lim _{n rightarrow infty}left(frac{1^2}{n^3+1 | vedclass

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(D) Let $L = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^2}{n^3+r^3}$.We can rewrite the sum as:$L = \lim _{n \rightarrow \infty} \sum_{r=1}^n \

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $L = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^2}{n^3+r^3}$.We can rewrite the sum as:$L = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^2}{n^3(1+(r/n)^3)} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{(r/n)^2}{1+(r/n)^3}$.Using the definition of the definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n f(\frac{r}{n}) = \int_0^1 f(x) dx$,we get:$L = \int_0^1 \frac{x^2}{1+x^3} dx$.Let $1+x^3 = t$,then $3x^2 dx = dt$,or $x^2 dx = \frac{dt}{3}$.When $x=0, t=1$ and when $x=1, t=2$.$L = \int_1^2 \frac{1}{t} \cdot \frac{dt}{3} = \frac{1}{3} [\log |t|]_1^2 = \frac{1}{3} (\log 2 - \log 1) = \frac{1}{3} \log 2$.Using the property $a \log b = \log b^a$,we have $\frac{1}{3} \log 2 = \log 2^{1/3} = \log \sqrt[3]{2}$.

$\lim _{n \rightarrow \infty}\left(\frac{1^2}{n^3+1^3}+\frac{2^2}{n^3+2^3}+\ldots+\frac{n^2}{n^3+n^3}\right)=$

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