Numerical integration — trapezoidal rule (composite), error

1. From first principles — one strip

HOW we approximate one strip $[x_i, x_{i+1}]$ of width $h=x_{i+1}-x_i$ :

Replace $f$ by the straight line $L(x)$ through the endpoints $(x_i,f(x_i))$ and $(x_{i+1},f(x_{i+1}))$ . The region under $L$ is a trapezium with parallel sides $f(x_i)$ and $f(x_{i+1})$ and height $h$ .

2. Composite rule — chain the strips

Derivation by summing: $\int_a^b f\,dx \approx \sum_{k=0}^{n-1}\frac{h}{2}\big(f(x_k)+f(x_{k+1})\big)$ Write out terms: $f(x_0)+f(x_1)\;+\;f(x_1)+f(x_2)\;+\dots+\;f(x_{n-1})+f(x_n)$ . Every interior node $x_1,\dots,x_{n-1}$ appears in two strips → counted twice. Endpoints appear once.

Figure — Numerical integration — trapezoidal rule (composite), error

3. The error — where does it come from?

Single-strip error (derive the order). On $[x_i,x_{i+1}]$ , Taylor-expand and integrate the difference. The exact result (from integrating the interpolation error of a line): $E_{\text{strip}} = -\frac{h^3}{12}f''(\xi_i),\qquad \xi_i\in(x_i,x_{i+1})$

Composite error. Sum over $n$ strips and use the mean of the $f''(\xi_i)$ values $\to f''(\xi)$ : $E_n=\sum -\frac{h^3}{12}f''(\xi_i)= -\frac{h^3}{12}\,n\,f''(\xi)$ Now $n=\dfrac{b-a}{h}$ , so $n h^3 = (b-a)h^2$ :

4. Worked examples

Worked example Example 1 —

$\int_0^1 x^2\,dx$ with $n=4$ Exact $=\frac13$ . $h=0.25$ , nodes $0,0.25,0.5,0.75,1$ , $f=x^2$ : $0,0.0625,0.25,0.5625,1$ . $T_4=\frac{0.25}{2}\big[0 + 2(0.0625+0.25+0.5625) + 1\big]$ — Why? endpoints weight 1, interiors weight 2. $=0.125\,[\,0 + 2(0.875)+1\,]=0.125(2.75)=0.34375$ . Error $=0.34375-0.3333=0.0104>0$ — overestimate, expected since $f''=2>0$ (convex). ✓ Check bound: $|E|\le \frac{(1-0)}{12}(0.25)^2(2)=0.0104$ . Equality because $f''$ constant. ✓

5. Common mistakes (steel-manned)

Recall Feynman: explain to a 12-year-old

Imagine a wiggly hill and you want the area under it. You can't measure a wiggle, but you CAN measure straight slopes. So you cut the hill into thin slices and pretend each slice's top is a straight ramp. The area under a ramp is easy (a slanted box). Add all the boxes — that's your answer. If the hill curves a lot, thin slices guess wrong by a little; making slices twice as thin makes the mistake four times smaller.

Flashcards

Composite trapezoidal formula

T_n=\frac{h}{2}\big[f(x_0)+2\sum_{k=1}^{n-1}f(x_k)+f(x_n)\big]

h=\frac{b-a}{n}

Node weights pattern

1,2,2,\dots,2,1

(endpoints 1, interiors 2)

Global error of composite trapezoidal

E_n=-\frac{(b-a)}{12}h^2 f''(\xi)

, i.e.

O(h^2)

What happens to error if you double

n

It falls by a factor of

\approx 4

(since

E\propto h^2=1/n^2

)

Trapezoidal rule is exact for polynomials up to degree

1 (linear), because error contains

f''

Sign of error when

f

is convex (

f''>0

)

Positive — rule overestimates (chord above curve)

Single-strip error term

-\frac{h^3}{12}f''(\xi)

on each strip

Why interior nodes get weight 2

Each interior node is shared by two adjacent strips

Number of nodes for

n

strips

n+1

Connections

Simpson's Rule — fits parabolas, error $O(h^4)$ , exact to degree 3
Riemann Sums — trapezoidal = average of left & right Riemann sums
Taylor's Theorem — source of the interpolation error term
Linear Interpolation — the line $L(x)$ on each strip
Richardson Extrapolation — combine $T_n,T_{2n}$ to kill the $h^2$ term → Romberg
Numerical Methods — parent index

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, kabhi-kabhi integral $\int_a^b f(x)\,dx$ nikalna mushkil hota hai kyunki antiderivative milta hi nahi (jaise $e^{-x^2}$ ), ya phir function sirf kuch data points pe pata hai. Toh hum area ko approximate karte hain. Trapezoidal rule ka core idea simple hai: poore interval ko $n$ patli-patli strips mein kaato (har strip ki width $h=\frac{b-a}{n}$ ), aur har strip ke upar curve ko seedhi line maan lo. Seedhi line ke neeche ka area ek trapezium hota hai, jiska area $=\frac{h}{2}(\text{left height}+\text{right height})$ .

Jab sab strips ko jodte ho, toh interior wale points dono taraf ke strips mein aate hain, isliye unka weight 2 ho jata hai, aur do endpoints ka weight 1. Yahi se formula banta hai: $T_n=\frac{h}{2}[f_0+2(f_1+\dots+f_{n-1})+f_n]$ . Pattern yaad rakho: $1,2,2,\dots,2,1$ .

Ab error ki baat. Seedhi line curve ka curvature (mudna) ignore karti hai, isliye error $f''$ pe depend karta hai: $E_n=-\frac{(b-a)}{12}h^2 f''(\xi)$ . Sabse important baat — error $h^2$ ke proportional hai. Matlab agar tum strips ko double kar do ( $n$ double, $h$ aadha), toh error chaar guna kam ho jata hai! Agar function convex hai ( $f''>0$ , cup-jaisa upar khula), toh chord curve ke upar rahegi, toh rule thoda zyada estimate karega. Yeh intuition exam mein sign-check ke liye bahut kaam aata hai.

Test yourself — Numerical Methods

Connections