Numerical Integration: Trapezoid and Simpson's Rule

By the end of this lesson, you will be able to apply the Trapezoid Rule and Simpson's Rule to approximate definite integrals, state the error bounds, and choose the appropriate method based on accuracy needed.

Why Numerical Integration?

Many functions have no closed-form antiderivative: $e^{-x^2}$, $\sin(x^2)$, $\dfrac{\sin x}{x}$. For these, numerical methods provide accurate approximations.

Trapezoid Rule

Replace each subinterval with a trapezoid (two sample points per subinterval).

$$T_n = \frac{\Delta x}{2}[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)]$$

Error bound: $|E_T| \leq \dfrac{M(b-a)^3}{12n^2}$ where $M = \max|f''(x)|$ on $[a,b]$.

Simpson's Rule

Use parabolas through three points per pair of subintervals (requires $n$ even).

$$S_n = \frac{\Delta x}{3}[f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + 4f(x_{n-1}) + f(x_n)]$$

Pattern of coefficients: $1, 4, 2, 4, 2, \ldots, 4, 2, 4, 1$.

Error bound: $|E_S| \leq \dfrac{M(b-a)^5}{180n^4}$ where $M = \max|f^{(4)}(x)|$ on $[a,b]$.

Simpson's Rule is far more accurate for smooth functions.

Example 1 — Trapezoid Rule

Approximate $\int_0^1 e^{-x^2},dx$ using $n = 4$.

$\Delta x = 0.25$; points: $0, 0.25, 0.5, 0.75, 1$

$f$ values: $1, 0.9394, 0.7788, 0.5698, 0.3679$

$T_4 = \frac{0.25}{2}[1 + 2(0.9394) + 2(0.7788) + 2(0.5698) + 0.3679]$

$= 0.125[1 + 1.8788 + 1.5576 + 1.1396 + 0.3679] = 0.125 \times 5.9439 \approx 0.7430$

Actual: $\approx 0.7468$ (error $\approx 0.004$)

Example 2 — Simpson's Rule

Same integral, $n = 4$.

$S_4 = \frac{0.25}{3}[1 + 4(0.9394) + 2(0.7788) + 4(0.5698) + 0.3679]$

$= \frac{0.25}{3}[1 + 3.7576 + 1.5576 + 2.2792 + 0.3679]$

$= \frac{0.25}{3}(8.9623) \approx 0.7469$

Error $\approx 0.0001$ — much better than Trapezoid with the same $n$.

Comparison: Trapezoid vs. Simpson's

Method	Error order	Good for
Trapezoid	$O(n^{-2})$	Quick approximation
Simpson's