Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your sample points.)
\[
\begin{array}{c}
\int_{1}^{4} \frac{x}{1+x^{5}} d x \\
\lim _{n \rightarrow \infty} \sum_{i=1}^{n}
\end{array}
\]
Answer
Finally, the integral can be expressed as a limit of Riemann sums as follows: \(\boxed{\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \frac{3(3i/n + 1)}{n((3i/n + 1)^5 + 1)}}\).
Step 1 ：First, we divide the interval from 1 to 4 into n equal subintervals. Each subinterval has a width of \(\Delta x = \frac{4-1}{n} = \frac{3}{n}\).
Step 2 ：The right endpoint of the i-th subinterval is given by \(x_i = 1 + i\Delta x = 1 + \frac{3i}{n}\).
Step 3 ：We then calculate the height of the rectangle for each subinterval using the function \(f(x_i) = \frac{x_i}{1 + x_i^5} = \frac{3i/n + 1}{(3i/n + 1)^5 + 1}\).
Step 4 ：The Riemann sum is the sum of the areas of these rectangles, which is given by \(\sum_{i=1}^{n} f(x_i)\Delta x = \sum_{i=1}^{n} \frac{3(3i/n + 1)}{n((3i/n + 1)^5 + 1)}\).
Step 5 ：Finally, the integral can be expressed as a limit of Riemann sums as follows: \(\boxed{\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \frac{3(3i/n + 1)}{n((3i/n + 1)^5 + 1)}}\).
