By the definition of the definite integral,th | vedclass

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(B) The given limit is $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^4}{r^5+n^5}$.We can rewrite this as $S = \lim _{n \rightarrow \infty}

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

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(B) The given limit is $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^4}{r^5+n^5}$.We can rewrite this as $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^4}{n^5( (r/n)^5 + 1 )} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{(r/n)^4}{(r/n)^5 + 1}$.Using the definition of the definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f(r/n) = \int_0^1 f(x) dx$.Here,$f(x) = \frac{x^4}{x^5 + 1}$.Thus,$S = \int_0^1 \frac{x^4}{x^5 + 1} dx$.Let $t = x^5 + 1$,then $dt = 5x^4 dx$,or $x^4 dx = \frac{dt}{5}$.When $x = 0$,$t = 1$. When $x = 1$,$t = 2$.$S = \int_1^2 \frac{1}{t} \cdot \frac{dt}{5} = \frac{1}{5} [\ln |t|]_1^2 = \frac{1}{5} (\ln 2 - \ln 1) = \frac{1}{5} \ln 2$.

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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