Definite Integrals as Area Under Curves
For a non-negative continuous function $f(x) \ge 0$, the definite integral represents the exact area enclosed between the curve, the $x$-axis, and vertical lines $x = a$ and $x = b$:
In real life, many boundary curves (e.g. riverbanks, coastlines, highway profiles) do not possess an explicit algebraic equation. Numerical approximation is required.
The Trapezoidal Rule Formula
Dividing the interval $[a, b]$ into $n$ equal subintervals (strips) of uniform width $h$:
Counting principle:
- $n$ strips require $n + 1$ ordinates (height values): $y_0, y_1, y_2, \dots, y_n$.
- The two outer boundaries ($y_0$ and $y_n$) appear once.
- All interior heights ($y_1, y_2, \dots, y_{n-1}$) are multiplied by 2.
Concavity and Over/Under Estimation Analysis
The Trapezoidal Rule replaces curved arches with straight chord segments connecting consecutive ordinates:
Exam Tip: If an exam asks whether your estimate is an overestimate or underestimate, sketch the curve and draw straight chords connecting the points. The visual position of the chords relative to the curve gives you the complete reasoning mark.
Unequal intervals, signed values and uncertainty
- Unequal spacings require $\sum_i (x_{i+1}-x_i)(y_i+y_{i+1})/2$; do not use a single h. The estimate integrates the piecewise-linear interpolant.
- A definite integral is signed. Total geometric area is $\int|f|$; distance travelled is $\int|v|$. Split at sign changes, including interpolated zeros when using tabulated data.
- Concavity gives an error direction for the signed integral if it holds on every strip. Changing concavity or noisy measurements can invalidate a single overall over/under claim. More strips do not remove measurement error.
An interval from $x = 2$ to $x = 14$ is divided into 4 equal strips for the Trapezoidal Rule. Find the strip width $h$ and list the $x$-values of all 5 ordinates.
Solution:
$h = \frac{b - a}{n} = \frac{14 - 2}{4} = \frac{12}{4} = 3$. [M1][A1]
$x$-values: $x_0 = 2$, $x_1 = 5$, $x_2 = 8$, $x_3 = 11$, $x_4 = 14$. [A1]
The width of an irregular lake was measured at 10-meter intervals as shown in the table:
| Distance $x$ (m) | 0 | 10 | 20 | 30 | 40 |
|---|---|---|---|---|---|
| Width $y$ (m) | 0 | 18 | 26 | 14 | 0 |
Use the Trapezoidal Rule to estimate the surface area of the lake.
Solution:
$h = 10$, $y_0 = 0, y_1 = 18, y_2 = 26, y_3 = 14, y_4 = 0$. [M1]
$\text{Area} \approx \frac{10}{2} \left[ (0 + 0) + 2(18 + 26 + 14) \right]$ [M1]
$= 5 \left[ 0 + 2(58) \right] = 5(116)$ [M1]
$= 580\text{ m}^2$. [A1]
Estimate $\int_1^5 \frac{12}{x} \, dx$ using the Trapezoidal Rule with $n = 4$ strips.
Solution:
$h = \frac{5 - 1}{4} = 1$. The $x$-values are $1, 2, 3, 4, 5$. [M1]
Ordinates: $y_0 = \frac{12}{1} = 12$, $y_1 = \frac{12}{2} = 6$, $y_2 = \frac{12}{3} = 4$, $y_3 = \frac{12}{4} = 3$, $y_4 = \frac{12}{5} = 2.4$. [A1]
$\text{Estimate} = \frac{1}{2} \left[ (12 + 2.4) + 2(6 + 4 + 3) \right]$ [M1]
$= 0.5 \left[ 14.4 + 2(13) \right] = 0.5(14.4 + 26) = 0.5(40.4) = 20.2$. [A2]
For the function $f(x) = \frac{12}{x}$ on $[1, 5]$ in Exercise 3:
- Find $f''(x)$ and determine the concavity of the curve on $[1, 5]$.
- Deduce whether the estimate $20.2$ is an overestimate or an underestimate of the true integral.
Solution:
(a) $f(x) = 12x^{-1} \implies f'(x) = -12x^{-2} \implies f''(x) = 24x^{-3} = \frac{24}{x^3}$. [M1][A1]
For $x \in [1, 5]$, $x^3 > 0 \implies f''(x) > 0$. The curve is concave up ($\cup$). [R1]
(b) Because the curve is concave up, the straight trapezoid chord segments lie above the curve. Therefore, $20.2$ is an overestimate. [A1]
The exact value of $\int_1^5 \frac{12}{x} \, dx$ is $12\ln(5) \approx 19.313$. Calculate the percentage error of the Trapezoidal Rule approximation from Exercise 3 ($20.2$).
Solution:
$\text{Percentage error} = \left|\frac{\text{Approx} - \text{Exact}}{\text{Exact}}\right| \times 100\%$ [M1]
$= \left|\frac{20.2 - 19.313}{19.313}\right| \times 100\% = \frac{0.887}{19.313} \times 100\%$ [M1]
$\approx 4.59\%$ (to 3 s.f.). [A1]
State two separate methods by which a surveyor could increase the accuracy of a land area estimated using the Trapezoidal Rule.
Solution:
1. Increase the number of strips $n$ (i.e. decrease the step size $h$ between measurements) so the chords closer approximate the true curve. [A2]
2. Use Simpson's rule (parabolic arches) or fit a higher-order regression curve to the surveyed coordinates. [A2]
A cross-section of a tunnel entrance is given by the table of heights $y$ (meters) measured from the base at $1.5\text{-meter}$ intervals:
Heights: $y = [0, 2.4, 3.6, 4.2, 4.0, 2.8, 0]$ at $x = [0, 1.5, 3.0, 4.5, 6.0, 7.5, 9.0]$. Estimate the cross-sectional area of the entrance.
Solution:
$h = 1.5$, $n = 6$ strips, outer ends $y_0 = 0$, $y_6 = 0$. [M1]
$\text{Area} \approx \frac{1.5}{2} \left[ (0 + 0) + 2(2.4 + 3.6 + 4.2 + 4.0 + 2.8) \right]$ [M1]
$= 0.75 \left[ 2(17.0) \right] = 0.75(34.0)$ [M1]
$= 25.5\text{ m}^2$. [A1]
For the tunnel entrance in Exercise 7, the roof is an arch that is concave down throughout. Explain whether $25.5\text{ m}^2$ is an underestimate or overestimate of the true entrance area.
Solution:
Since the roof arch is concave down ($\cap$), straight line segments between survey points lie entirely underneath the arch. [R2]
Therefore, the Trapezoidal Rule calculation of $25.5\text{ m}^2$ is an underestimate. [A1]
Find the exact value of the definite integral $\int_0^3 (3x^2 - 4x + 5) \, dx$ using analytical integration.
Solution:
$\int (3x^2 - 4x + 5) \, dx = \left[ \frac{3x^3}{3} - \frac{4x^2}{2} + 5x \right]_0^3 = \left[ x^3 - 2x^2 + 5x \right]_0^3$ [M1][A1]
Evaluate at upper limit $x = 3$: $3^3 - 2(3^2) + 5(3) = 27 - 18 + 15 = 24$. [M1]
Evaluate at lower limit $x = 0$: $0$.
Exact integral $= 24 - 0 = 24$. [A1]
A speed-time graph has recorded velocities $v$ (m/s) at 2-second intervals: $t = [0, 2, 4, 6]$, $v = [0, 8, 14, 18]$. Use the Trapezoidal Rule to estimate the total distance travelled over the 6 seconds.
Solution:
Distance is the integral of velocity: $s = \int_0^6 v(t) \, dt$. [M1]
$h = 2$, $n = 3$ strips. [M1]
$\text{Distance} \approx \frac{2}{2} \left[ (0 + 18) + 2(8 + 14) \right] = 1 \left[ 18 + 2(22) \right] = 18 + 44 = 62\text{ meters}$. [A2]
At t0,1,3seconds, velocities are2,4,−2m/s. Under linear interpolation find displacement and total distance.
Trapezoids:1(2+4)/2+2(4−2)/2=5m displacement. Zero occurs t=7/3. Positive area3+8/3andnegative magnitude2/3give distance19/3≈6.33m.
Use one trapezoid for $f(x)=x^2-1$ on[0,2]. Compare signed integral and total area.
Trapezoid2(−1+3)/2=2; exact integral2/3. Convexity gives an overestimate of the signed integral. Exact total area is2, after splitting at1; the coincidence with the trapezoid value is not a general rule.
Core Formulas (Topic 5.7 & 5.8)
| Trapezoidal Rule | $$\int_a^b y \, dx \approx \frac{h}{2}\left[(y_0 + y_n) + 2(y_1 + y_2 + \dots + y_{n-1})\right]$$ |
| Step Size $h$ | $$h = \frac{b - a}{n} \quad (n = \text{number of strips})$$ |
| Power Integration Rule | $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \ne -1)$$ |
| Concavity Rule | $$f''(x) > 0 \implies \text{Overestimate}, \quad f''(x) < 0 \implies \text{Underestimate}$$ |
Top 5 IB Exam Traps to Avoid
- Trap 1: Confusing number of strips $n$ with number of points: A table with 7 points has $n = 6$ strips! Using $n = 7$ in $h = \frac{b-a}{n}$ destroys the calculation.
- Trap 2: Doubling the outer ordinates: Only the interior ordinates ($y_1$ to $y_{n-1}$) are multiplied by 2. $y_0$ and $y_n$ appear without the factor of 2.
- Trap 3: Omitting the factor of $\frac{h}{2}$: Forgetting to divide by 2 or forgetting $h$ before the bracket.
- Trap 4: Inverting the concavity reasoning: Concave up curves hold water like a cup ($\cup$); straight chords go over the bowl $\implies$ overestimate.
- Trap 5: Negative regions in area questions: Definite integrals calculate net signed area. If computing geometric land area, ensure sections below the axis are treated positively.
5-Question Concept Check
Score: - / 51. If an interval $[0, 10]$ is divided into 5 strips, what is the step size $h$?
2. How many ordinates (height values) are required for an approximation using 8 strips?
3. When approximating area under a strictly concave down curve ($f''(x) < 0$), the Trapezoidal Rule produces an:
4. What is $\int_1^3 4x \, dx$?
5. In the Trapezoidal Rule formula, which ordinates are multiplied by a factor of 2?