Infi-2 4

\begin{matrix} Let f (x) = \cosh x \\ Calculate revolution surface area on [0, 1] \\ A (f) = 2 π \int_{0}^{1} f (x) \sqrt{1 + (f^{'} (x))^{2}} d x = 2 π \int_{0}^{1} f^{2} (x) d x = 2 π \int_{0}^{1} \frac{e^{2 x} + 2 + e^{- 2 x}}{4} d x = \\ = \frac{π}{2} (\frac{e^{2 x}}{2} + 2 x - \frac{e^{- 2 x}}{2}) |_{0}^{1} = \frac{π}{2} (\frac{e^{2}}{2} + 2 - \frac{e^{- 2}}{2}) \end{matrix}

\begin{matrix} lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n}{n^{2} + k^{2}} = lim_{n \to \infty} \frac{1}{n} \cdot \frac{n^{2}}{n^{2} + k^{2}} = lim_{λ (P) \to 0} S (f, P, C) \\ f (c_{k}) = f (\frac{k}{n}) = \frac{n^{2}}{n^{2} + k^{2}} = \frac{1}{1 + \frac{k^{2}}{n^{2}}} = \frac{1}{1 + {(\frac{k}{n})}^{2}} \\ ⟹ f (x) = \frac{1}{1 + x^{2}} \\ ⟹ lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n}{n^{2} + k^{2}} = \int_{0}^{1} f (x) d x = \arctan (x) |_{0}^{1} = \frac{π}{4} \end{matrix}

\begin{matrix} \sum_{k = 1}^{n} \frac{k}{n^{2}} \sin (\frac{k}{n}) = \sum_{i = 1}^{n} \frac{1}{n} \cdot \frac{k}{n} \sin (\frac{n}{k}) = \int_{0}^{1} x \sin x d x = - x \cos x |_{0}^{1} + \int_{0}^{1} \cos x d x = \\ = - \cos 1 + \sin 1 \end{matrix}

\begin{matrix} \sum_{k = 1}^{n} \frac{k}{n^{2}} \sqrt[n]{e^{k}} = \int_{0}^{1} x e^{x} d x = (x - 1) e^{x} |_{0}^{1} = 1 \end{matrix}

\begin{matrix} {(\int_{α (x)}^{β (x)} f (t) d t)}^{'} = f (β (x)) β^{'} (x) - f (α (x)) α^{'} (x) \end{matrix}

\begin{matrix} \int_{x}^{x^{3}} \sin (t) \cos (t^{2}) e^{t} d t = \sin (x^{3}) \cos (x^{6}) e^{x^{3}} 3 x^{2} - \sin (x) \cos (x^{2}) e^{x} \end{matrix}

\begin{matrix} lim_{x \to 0} \frac{(\int_{0}^{x^{2}} \sin (t^{2}) d t)}{x^{6}} \overset{L}{=} lim_{x \to 0} \frac{2 x \sin (x^{4})}{6 x^{5}} = \frac{1}{3} lim_{x \to 0} \frac{\sin (x^{4})}{x^{4}} = \frac{1}{3} \end{matrix}

\begin{matrix} \int_{e^{- 1}}^{e} \frac{\ln (x) \sin (\ln (x) + 1)}{x} d x = {\begin{matrix} t = \ln x + 1 ⟹ d t = \frac{d x}{x} \\ t (e) = 2 \\ t (e^{- 1}) = 0 \end{matrix}} = \\ = \int_{0}^{2} (t - 1) \sin t d t d x = {\begin{matrix} f (t) = t - 1 ⟹ f^{'} (t) = 1 \\ g^{'} (t) = \sin t ⟹ g (t) = - \cos t \end{matrix}} = \\ = (1 - t) \cos t |_{0}^{2} + \sin t |_{0}^{2} = - \cos (2) - 1 + \sin (2) \end{matrix}

\begin{matrix} m \leq f (x) \leq M \\ m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a) \\ f (c) = \frac{(\int_{a}^{b} f (x) d x)}{b - a} \end{matrix}

\begin{matrix} Prove that the following equation has a unique solution on [- 1, 1] \\ x = \int_{0}^{x} \sin^{100} t d t \\ g (x) = x - \int_{0}^{x} \sin^{100} t d t \\ g (- 1) = - 1 + \underset{\leq 1}{\underset{⏟}{\int_{- 1}^{0} \sin^{100} t d t}} \leq 0 \\ g (1) = 1 - \underset{\geq - 1}{\underset{⏟}{\int_{0}^{1} \sin^{100} t d t}} \geq 0 \\ ⟹ \exists c \in [- 1, 1] : g (c) = 0 \\ g^{'} (x) = 1 - \sin^{100} x > 0 ⟹ \exists! c \in [- 1, 1] : g (c) = 0 \end{matrix}