Let's go through the problems step by step: --- ### (a) Find \(a, b, c, d\) so the quadrature formula has degree of precision 3 The formula is: \[ \int_{-1}^1 f(x)dx = a f(-1) + b f(1) + c f'(-1) + d f'(1) \] It should be **exact for all polynomials of degree ≤ 3**. #### Let's impose exactness for \(f(x) = 1, x, x^2, x^3\): --- #### 1. For \(f(x) = 1\): \[ \int_{-1}^1 1\,dx = 2 \] \[ a \cdot 1 + b \cdot 1 + c \cdot 0 + d \cdot 0 = a + b \] So, \[ a + b = 2 \] --- #### 2. For \(f(x) = x\): \[ \int_{-1}^1 x\,dx = 0 \] \[ a(-1) + b(1) + c \cdot 0 + d \cdot 0 = -a + b \] So, \[ -a + b = 0 \implies b = a \] --- #### 3. For \(f(x) = x^2\): \[ \int_{-1}^1 x^2\,dx = \frac{2}{3} \] \[ f(-1) = 1,\, f(1) = 1,\, f'(-1) = -2,\, f'(1) = 2 \] So, \[ a \cdot 1 + b \cdot 1 + c(-2) + d(2) = a + b - 2c + 2d = \frac{2}{3} \] --- #### 4. For \(f(x) = x^3\): \[ \int_{-1}^1 x^3\,dx = 0 \] \[ f(-1) = -1,\, f(1) = 1,\, f'(-1) = 3(-1)^2 = 3,\, f'(1) = 3(1)^2 = 3 \] So, \[ a(-1) + b(1) + c(3) + d(3) = -a + b + 3c + 3d = 0 \] --- Now, let's collect all equations: 1. \(a + b = 2\) 2. \(-a + b = 0\) 3. \(a + b - 2c + 2d = \frac{2}{3}\) 4. \(-a + b + 3c + 3d = 0\) From (2): \(b = a\) Plug into (1): \(2a = 2 \implies a = 1, b = 1\) Now substitute \(a = b = 1\) into (3) and (4): (3): \(1 + 1 - 2c + 2d = \frac{2}{3} \implies 2 - 2c + 2d = \frac{2}{3}\) \[ -2c + 2d = \frac{2}{3} - 2 = -\frac{4}{3} \] \[ -c + d = -\frac{2}{3} \] (4): \(-1 + 1 + 3c + 3d = 0 \implies 3c + 3d = 0 \implies c + d = 0\) Now solve: - \(c + d = 0 \implies d = -c\) - \(-c + d = -\frac{2}{3}\) Substitute \(d = -c\) into the second equation: \[ -c + (-c) = -\frac{2}{3} \implies -2c = -\frac{2}{3} \implies c = \frac{1}{3} \] \[ d = -c = -\frac{1}{3} \] **Final answer:** \[ \boxed{ a = 1,\quad b = 1,\quad c = \frac{1}{3},\quad d = -\frac{1}{3} } \] --- ### (b) Is it true that \(\sum_{i=1}^{n} w_i = 1\) for quadrature on \([0,1]\)? **No, not always.** **Justification:** If the quadrature rule integrates the constant function \(f(x) = 1\) exactly on \([0,1]\), then: \[ \int_0^1 1\,dx = 1 = \sum_{i=1}^n w_i \cdot 1 = \sum_{i=1}^n w_i \] So if the quadrature rule has at least degree of precision 0 (i.e., is exact for constants), then \(\sum w_i = 1\). However, for some quadrature formulas (such as some that include derivatives, or are not based on interpolating polynomials through points in \([0,1]\)), this may not be true. But **for standard polynomial interpolation quadrature formulas on \([0,1]\), yes, \(\sum_{i=1}^n w_i = 1\)**. --- ### (c) Prove \(M_k(f) \leq \int_a^b f(x)dx \leq T_k(f)\) for \(f''(x) \geq 0\) (convex function). #### **Proof:** - The **midpoint rule** underestimates the integral for convex functions (\(f'' \geq 0\)). - The **trapezoidal rule** overestimates it. For a single interval \([x_0, x_1]\): - Midpoint: \(M = h f\left(\frac{x_0 + x_1}{2}\right)\) - Trapezoid: \(T = \frac{h}{2}[f(x_0) + f(x_1)]\) - Actual integral: \(I = \int_{x_0}^{x_1} f(x)dx\) For convex \(f\), by **Jensen's inequality**: \[ f\left(\frac{x_0 + x_1}{2}\right) \leq \frac{f(x_0) + f(x_1)}{2} \] So, \[ M \leq T \] And using the properties of the midpoint and trapezoid rule error terms (since \(f'' \geq 0\)), the error for the midpoint rule is negative (underestimate), and for the trapezoid rule is positive (overestimate): \[ M \leq I \leq T \] For **composite rules** (sum over all subintervals), the same inequalities hold by linearity, so: \[ M_k(f) \leq \int_a^b f(x)dx \leq T_k(f) \] --- ### **Summary of Answers** **(a)** \[ \boxed{a = 1,\quad b = 1,\quad c = \frac{1}{3},\quad d = -\frac{1}{3}} \] **(b)** Yes, \(\sum w_i = 1\) for quadrature based on polynomial interpolation on \([0,1]\), because the rule is exact for constants. **(c)** For convex \(f\) (\(f'' \geq 0\)), the composite midpoint rule underestimates, and the composite trapezoid rule overestimates the integral: \[ M_k(f) \leq \int_a^b f(x)dx \leq T_k(f) \]