The obtained limit is identical to the limit that was specified. Therefore, the right answer is option C.
Generally, the equation for the limit is mathematically given as
[tex]$\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{4}{n} \sqrt{1+\frac{4 i}{n}}$.[/tex]
The goal is to locate the zone whose area corresponds to the value supplied by the limit.
The definite integral of a function is what is used to compute the area of that function that is underneath its graph.
The limit,
[tex]\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f(a+i \cdot \Delta x) \Delta x[/tex]
where[tex]$\Delta x=\frac{b-a}{n}$[/tex] and [tex]$x_{i}=a+i \Delta x$[/tex] for the interval $[a, b]$, is equivalent to the integral
[tex]\int_{a}^{b} f(x) d x .[/tex]
The given limit can also be written as
[tex]\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \sqrt{1+\left(\frac{4}{n}\right)} i \cdot \frac{4}{n} .[/tex]
In this limit, [tex]$\Delta x=\frac{4}{n}$[/tex]. It can be observed that[tex]$f(a+i \Delta x)=\sqrt{1+\left(\frac{4}{n}\right)} i$[/tex] which implies that [tex]$a=1$ and $f(x)=\sqrt{x}$.[/tex]
Solve the [tex]$\Delta x=\frac{b-a}{n}$[/tex]equation for as follows:
[tex]\begin{aligned}\frac{4}{n} &=\frac{b-1}{n} \\4 &=b-1 \\5 &=b\end{aligned}[/tex]
Therefore, the specified limit may be expressed as an integral as follows:
[tex]\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \sqrt{1+\left(\frac{4}{n}\right) i} \cdot \frac{4}{n}=\int_{1}^{5} \sqrt{x} d x[/tex]
Therefore, the limit that has been provided designates the area of the graph of "sqrt(x)" on the interval.[1,5]
However, none of the available choices are compatible with this choice. So, consider
[tex]a=0, f(x)=\sqrt{1+x}$ and $\Delta x=\frac{4}{n}$.[/tex]
Find the value of $b$ as:
[tex]$\begin{aligned}\frac{4}{n} &=\frac{b-0}{n} \\4 &=b\end{aligned}$[/tex]
Find the value of x_{i} as:
[tex]\begin{aligned}&x_{i}=0+\frac{4}{n} i \\&x_{i}=\frac{4}{n} i\end{aligned}[/tex]
$
Thus, the integral [tex]$\int_{0}^{4} \sqrt{1+x} d x$[/tex] can be expressed using the equation [tex]$\int_{a}^{b} f(x) d x=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(x_{i}\right) \Delta x$[/tex]
[tex]\begin{aligned}\int_{0}^{4} \sqrt{1+x} d x &=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \sqrt{1+\frac{4 i}{n} \frac{4}{n}} \\&=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{4}{n} \sqrt{1+\frac{4 i}{n}}\end{aligned}[/tex]
In conclusion, The obtained limit is identical to the limit that was specified. Therefore, the right answer is option C.
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