By the definition of the definite integral,th | vedclass

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Question

(B) The given limit is $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^2}{r^3+n^3}$.We can rewrite this as $S = \lim _{n \rightarrow \infty}

Vedclass · Accepted Answer

$\log \sqrt[3]{2}$

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(B) The given limit is $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^2}{r^3+n^3}$.We can rewrite this as $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^2}{n^3( (r/n)^3 + 1 )}$.Multiplying and dividing by $n$,we get $S = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{(r/n)^2}{(r/n)^3 + 1}$.Using the definition of the definite integral $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f(\frac{r}{n}) = \int_{0}^{1} f(x) dx$,we have $S = \int_{0}^{1} \frac{x^2}{x^3+1} dx$.Let $u = x^3+1$,then $du = 3x^2 dx$,which implies $x^2 dx = \frac{du}{3}$.When $x=0$,$u=1$. When $x=1$,$u=2$.Thus,$S = \int_{1}^{2} \frac{1}{u} \cdot \frac{du}{3} = \frac{1}{3} [\ln |u|]_{1}^{2} = \frac{1}{3} (\ln 2 - \ln 1) = \frac{1}{3} \ln 2 = \ln 2^{1/3} = \log \sqrt[3]{2}$.

By the definition of the definite integral,the value of $\lim _{n \rightarrow \infty}\left[\frac{1^2}{1^3+n^3}+\frac{2^2}{2^3+n^3}+\ldots+\frac{n^2}{n^3+n^3}\right]=$

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