Rozwiąż t = limit (as n approaches infty) of sum_k=1^nf(1+k*4/n)*4/n=

$$t = \lim _ { n \rightarrow \infty } \sum _ { k = 1 } ^ { n } f ( 1 + k \cdot \frac { 4 } { n } ) \cdot \frac { 4 } { n } =$$

Rozwiąż względem f

$$\left\{\begin{matrix}f=\frac{nt}{4\left(3n+2\right)}\text{, }&n\neq -\frac{2}{3}\text{ and }n\neq 0\\f\in \mathrm{R}\text{, }&n=-\frac{2}{3}\text{ and }t=0\end{matrix}\right,$$

Rozwiąż względem t

$$t=12f$$