# Chapter B: Appendix B: Estimating Integrals

> When the antiderivative is impossible or painful: strips, the trapezoidal rule, Simpson’s rule — and honest error bounds.

**Canonical:** https://calculus.kitzo.work/chapter/appendix-b  
**Level:** Advanced  
**Estimated time:** 35 minutes  
**Exercises:** 8 (auto-marked; answers stay server-side)

## Topics
- Estimating integrals using strips
- Evenly-spaced partitions
- The trapezoidal rule
- Simpson’s rule
- The error in our approximations

**Why this matters**

Real life rarely grants closed-form antiderivatives. Estimation is how mathematics still gets answers — with a guarantee attached. This appendix collects the three classic ways and their error terms.

## 1 · Estimating integrals using strips

Divide [a,b] into n subintervals of width h = (b−a)/n. Approximate the integral by rectangles — left, right, or midpoint. Errors shrink as 1/n (first-order methods), more slowly than Taylor estimates but always available.

## 2 · The trapezoidal rule

Replace each rectangle by a trapezoid and average left and right ends:

Tₙ = h/2 · [f(x₀) + 2Σ_{i=1}^{n−1} f(xᵢ) + f(xₙ)].

Second-order accurate: error O(1/n²).

## 3 · Simpson’s rule — parabola-fitting

Fit a parabola through triples of points (n must be even):

Sₙ = (h/3) · [f(x₀) + 4Σ odd i + 2Σ even i (excluding ends) + f(xₙ)]. Works beautifully on cubics (exact!) and usually rivals trapezoids with far fewer strips.

**Worked example 1**

Estimate ∫₀¹ e^{−x²}dx with Simpson, n=2 (h=0.5): values are 1, e^{−0.25}≈0.779, e^{−1}≈0.368. Simpson: 0.5/3·[1 + 4·0.779 + 0.368] ≈ 0.7477. The true value is 0.7468 — error 0.001. Trapezoidal with the same two strips gives 0.7533, with more error. Simpson wins.

## 4 · The error in our approximations

For smooth f, trapezoidal error ≤ −(b−a)h²/12 · f″(ξ), Simpson error ≤ −(b−a)h⁴/180 · f⁽⁴⁾(ξ). The h⁴ in Simpson explains its reputation for accuracy per strip.

**Watch out**

Simpson’s rule demands an even number of strips — an odd n silently breaks the pairing. And on non-smooth functions (sharp corners, kinks), both rules lose their high order: refine near the kink, and don’t trust an error bound that assumed 4th-order smoothness you don’t have.

**Chapter formula sheet**

Tₙ = (h/2)[f(x₀) + 2Σmid + f(xₙ)]  error ≈ −(b−a)h²/12 · f″(ξ)

Sₙ = (h/3)[f(x₀) + 4Σodd + 2Σeven + f(xₙ)]  error ≈ −(b−a)h⁴/180 · f⁽⁴⁾(ξ)

Exact for: Simpson on polynomials degree ≤ 3

Choose n for desired digits: more strips are cheap, error bounds are not

## Common mistakes

- Misidentifying h: it’s (b−a)/n, not 1/n — scaling matters.

- Applying Simpson with odd n — it needs an even number of subintervals.

- Trusting a trapezoid on a sharp peak; refine where f curves, not where it’s flat.

- Quoting an estimate without its error bound — “≈ 0.75” is a claim; “0.75 ± 0.01” is an answer.

## Practice exercises

1. The trapezoidal rule uses:

- A. trapezoids over each subinterval
- B. parabolas over each subinterval
- C. midpoint rectangles only
- D. exact antiderivatives

2. Simpson’s rule (paired strips) uses:

- A. parabola fits over pairs of strips
- B. single rectangles
- C. trapezoids
- D. Taylor series

3. Trapezoidal estimate of ∫₀¹ x² dx with n=2 (h=0.5) is: values 0, 0.25, 1 → (h/2)[0 + 2·0.25 + 1]. Compute it.

4. Simpson with n=2 for ∫₀¹ x² dx is exact. Recall its value.

5. Simpson needs which parity of n?

- A. n even
- B. n odd
- C. any n
- D. n = 1

6. The trapezoidal error scales like:

- A. h²
- B. h⁴
- C. h
- D. constant

7. For ∫₀¹ e^{−x²}dx, Simpson (n=2) gave ≈ 0.7477 and the true value is 0.7468. The error is:

8. Why does Simpson beat trapezoidal on smooth integrands?

- A. its error term is O(h⁴) versus O(h²)
- B. it uses more memory
- C. it requires antiderivatives
- D. it averages left and right

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Practice interactively: https://calculus.kitzo.work/chapter/appendix-b

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