$\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 r^3}{r^4+n^4}=p$
$$
e^p=\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 r^3}{n^4 \cdot\left[1+\frac{r^4}{n^4}\right]} \Rightarrow \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 \frac{r^3}{n^3}}{\left[n\left[1+\frac{r^4}{n^4}\right]\right.}
$$
Let $x=\frac{r}{n}$ and $d x=\frac{1}{n}$, so limit value will be from ' 0 ' to 1 .
So,
$$
\begin{aligned}
4 \int_0^1 \frac{x^3}{1+x^4} d x & \Rightarrow 1+x^4=t \\
4 x^3 d x=d t & \Rightarrow \int_1^2 \frac{d t}{t}
\end{aligned}
$$
Now, $x=0$, then $t=1 x=1$, then $t=2$
$$
\begin{array}{ll}
& {[\ln t]_1^2} \\
& \\
\text { So, } & \ln 2-\ln (1) \Rightarrow \ln 2=P \\
e^p=e^{\ln 2} \Rightarrow 2
\end{array}
$$