Infi-2 3

\begin{matrix} \int_{0}^{1} x d x = {\begin{matrix} Δ x_{i} = \frac{1}{m} \\ x_{i} = \frac{i}{m} \end{matrix}} = lim_{m \to \infty} \sum_{i = 1}^{m} \frac{f (\frac{i}{m})}{m} = lim_{m \to \infty} \sum_{i = 1}^{m} \frac{i}{m^{2}} = lim_{m \to \infty} \frac{1}{m^{2}} \sum_{i = 1}^{m} i = \\ = lim_{m \to \infty} \frac{1}{m^{2}} \cdot \frac{m (m + 1)}{2} = lim_{m \to \infty} \frac{m^{2} + m}{2 m^{2}} = \frac{1}{2} \end{matrix}

\begin{matrix} \int_{0}^{5} (5 - x) d = {\begin{matrix} δ x_{i} = \frac{5}{m} \\ x_{i} = \frac{5 i}{m} \end{matrix}} = lim_{m \to \infty} \sum_{i = 1}^{m} (5 - \frac{5 i}{m}) \cdot \frac{5}{m} = \\ = lim_{m \to \infty} \sum_{i = 1}^{m} \frac{25}{m} - \sum_{i = 1}^{m} \frac{25 i}{m^{2}} = 25 - lim_{m \to \infty} \frac{25}{m^{2}} \sum_{i = 1}^{m} i = \frac{25}{2} \end{matrix}

\begin{matrix} \int_{0}^{1} x^{2} d x = {\begin{matrix} Δ x_{i} = \frac{1}{m} \\ x_{i} = \frac{i}{m} \end{matrix}} = lim_{m \to \infty} \sum_{i = 1}^{m} \frac{i^{2}}{m^{3}} = lim_{m \to \infty} \frac{1}{m^{3}} \sum_{i = 1}^{m} i^{2} = \\ = lim_{m \to \infty} \frac{m (m + 1) (2 m + 1)}{6 m^{3}} = lim_{m \to \infty} \frac{2 m^{3} + 3 m^{2} + m}{m^{3}} = \frac{1}{3} \end{matrix}

\begin{matrix} \int_{3}^{5} x d = {\begin{matrix} Δ x_{i} = 3 + \frac{2}{m} \\ x_{i} = \frac{2 i}{m} \end{matrix}} = lim_{m \to \infty} \sum_{i = 1}^{m} \frac{6}{m} + \frac{4 i}{m^{2}} = lim_{m \to \infty} 6 \sum_{i = 1}^{m} \frac{1}{m} + \frac{4 m^{2} + 4 m}{2 m^{2}} = \\ = 6 + 2 = 8 \end{matrix}

\begin{matrix} \int_{0}^{1} \sqrt{x} d x = {\begin{matrix} Δ x_{i} = \frac{1}{m} \\ x_{i} = \frac{i}{m} \end{matrix}} = lim_{m \to \infty} \sum_{i = 1}^{m} \sqrt{\frac{i}{m}} \frac{1}{m} = lim_{m \to \infty} \frac{1}{m \sqrt{m}} \sum_{i = 1}^{m} \sqrt{i} = ? ? ? \\ \int_{0}^{1} \sqrt{x} d x = {\begin{matrix} Δ x_{i} = \frac{2 i - 1}{m^{2}} \\ x_{i} = \frac{i^{2}}{m^{2}} \end{matrix}} = lim_{m \to \infty} \sum_{i = 1}^{m} \frac{i (2 i - 1)}{m^{3}} = lim_{m \to \infty} \frac{2}{m^{3}} \sum_{i = 1}^{m} i^{2} - \frac{1}{m^{3}} \sum_{i = 1}^{m} i = \\ = lim_{m \to \infty} \frac{2 (2 m^{3} + 3 m^{2} + m)}{6 m^{3}} - \frac{m^{2} + m}{2 m^{3}} = lim_{m \to \infty} \frac{4 m^{3} + 3 m^{2} - m}{6 m^{3}} = \frac{2}{3} \end{matrix}

\begin{matrix} D (x) = {\begin{matrix} 0 & x \in Q \\ 1 & x \notin Q \end{matrix} \\ x_{i} \in Q ⟹ \sum_{i = 1}^{m} \frac{1}{m} \cdot 0 \to 0 \\ x_{i} \notin Q ⟹ \sum_{i = 1}^{m} \frac{1}{m} \cdot 1 \to 1 \\ ⟹ D (x) is not Riemann-integrable \end{matrix}

\begin{matrix} f (x) = {\begin{matrix} \frac{1}{x^{2}} & x > 0 \\ 0 & x = 0 \end{matrix} on [0, 1] \\ f is not bounded at 0 ⟹ f is not integrable on [0, 1] \end{matrix}

\begin{matrix} \int_{1}^{4} \frac{1}{x^{3}} d x = {\begin{matrix} Δ x_{i} = 4^{i / n} (1 - \frac{1}{4^{1 / n}}) \\ x_{i} = 4^{i / n} \end{matrix}} = lim_{n \to \infty} \sum_{i = 1}^{n} \frac{1}{4^{3 i / n}} \cdot 4^{i / n} (1 - \frac{1}{4^{1 / n}}) = \\ = lim_{n \to \infty} (1 - \frac{1}{4^{1 / n}}) \cdot \sum_{i = 1}^{n} \frac{1}{4^{2 i / n}} = lim_{n \to \infty} (1 - \frac{1}{4^{1 / n}}) \cdot \frac{1}{4^{2 / n}} \cdot \frac{(\frac{1}{4^{2}} - 1)}{\frac{1}{4^{2 / n}} - 1} = \\ lim_{n \to \infty} \frac{\frac{1}{4^{2 / n}} (1 - \frac{1}{16})}{1 + 4^{1 / n}} = \frac{15}{32} \end{matrix}