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Determine a region whose area is equal to the given limit. Do not evaluate the limit.

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

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 function 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 standard 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

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