\[\lim_{n\to\infty}\sum_{i=1}^{n}\frac{\pi}{4n}{tan\left(\frac{i\pi}{4n}\right)} \]

The purpose of this article is to find the **region** having an **area under the curve** that is represented by a given **limit**.

The basic concept behind this guide is the use of the **Limit Function** to determine an **area of the region**. The **area of a region** that covered the space above the $x-axis$ and the below the **curve of given function** $f$ **integrable** on $a$ to $b$ is calculated by **integrating the curve functio**n over a **limit interval**. The function is expressed as follows:

\[\int_{a}^{b}{f(x)dx} \]

The **area of the region** enclosed by $x-axis$ and **curve function** $f$ is expressed in **limit form** as follows:

\[\int_{a}^{b}{f(x)dx}=\lim_{n\to\infty}\sum_{i=1}^{n}f{(x_i)}∆x \]

Where:

\[x_i=a+i ∆x \]

So:

\[\int_{a}^{b}{f(x)\ dx}\ =\lim_{n\to\infty}\sum_{i=1}^{n} f(a+i∆x) ∆x \]

Here:

\[∆x = \frac{b-a}{n} \]

## Expert Answer

Given **Function** is:

\[\int_{a}^{b}{\ f(x)\ \ dx}\ =\ \lim_{n\to\infty} \sum_{i\ =\ 1}^{n}{\ \frac{\pi}{4n}}{\ tan\ \left(\frac{i\pi}{4n}\right)} \]

We know that the **standard form** for an **area of the region**:

\[\int_{a}^{b}{f(x)\ dx}\ =\lim_{n\to\infty}\sum_{i=1}^{n} f(a+i∆x) ∆x \]

Comparing the given function with the **s****tandard function**, we find the value of each component as follows:

\[a\ +\ i\ ∆x = \frac{i\pi}{4n} \]

Hence:

\[a\ =\ 0 \]

\[∆x = \frac{\pi}{4n} \]

As we know:

\[∆x = \frac{b-a}{n}=\frac{\pi}{4n} \]

\[\frac{b-0}{n}\ =\ \frac{\pi}{4n} \]

\[b\ =\ \frac{\pi}{4} \]

Let’s consider:

\[f(x)\ =\ tan\ (x) \]

So:

\[\lim_{n\to\infty}\sum_{i=1}^{n}\frac{\pi}{4n}{tan\left(\frac{i\pi}{4n}\right)}\ =\ \int_{a}^{b}{\ f(x)\ dx} \]

Substituting the values on the left-hand side of the above expression:

\[\lim_{n\to\infty}\sum_{i=1}^{n}\frac{\pi}{4n}{tan\left(\frac{i\pi}{4n}\right)}\ =\ \int_{0}^{\frac{\pi}{4}}{\ tan\ (x)\ dx\ =\ 0.346} \]

The **equation for the curve** is:

\[f(x)\ =\ tan\ (x) \]

The **interval** for $x-axis$ is:

\[x\ \in\ \left[0,\ \frac{\pi}{4}\right] \]

It is represented by the following graph:

Figure 1

## Numerical Result

The **region**, having an **area** defined by the given **limit,** is equal to the region below the following **curve function** and above $x-axis$ for the given **interval**, as follows:

\[f(x)\ =\ tan(x),\ \ x\ \in\ \left[0,\ \frac{\pi}{4}\right] \]

Figure 1

## Example

Find an expression for the **region** having an **area** equal to the following **limit**:

\[\lim_{n\to\infty}\ \sum_{i\ =\ 1}^{n}{\ \frac{2}{n}}\ {\left(5\ +\ \frac{2i}{n}\right)} \]

**Solution**

Given **Function** is:

\[\int_{a}^{b}{\ f(x)\ dx}\ =\ \lim_{n\to\infty}\ \sum_{i\ =\ 1}^{n}{\ \frac{2}{n}}{\ \left(5\ +\ \frac{2i}{n}\right)} \]

We know that the **standard form** for an **area of the region**:

\[\int_{a}^{b}{f(x)\ dx}\ =\lim_{n\to\infty}\sum_{i=1}^{n} f(a+i∆x) ∆x \]

Comparing the given function with the **standard function**, we find the value of each component as follows:

\[a\ +\ i∆x = 5 + i \frac{2}{n} \]

Hence:

\[a\ =\ 5 \]

\[∆x =\frac{2}{n} \]

As we know:

\[∆x = \frac{b-a}{n} \]

\[\frac{b-5}{n}\ =\ \frac{2}{n} \]

\[b\ =\ 7 \]

Let’s consider:

\[f(x)\ =\ 5\ +\ x \]

So:

\[ \lim_{n\to\infty}\ \sum_{i\ =\ 1}^{n}{\ \frac{2}{n}}\ {\left(5\ +\ \frac{2i}{n}\right)}\ =\ \int_{a}^{b}{\ f(x)\ dx} \]

Substituting the values on the left-hand side of the above expression:

\[ \lim_{n\to\infty}\ \sum_{i\ =\ 1}^{n}{\ \frac{2}{n}}\ {\left(5\ +\ \frac{2i}{n}\right)}\ =\ \int_{5}^{7}{\ (5\ +\ x)\ dx} \]

The **equation for the curve** is:

\[ f(x)\ =\ 5\ +\ x \]

The **interval** for $x-axis$ is:

\[ x\ \in\ \left[5,\ 7\right] \]

*Image/Mathematical drawings are created in Geogebra*