Determine which of the following regions has an area equal to the given limit without evaluating the limit
\[
\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{\pi}{4 n} \tan \left(\frac{i \pi}{4 n}\right)
\]
A. The area of the region under the graph of $y=\sec (x)$ on the interval $[0, \pi / 4]$.
B. The area of the region under the graph of $y=\tan (x)$ on the interval $[0, \pi / 8]$.
C. The area of the region under the graph of $y=\sec (x)$ on the interval $[0, \pi / 8]$.
D. The area of the region under the graph of $y=\tan (x)$ on the interval $[0, \pi / 4]$.

Step 1 ：The given limit is a Riemann sum for the definite integral of the function \(f(x) = \frac{\pi}{4} \tan(x)\) from 0 to \(\frac{\pi}{4}\). This is because the limit of the sum of \(f(x_i)\Delta x\) as \(n \rightarrow \infty\) is the definition of the definite integral of \(f(x)\) from \(a\) to \(b\), where \(x_i\) are points in the partition of the interval \([a, b]\) and \(\Delta x = \frac{b - a}{n}\). In this case, \(a = 0\), \(b = \frac{\pi}{4}\), \(f(x_i) = \frac{\pi}{4} \tan(x_i)\), and \(\Delta x = \frac{\pi}{4n}\). Therefore, the limit is equal to the area under the graph of \(y = \frac{\pi}{4} \tan(x)\) on the interval \([0, \pi / 4]\).

Step 2 ：The area under the graph of \(y = \frac{\pi}{4} \tan(x)\) on the interval \([0, \pi / 4]\) can be calculated as \(-\frac{\pi}{4} \log\left(\frac{\sqrt{2}}{2}\right)\).

Step 3 ：Comparing the options, we find that the area corresponds to the region under the graph of \(y=\tan (x)\) on the interval \([0, \pi / 4]\).

Step 4 ：Final Answer: \(\boxed{\text{D. The area of the region under the graph of } y=\tan (x) \text{ on the interval } [0, \pi / 4]}\)