Numerical integration and bracketed roots

What this concept is

The composite trapezoidal rule joins sampled endpoints with straight lines. For n equal intervals of width h=(b−a)/n, T=h[f(a)/2+sum of interior f values+f(b)/2]. For continuous f with opposite endpoint signs, bisection repeatedly retains the half interval containing a sign change.

Why it matters here

Executable approximations must report limits honestly. A small residual need not imply a small root-position error; a narrow valid bracket does.

Concrete example

For x² on [0,1], one trapezoid gives 1/2 versus exact 1/3. The smooth-function error bound is M(b−a)³/(12n²), where M bounds |f″|. Here M=2 and error is exactly 1/(6n²). For x²−2 on [1,2], retaining a sign bracket and returning its midpoint after k halvings bounds position error by 1/2^(k+1).

Common confusion / misconception

A sign change across a discontinuity, such as 1/x around zero, does not guarantee a root. A checker cannot establish continuity of an arbitrary callback. Two close quadrature estimates can both miss a narrow spike.

How to use it

Derive the interval width bound, impose an iteration budget, and stop with failure if tolerance cannot be reached. Check finite values. In the lab, bisection promises absolute position tolerance only under the caller’s continuity contract and reliable function signs; exact floating-point zero is treated as a computed root. Use independent analytic answers to calibrate, then run changed cases.

Check yourself

How many subintervals suffice for trapezoid absolute error ≤0.001 for x² on [0,1]? Answer: n≥13. Which assumptions support the bound?

Mini drill or application

Complete the numerical lab. Compare n=10,20,40 errors for x²; ratios should be near 4 before roundoff dominates. Explain why those observations alone do not prove a universal convergence rate.

Read this only if stuck

OpenStax selective support. Use the reading guide for the section sequence and practice contract. Record your reasoning before consulting solutions.