GUIDING QUESTION: How can I compute an integral numerically?
Motivation The definite integral $$\int_a^b f(x) dx$$ computes the area under the graph of $f$, between $x = a$ and $x = b$. How can I compute it? The Riemann integral is defined as a limit of Riemann sums Riemann sum.Guiding philosophy: We may approximate this limit numerically by taking a finite sum.
Quadrature rules So we set out to find nodes $x_0, x_1, \ldots, x_n \in [a, b]$ and weights $w_0, w_1, \ldots, w_n$ to approximate $$\int_a^b f(x) dx \color{var(--emphColor)}{\approx} \displaystyle \sum_j w_j f_j,$$ with $f_j = f(x_j)$. Déjà vu? $$f'(x_i) \approx \sum_j \alpha_j f_j$$ In the last module, we used linear combinations of function values to approximate the function's derivative. Here, we'll find which linear combinations of function values are good approximations to the function's definite integral! Via interpolant We'll also use the interpolating polynomial here. Let $x_0, x_1, \ldots, x_n$ denote $\bbox[3pt, border: 3pt solid var(--emphColor)]{n + 1}$ evenly-spaced nodes on the interval $[a, b]$. Let $p(x)$ denote the polynomial of degree at most $n$ interpolating $f$ at the given $x_i$. We'll approximate \begin{equation*} \int_a^b f(x) dx \color{var(--emphColor)}{\approx} \int_a^b p(x) dx. \end{equation*} We'll restrict our attention to equidistant nodes for simplicity. Newton-Cotes quadrature The nodes are given, so we need only find the right weights $\color{var(--emphColor)}{w_j}$. We'll obtain these by integrating the interpolating polynomial $p$ analytically. Newton-Cotes quadrature In the Lagrange basis, $p(x) = \sum_j f_j L_j(x)$, so \begin{align*} \int_a^b f(x) dx &\approx \int_a^b p(x) dx = \int_a^b \sum_{j = 0}^n f_j L_{j}(x) dx \\ \end{align*} \begin{align} & \,\,= \sum_{j = 0}^n \bigg( \int_a^b L_{j}(x)dx \bigg) f_j \\ & \,\,= \displaystyle \sum_j \color{var(--emphColor)}{w_j} f_j, \end{align} with $\color{var(--emphColor)}{w_j} = \int_a^b L_j(x) dx$. When deriving finite difference formulas, we obtained \begin{equation*} f'(x_i) \approx p'(x_i) = \sum_{j = 0}^n \color{var(--emphColor)}{d_{ij}} f_j, \end{equation*} with $\color{var(--emphColor)}{d_{ij}} = L_{j}'(x_i)$. We he an analogous statement here! \begin{equation*} \int_{a}^{b} f(x) dx \approx \int_a^b p(x) dx = \sum_{j = 0}^n \color{var(--emphColor)}{w_j} f_j, \end{equation*} with $\color{var(--emphColor)}{w_j} = \int_{a}^{b} L_{j}(x) dx$. When computing derivatives or integrals numerically, we approximate the desired quantity by a weighted sum of function values, and performing the appropriate operation on the Lagrange basis polynomials gives the right weight. When the nodes $x_j$ are given and the weights $\color{var(--emphColor)}{w_j}$ are defined by integrals of Lagrange basis polynomials, the quadrature rule $$\int_a^b f(x) dx {\approx} \displaystyle \sum_j \color{var(--emphColor)}{w_j} f_j$$ is a Newton-Cotes (NC) rule. The rule is said to be closed if $x_0 = a$ and $x_n = b$. Otherwise, the rule is said to be open. Since we are considering equidistant nodes only, the nodes for a closed rule are When the rule is open, the nodes are General Newton-Cotes To specify an NC rule, we must Decide whether it is open or closed Determine $\bbox[3pt, border: 3pt solid var(--emphColor)]{n + 1}$, the number of nodes Compute $w_j = \int_a^b L_{n,j}(x)dx$, where $L_{j}$ denotes the $j$th Lagrange basis polynomial. (Closed NC rule, $\bbox[3pt, border: 3pt solid var(--emphColor)]{n + 1} = 3$) As usual, let $h = x_{i+1} - x_i$. The rule is closed, so $x_0 = a$, $x_1 = (a + b)/2$, $x_2 = b$, and $h = (b-a) / 2$. By definition, $$\color{var(--emphColor)}{w_0} = \int_a^b L_{0}(x) dx = \int_a^b \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0 - x_2)} dx.$$ The substitution $$\bbox[3pt, border: 3pt solid var(--emphColor)]{x = a + th}$$ is very helpful when computing the weights! (Closed NC rule, $\bbox[3pt, border: 3pt solid var(--emphColor)]{n + 1} = 3$) The rule is closed, so $h = x_{i+1} - x_i = (b-a) / 2$ and therefore $$x_0 = a, \quad x_1 = (a + b)/2, \quad \text{and} \quad x_2 = b.$$ By definition, $$\color{var(--emphColor)}{w_0} = \frac{h}{2} \int_0^2 (t-1)(t-2) dt$$ (Closed NC rule, $\bbox[3pt, border: 3pt solid var(--emphColor)]{n + 1} = 3$) The rule is closed, so $h = x_{i+1} - x_i = (b-a) / 2$ and therefore $$x_0 = a, \quad x_1 = (a + b)/2, \quad \text{and} \quad x_2 = b.$$ By definition, $$\color{var(--emphColor)}{w_0} = \frac{h}{2} \bigg(\frac{t^3}{3} - \frac{3}{2}t^2 + 2t \bigg|_{t=2}\bigg)$$ (Closed NC rule, $\bbox[3pt, border: 3pt solid var(--emphColor)]{n + 1} = 3$) The rule is closed, so $h = x_{i+1} - x_i = (b-a) / 2$ and therefore $$x_0 = a, \quad x_1 = (a + b)/2, \quad \text{and} \quad x_2 = b.$$ By definition, $$\color{var(--emphColor)}{w_0} = \frac{h}{3}$$ (Closed NC rule, $\bbox[3pt, border: 3pt solid var(--emphColor)]{n + 1} = 3$) The rule is closed, so $h = x_{i+1} - x_i = (b-a) / 2$ and therefore $$x_0 = a, \quad x_1 = (a + b)/2, \quad \text{and} \quad x_2 = b.$$