91Ó°ÊÓ

The given limit is similar to the formula for a right Riemann sum: \(\lim _{n \rightarrow\infty} \sum_{i=1}^{n} \frac{\pi}{4 n} \tan \frac{i \pi}{4 n}\) To rewrite it as an integral, note that \(\frac{\pi}{4n}\) is the width of a partition (or a "small step"), and \(\tan(\frac{i\pi}{4n})\) represents the value of a function at a certain point in that partition. Intuitively, this sum is calculating an approximation of the area under the curve of a function. The integral representing the area under the curve of the function f(x) would be: \(\int_{a}^{b} f(x) dx\) Now we just need to find the function \(f(x)\), and the limits of integration, \(a\) and \(b\).

To find the function and the limits of integration, observe the terms in the sum. In the given sum: \(\lim _{n \rightarrow\infty} \sum_{i=1}^{n} \frac{\pi}{4 n} \tan \frac{i \pi}{4 n}\) The function should derive from the factor \(\tan(\frac{i\pi}{4n})\). Since \(i\) in this sum is analogous to replacing every \(x\) in our function, the function \(f(x)\) is given by: \(f(x) = \tan (\frac{\pi x}{4})\) Now let's find the limits of integration. The given sum starts from \(i=1\) and ends with \(i=n\). We can find the corresponding \(x\) values for these values of \(i\) considering that: \(x = \frac{i}{n}\) For \(i = 1\), we have \(x = \frac{1}{n}\), and as \(n \to \infty\), \(x \to 0\). For \(i = n\), we have \(x = \frac{n}{n} = 1\). Hence, the limits of integration are \(a = 0\) and \(b = 1\).

Now that we have found the function \(f(x)\) and the limits of integration \(a\) and \(b\), we can rewrite the given limit as an integral: \(\lim _{n \rightarrow\infty} \sum_{i=1}^{n} \frac{\pi}{4 n} \tan \frac{i \pi}{4 n} = \int_{0}^{1} \tan \frac{\pi x}{4} dx\) This integral represents the area of the region we seek. We were told not to evaluate the limit, so we stop here. The region whose area is equal to the given limit is the area under the curve \(y = \tan (\frac{\pi x}{4})\) on the interval \([0, 1]\).