By the definition of the definite integral,th | vedclass

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(B) The given limit can be expressed as a Riemann sum: $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^4}{r^5+n^5} = \lim _{n \rightarrow \in

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$\frac{1}{5} \log 2$

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(B) The given limit can be expressed as a Riemann sum: $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^4}{r^5+n^5} = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^4}{n^5( (r/n)^5 + 1 )} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{(r/n)^4}{(r/n)^5 + 1}$. By the definition of the definite integral,this is equal to $\int_0^1 \frac{x^4}{x^5+1} dx$. Let $t = x^5 + 1$,then $dt = 5x^4 dx$,or $x^4 dx = \frac{dt}{5}$. When $x=0, t=1$. When $x=1, t=2$. Thus,the integral becomes $\int_1^2 \frac{1}{t} \cdot \frac{dt}{5} = \frac{1}{5} [\log |t|]_1^2 = \frac{1}{5} (\log 2 - \log 1) = \frac{1}{5} \log 2$.

By the definition of the definite integral,the value of $\lim _{n \rightarrow \infty}\left(\frac{1^4}{1^5+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\ldots+\frac{n^4}{n^5+n^5}\right)$ is

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