Numerical Methods intro

Estimate an integral with the trapezoidal rule

Approximate ∫₀¹ e^(−x²) dx using the trapezoidal rule with 4 sub-intervals, then bound the error and compare to the true value.

concept: Trapezoidal Rule

#integration #quadrature #trapezoidal #numerical-analysis

Problem

Approximate the integral

I = \int_{0}^{1} e^{- x^{2}} d x

using the composite trapezoidal rule with $n = 4$ sub-intervals. Then bound the error and compare to the true value.

Approach

The composite trapezoidal rule on $[a, b]$ with $n$ equal sub-intervals is

T_{n} = \frac{h}{2} [f (x_{0}) + 2 k = 1 \sum n - 1 f (x_{k}) + f (x_{n})], h = \frac{b - a}{n}

Geometrically: connect adjacent sample points with straight chords, sum the areas of the resulting trapezoids. Endpoints get weight 1, interior points get weight 2.

Walkthrough

Step 1 — Set up the grid

$a = 0$ , $b = 1$ , $n = 4$ , so $h = 0.25$ .

Sample points: $x_{0}, x_{1}, x_{2}, x_{3}, x_{4} = 0, 0.25, 0.5, 0.75, 1$ .

Step 2 — Evaluate

f

at each grid point

With $f (x) = e^{- x^{2}}$ :

f(0.00) = e^0       = 1.000000
f(0.25) = e^-0.0625 ≈ 0.939413
f(0.50) = e^-0.25   ≈ 0.778801
f(0.75) = e^-0.5625 ≈ 0.569783
f(1.00) = e^-1      ≈ 0.367879

Step 3 — Apply the formula

Sum the interior values (weight 2) and the endpoints (weight 1):

T_{4} = \frac{0.25}{2} [1 + 2 (0.939413 + 0.778801 + 0.569783) + 0.367879]

T_{4} = 0.125 \cdot [1 + 2 \cdot 2.287997 + 0.367879]

T_{4} = 0.125 \cdot 5.943873

T_{4} \approx 0.742984

Step 4 — Bound the error

The composite trapezoidal rule has the error bound

∣ E_{T} ∣ \leq \frac{( b - a ) ^{3}}{12 n ^{2}} \cdot [a, b] max ∣ f^{''} (x) ∣

Compute $f^{''}$ :

f (x) = e^{- x^{2}}, f^{'} (x) = - 2 x e^{- x^{2}}, f^{''} (x) = (4 x^{2} - 2) e^{- x^{2}}

On $[0, 1]$ , evaluate at the candidates: $f^{''} (0) = - 2$ and $f^{''} (1) = 2 e^{- 1} \approx 0.736$ . The maximum magnitude is $2$ , at $x = 0$ .

∣ E_{T} ∣ \leq \frac{1 ^{3}}{12 \cdot 16} \cdot 2 = \frac{2}{192} \approx 0.01042

Step 5 — Compare to the true value

The true value is

I = \frac{π}{2} erf (1) \approx 0.746824

Our trapezoidal estimate was $T_{4} \approx 0.742984$ . The actual error is

∣ I - T_{4} ∣ \approx ∣0.746824 - 0.742984∣ = 0.003840

which is comfortably below the bound of $0.01042$ .

Answer

$T_{4} \approx 0.742984$

Actual error $\approx 0.00384$ , within the predicted bound of $\approx 0.01042$ .

Remarks

The error scales as $O (h^{2}) = O (1/ n^{2})$ . Doubling $n$ shrinks the bound by a factor of 4. To get error below $1 0^{- 6}$ on this integral, you'd need roughly $n \approx 400$ sub-intervals — workable but not great.

For this $O (h^{2})$ price, Simpson's rule gives you $O (h^{4})$ accuracy for a few extra multiplications per sample. On smooth integrands like $e^{- x^{2}}$ , Gaussian quadrature does even better — 6–8 nodes will match what trapezoidal needs hundreds of nodes for.