Question

The function is given $f: [0, b] \rightarrow \mathbb{R}$ with $f(x) = e^x$, $b > 0$ and $N \in \mathbb{N}$. a) Determine the equidistant decomposition $Z_N = \{x_0, x_1, ..., x_N\} = \{0, \frac{b}{N}, \frac{2b}{N}, \frac{3b}{N}, ..., b\}$ Sub sum belonging to $[0, b]$. $\underline{S}_{Z_N} = \sum_{i=1}^{N} \left( \min_{x \in [x_{i-1}, x_i]} f(x) \right) \cdot (x_i - x_{i-1})$. For control: $\underline{S}_{Z_N} = (e^b - 1) \cdot \frac{b}{N} \cdot \frac{1}{e^{\frac{b}{N}} - 1}$ b) Determine (without using knowledge about the antiderivative of "f) $\lim_{n \to \infty} \underline{S}_{Z_N}$

          The function is given $f: [0, b] \rightarrow \mathbb{R}$ with $f(x) = e^x$, $b > 0$ and $N \in \mathbb{N}$.
a) Determine the equidistant decomposition
$Z_N = \{x_0, x_1, ..., x_N\} = \{0, \frac{b}{N}, \frac{2b}{N}, \frac{3b}{N}, ..., b\}$
Sub sum belonging to $[0, b]$.
$\underline{S}_{Z_N} = \sum_{i=1}^{N} \left( \min_{x \in [x_{i-1}, x_i]} f(x) \right) \cdot (x_i - x_{i-1})$.
For control:
$\underline{S}_{Z_N} = (e^b - 1) \cdot \frac{b}{N} \cdot \frac{1}{e^{\frac{b}{N}} - 1}$
b) Determine (without using knowledge about the antiderivative of "f)
$\lim_{n \to \infty} \underline{S}_{Z_N}$

$The function is given f: [0, b] →ℝ with f(x) = e^x, b > 0 and N ∈ℕ. a) Determine the equidistant decomposition ZN = {x0, x1, ..., xN} = {0, (b)/(N), (2b)/(N), (3b)/(N), ..., b} Sub sum belonging to [0, b]. SZN = ∑i=1^N( minx ∈ [xi-1, xi] f(x) ) · (xi - xi-1). For control: SZN = (e^b - 1) ·(b)/(N)·(1)/(e^(b)/(N) - 1) b) Determine (without using knowledge about the antiderivative of "f) limn →∞SZN$

Added by Evan Y.

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