Area Between Curves and the Trapezoidal Rule

Key Terms

Area between two curves = ∫_a^b [f(x) − g(x)] dx where f is the upper curve; Find limits by solving f(x) = g(x) for their intersection x-values; If curves switch position, split at the crossing point
Trapezoidal rule: approximates integrals using trapezoids instead of rectangles; More strips (larger n) ⇒ more accurate approximation

Area Between Two Curves:
A = ∫_a^b [f(x) − g(x)] dx (where f(x) ≥ g(x) on [a,b])

Trapezoidal Rule (n strips, strip width w = (b−a)/n):
∫_a^b f(x) dx ≈ (w/2)[f(x₀) + 2f(x₁) + 2f(x₂) + … + 2f(x_n−1) + f(x_n)]

Key pattern: first + last with coefficient 1; all middle terms with coefficient 2

Worked Example 1: Find the area enclosed between y = x² and y = x + 2.

Step 1: Find intersections: x² = x + 2 ⇒ x² − x − 2 = 0 ⇒ (x−2)(x+1) = 0 ⇒ x = −1, 2
Step 2: Determine upper curve on [−1,2]: at x=0, y=x+2=2 and y=x²=0. So y=x+2 is on top.
Step 3: A = ∫₋₁² [(x+2) − x²] dx
= [x²/2 + 2x − x³/3]₋₁²
= (2 + 4 − 8/3) − (1/2 − 2 + 1/3)
= (18/3 − 8/3) − (3/6 − 12/6 + 2/6)
= 10/3 − (−7/6) = 10/3 + 7/6 = 20/6 + 7/6 = 27/6 = 9/2 units²

Worked Example 2: Use the trapezoidal rule with n = 4 strips to estimate ∫₀² e^x dx.

w = (2−0)/4 = 0.5 x-values: 0, 0.5, 1, 1.5, 2
f(0)=1, f(0.5)=e^0.5≈1.6487, f(1)=e≈2.7183, f(1.5)=e^1.5≈4.4817, f(2)=e²≈7.3891
≈ (0.5/2)[1 + 2(1.6487) + 2(2.7183) + 2(4.4817) + 7.3891]
= 0.25[1 + 3.2974 + 5.4366 + 8.9634 + 7.3891]
= 0.25 × 26.0865 = 6.521
(Exact: e²−1 ≈ 6.389. The rule overestimates because e^x is concave up.)

Hot Tip: In the trapezoidal rule, count your terms carefully. With n strips, you have n+1 function values (x₀ to x_n). The first and last have coefficient 1; the n−1 middle values all have coefficient 2.

Area Between Two Curves: The Concept

When two curves enclose a region, we can find its area by subtracting the "lower" curve from the "upper" curve:

A = ∫_a^b [f_upper(x) − f_lower(x)] dx

Think of it this way: the area under the top curve minus the area under the bottom curve leaves exactly the area between them.

Always identify which curve is on top by testing a point inside the region (e.g., x = 0 in the interval [−1, 2]).

Finding the Limits of Integration

The limits a and b are the x-coordinates where the two curves intersect. To find them, set f(x) = g(x) and solve.

Example: y = x² and y = 4x − x²
x² = 4x − x² ⇒ 2x² − 4x = 0 ⇒ 2x(x−2) = 0 ⇒ x = 0, 2
Check at x=1: f=1, g=3. So g is on top: A = ∫₀²[(4x−x²) − x²]dx = ∫₀²(4x−2x²)dx

What if curves switch position? If f and g swap which is on top within [a,b], find the crossing point c and split:
A = ∫_a^c[f−g]dx + ∫_c^b[g−f]dx

The Trapezoidal Rule: Derivation and Use

Instead of rectangles (Riemann sums), the trapezoidal rule uses trapezoids to approximate the area under a curve. A trapezoid connecting (x_i, f(x_i)) to (x_i+1, f(x_i+1)) has area:

w/2 × [f(x_i) + f(x_i+1)]

Summing all n trapezoids and simplifying (adjacent function values appear twice, except the endpoints):

∫_a^b f(x)dx ≈ (w/2)[f(x₀) + 2f(x₁) + 2f(x₂) + … + 2f(x_n−1) + f(x_n)]

where w = (b−a)/n is the strip width, and x_i = a + i·w.

Accuracy:

For a concave up function (f″(x) > 0): the trapezoidal rule overestimates.
For a concave down function (f″(x) < 0): the trapezoidal rule underestimates.
Increasing n improves accuracy. Doubling n roughly quarters the error.

Exam technique: Set up a table of x and f(x) values before applying the formula. This organises your work and makes it easy to apply the 1, 2, 2, ..., 2, 1 pattern of coefficients.

When to Use Each Method

Situation	Method
Antiderivative can be found	Exact integral (FTC)
Function given as a table of values	Trapezoidal rule
Function is complex / no closed form antiderivative	Trapezoidal rule (or technology)
Want to estimate accuracy	Compare trapezoidal estimate to exact

Common Errors and Exam Strategy

Area between curves errors:

Getting upper/lower curve backwards — always verify with a test point.
Using incorrect intersection points as limits.
Forgetting to handle regions where curves switch position.

Trapezoidal rule errors:

Miscounting the number of terms (n strips ⇒ n+1 function values).
Applying coefficient 2 to the endpoint values (should be 1).
Using wrong strip width: w = (b−a)/n, NOT (b−a)/(n+1).

Mastery Practice

See Answers ➔

Find the area enclosed between y = x + 4 and y = x² − 2.
Use the trapezoidal rule with n = 4 to estimate ∫₁³ √x dx. Set up a table of values.
Find the area enclosed between y = x² and y = √x.
Use the trapezoidal rule with n = 3 to estimate ∫₀π sin x dx. Compare your answer to the exact value of 2.
Find the area of the region enclosed between y = x and y = x³ for x ∈ [0,1] and x ∈ [−1,0]. Find the total area for x ∈ [−1,1].
A region is bounded by y = e^x, y = e^−x, x = 0 and x = 1. Find the exact area of this region.
The table below gives values of f(x). Use the trapezoidal rule to estimate ∫₀⁶ f(x) dx:

x 0 1 2 3 4 5 6

f(x) 2 5 9 11 10 7 3
Find the area between y = sin x and y = cos x for x ∈ [0, π/2]. Identify which function is on top and verify with a test point.
Two curves are defined by y = x² − 2x and y = 4 − x². Find the area of the region enclosed between them, showing all working including finding the intersection points.
A river's cross-section has depths (in metres) measured at 1-metre intervals across its 6-metre width: 0, 1.2, 2.5, 3.1, 2.8, 1.9, 0. Use the trapezoidal rule to estimate the cross-sectional area of the river, then estimate the flow rate (m³/s) if water moves at 2.5 m/s.