Worked Examples

We compare all four methods on the same problem: integrating $f(x) = e^x$ over $[0, 2]$. The exact value is known analytically, so we can compute a precise relative error for each approximation and observe how accuracy improves as we use more nodes or a higher-degree rule.

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Setup

Problem: Approximate $\displaystyle\int_0^2 e^x\,dx$ using (i) the basic Trapezoidal Rule, (ii) the Composite Trapezoidal Rule with $m = 2$, (iii) the Composite Trapezoidal Rule with $m = 3$, and (iv) Simpson’s Rule. Compute the relative error for each.

The exact value is:

$\int_0^2 e^x\,dx = \left[e^x\right]_0^2 = e^2 - e^0 = e^2 - 1 \approx 6.389$

The relative error is defined as:

$\text{Relative Error} = \left|\frac{\text{Exact} - \text{Approximate}}{\text{Exact}}\right| \times 100\%$

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(i) Basic Trapezoidal Rule

Step 1 — Identify parameters

$a = 0, \quad b = 2, \quad f(x) = e^x$

Step 2 — Evaluate $f$ at both endpoints

$f(0) = e^0 = 1, \qquad f(2) = e^2 \approx 7.389$

Step 3 — Apply the Trapezoidal Rule

$I_1 = \frac{b - a}{2}\left[f(a) + f(b)\right] = \frac{2 - 0}{2}\left[f(0) + f(2)\right] = 1 \cdot \left[1 + 7.389\right] = \boxed{8.389}$

Step 4 — Compute the relative error

$\text{Relative Error} = \left|\frac{6.389 - 8.389}{6.389}\right| \times 100\% \approx 31.30\%$

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(ii) Composite Trapezoidal Rule ($m = 2$)

Step 1 — Compute the step size and nodes

$h = \frac{b - a}{m} = \frac{2 - 0}{2} = 1$

$x_0 = 0, \qquad x_1 = 0 + 1 = 1, \qquad x_2 = 1 + 1 = 2$

Step 2 — Evaluate $f$ at all nodes

$f(x_0) = f(0) = 1, \qquad f(x_1) = f(1) = e \approx 2.718, \qquad f(x_2) = f(2) = e^2 \approx 7.389$

Step 3 — Apply the Composite Trapezoidal formula

$C_{1,2} = \frac{h}{2}\left[f(x_0) + 2f(x_1) + f(x_2)\right] = \frac{1}{2}\left[1 + 2(2.718) + 7.389\right]$

$= \frac{1}{2}\left[1 + 5.436 + 7.389\right] = \frac{1}{2}(13.825) = \boxed{6.913}$

Step 4 — Compute the relative error

$\text{Relative Error} = \left|\frac{6.389 - 6.913}{6.389}\right| \times 100\% \approx 8.20\%$

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(iii) Composite Trapezoidal Rule ($m = 3$)

Step 1 — Compute the step size and nodes

$h = \frac{2 - 0}{3} = \frac{2}{3}$

$x_0 = 0, \qquad x_1 = \frac{2}{3}, \qquad x_2 = \frac{4}{3}, \qquad x_3 = 2$

Step 2 — Evaluate $f$ at all nodes

$f(x_0) = e^0 = 1, \qquad f(x_1) = e^{2/3} \approx 1.948$

$f(x_2) = e^{4/3} \approx 3.794, \qquad f(x_3) = e^2 \approx 7.389$

Step 3 — Apply the Composite Trapezoidal formula

$C_{1,3} = \frac{h}{2}\left[f(x_0) + 2f(x_1) + 2f(x_2) + f(x_3)\right]$

$= \frac{2/3}{2}\left[1 + 2(1.948) + 2(3.794) + 7.389\right]$

$= \frac{1}{3}\left[1 + 3.896 + 7.588 + 7.389\right] = \frac{1}{3}(19.873) = \boxed{6.624}$

Step 4 — Compute the relative error

$\text{Relative Error} = \left|\frac{6.389 - 6.624}{6.389}\right| \times 100\% \approx 3.68\%$

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(iv) Simpson’s Rule

Step 1 — Identify parameters

$a = 0, \qquad b = 2, \qquad \frac{a+b}{2} = 1$

Step 2 — Evaluate $f$ at all three nodes

$f(0) = 1, \qquad f(1) = e \approx 2.718, \qquad f(2) = e^2 \approx 7.389$

Step 3 — Apply Simpson’s Rule

$I_2 = \frac{b - a}{6}\left[f(a) + 4f\!\left(\frac{a+b}{2}\right) + f(b)\right]$

$= \frac{2 - 0}{6}\left[f(0) + 4f(1) + f(2)\right] = \frac{1}{3}\left[1 + 4(2.718) + 7.389\right]$

$= \frac{1}{3}\left[1 + 10.873 + 7.389\right] = \frac{1}{3}(19.262) = \boxed{6.421}$

Step 4 — Compute the relative error

$\text{Relative Error} = \left|\frac{6.389 - 6.421}{6.389}\right| \times 100\% \approx 0.50\%$

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Comparison

Method	Nodes	Approximate Value	Relative Error
Exact	—	$6.389$	$0.00\%$
Trapezoidal $I_1$	2	$8.389$	$31.30\%$
Composite $C_{1,2}$ ($m=2$)	3	$6.913$	$8.20\%$
Composite $C_{1,3}$ ($m=3$)	4	$6.624$	$3.68\%$
Simpson’s $I_2$	3	$6.421$	$0.50\%$

Note

Three observations stand out from this table. First, increasing $m$ in the Composite Trapezoidal Rule steadily reduces the error: going from $m=2$ to $m=3$ roughly halves the error from 8.20% to 3.68%, consistent with the $O(h^2)$ convergence rate. Second, Simpson’s Rule uses the same three nodes as the Composite Trapezoidal Rule with $m=2$, yet its relative error is 0.50% versus 8.20%, a more than 16-fold improvement from using a parabola instead of a line. Third, even the single-trapezoid rule ($I_1$) provides a rough estimate quickly, which is useful for sanity-checking or as a starting point before applying more refined methods.