Quadrature - Derivation and Formulae

It is a Numerical tool that is used to find the area under a curve without performing a symbolic integration. Some of the basic Quadratures are as follows.

1. Left rectangle rule

$$Q(f)=f(a)(b-a)$$

2. Right rectangle rule

$$Q(f)=f(b)(b-a)$$

3. Midpoint rule

$$Q(f)=f\left(\frac{a+b}{2}\right)(b-a)$$

4. Trapezoid rule

$$Q(f)=\frac{f(a)+f(b)}{2}\left(b-a\right)$$

5. Kepler / Simpson’s Rule

$$Q(f)=\frac{b-a}{6}\left[f(a)+4f\left(\frac{a+b}{2}\right)+f(b)\right]$$

Derivations

Left/Right Rectangle Rule

Idea: Approximate $f$ by constant polynomial $f(a)$ on $[a,b]$

$$p_0(x)=f(a)$$ $$\boxed{{\int }_a^{b}f(x)dx\approx (b-a)f(a)}$$

And by constant $f(b)$ for Right Rectangle Rule

Midpoint Rule

Step 1: interpolate $f$ at midpoint $m=\frac{a+b}{2}$ with constant polynomial:

$$p_0(x)=f(m)$$

Step 2: Integrate the constant:

$$I(f)\approx {\int }_a^{b}f(m)dx=f(m)\cdot (b-a)$$

Result:

$$\boxed{{\int }_a^{b}f(x)dx\approx (b-a)f\left(\frac{a+b}{2}\right)}$$

Trapezoid Rule

Interpolate $f$ using Lagrange basis polynomials at points $(a,f(a))$ and $(b,f(b))$:

$$p_1(x)=f(a)\cdot L_0(x)+f(b)\cdot L_1(x)$$

where:

• $L_0(x)=\frac{x-b}{a-b}$ (equals $1$ at $\mathrm{x}=\mathrm{a},0$ at $\mathrm{x}=\mathrm{b}$ )
• $L_1(x)=\frac{x-a}{b-a}$ (equals $0$ at $\mathrm{x}=\mathrm{a},1$ at $\mathrm{x}=\mathrm{b}$ ) So,

$$p_1(x)=f(a)\frac{x-b}{a-b}+f(b)\frac{x-a}{b-a}$$ $$I(f)\approx {\int }_a^b{p}_1(x)dx$$

It can be calculated

• ${\int }_a^b\frac{x-b}{a-b}dx=\frac{b-a}{2}$

• ${\int }_a^b\frac{x-a}{b-a}dx=\frac{b-a}{2}$

  $$\boxed{I(f)\approx \frac{b-a}{2}[f(a)+f(b)]}$$

Simpson’s/Kepler’s Rule

Step 1: Interpolate f through three points: $(a,f(a))$, $(m,f(m))$, $(b,f(b))$ where $m=\frac{a+b}{2}$

Step 2: The quadratic polynomial is:

$$p_2(x)=f(a)L_0(x)+f(m)L_1(x)+f(b)L_2(x)$$

Step 3: Key insight - for equidistant points:

• ${\int }_a^b{L}_0(x)dx=\frac{b-a}{6}$
• ${\int }_a^b{L}_1(x)dx=\frac{4(b-a)}{6}$
• ${\int }_a^b{L}_2(x)dx=\frac{b-a}{6}$

Result:

$$\boxed{{\int }_a^{b}f(x)dx\approx \frac{b-a}{6}[f(a)+4f(m)+f(b)]}$$