below = |∫₀² (x² − 2x) dx| = |[x³/3 − x²]₀²| = |8/3 − 4| = |-4/3| = 4/3

Step 4: Area above = ∫₂³ (x² − 2x) dx = [x³/3 − x²]₂³ = (9 − 9) − (8/3 − 4) = 0 + 4/3 = 4/3

Total = 4/3 + 4/3 = 8/3 square units

Practice

Question 2: Find the area enclosed between the curves y = x² and y = 2x + 3.

Step 1: Find intersections: x² = 2x + 3 → x² − 2x − 3 = 0 → (x − 3)(x + 1) = 0 → x = −1 or x = 3

Step 2: Check: At x = 0, line = 3, parabola = 0. Line is above.

Step 3: Area = ∫₋₁³ [(2x + 3) − x²] dx = ∫₋₁³ (2x + 3 − x²) dx

Step 4: = [x² + 3x − x³/3]₋₁³ = (9 + 9 − 9) − (1 − 3 + 1/3) = 9 − (−5/3) = 9 + 5/3 = 32/3

Area = 32/3 square units ≈ 10.67 square units

Practice

Question 3: The curve y = 4x − x² meets the x-axis at the origin and at point A. Find the area enclosed between the curve and the x-axis.

Step 1: Find point A: 4x − x² = 0 → x(4 − x) = 0 → x = 0 or x = 4. So A = (4, 0)

Step 2: Check: At x = 2, y = 8 − 4 = 4 > 0. Curve is above x-axis.

Step 3: Area = ∫₀⁴ (4x − x²) dx = [2x² − x³/3]₀⁴

Step 4: = (32 − 64/3) − 0 = 96/3 − 64/3 = 32/3

Area = 32/3 square units ≈ 10.67 square units

Summary

Key Formulas

SituationFormula

Area under curve (y ≥ 0)

\int_{a}^{b} y d x

Area when y < 0

- \int_{a}^{b} y d x

or

\int_{a}^{b} y d x

Area between curves

\int_{a}^{b} (top - bottom) d x

Using shapesArea under curve ± Area of triangle/trapezoid

Exam Checklist

Find intersection points / roots first

Check which curve is on top at a test point

Split integrals where curves cross or touch x-axis

Take absolute values for areas below x-axis

Consider using triangles/trapezoids for straight lines

Always give exact answers unless told otherwise