If lim _{n rightarrow infty} sum_{r=1}^n frac | vedclass

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(C) Given the limit expression: $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 r^3}{r^4+n^4}=p$. We can rewrite the sum as: $\lim _{n \rightarrow

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(C) Given the limit expression: $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 r^3}{r^4+n^4}=p$. We can rewrite the sum as: $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 r^3}{n^4 \left(1+\frac{r^4}{n^4}\right)} = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 (r/n)^3}{1+(r/n)^4} \cdot \frac{1}{n}$. Using the definition of a definite integral as the limit of a sum, let $x = \frac{r}{n}$ and $dx = \frac{1}{n}$. As $n \rightarrow \infty$, the sum becomes the integral from $0$ to $1$: $p = \int_0^1 \frac{4x^3}{1+x^4} dx$. Let $t = 1+x^4$, then $dt = 4x^3 dx$. When $x=0$, $t=1$. When $x=1$, $t=2$. Thus, $p = \int_1^2 \frac{dt}{t} = [\ln t]_1^2 = \ln 2 - \ln 1 = \ln 2$. Therefore, $e^p = e^{\ln 2} = 2$.

If $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 r^3}{r^4+n^4}=p$, then $e^p=$

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