Question

$\displaystyle \rm \red{\lim_{n \to \infty } \sum_{k = 1}^{n} \bigg( 5 + \frac{3k}{n} \bigg) \frac{3}{n} }$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\displaystyle\rm {\lim_{n \to \infty } \sum_{k = 1}^{n} \bigg( 5 + \frac{3k}{n} \bigg) \frac{3}{n} }$

can be rewritten as

$\rm \: =  \: 3\displaystyle\rm {\lim_{n \to \infty } \sum_{k = 1}^{n} \bigg( 5 + \frac{3k}{n} \bigg) \frac{1}{n} }$

Now, using Limit as sum of definite integrals

$\rm \: \dfrac{1}{n} \: changes \: to \: dx$

$\rm \: \dfrac{k}{n} \: changes \: to \: x$

$\rm \: \displaystyle \sf \lim_{n \to \infty} \sum \: changes \: to \: \int$

$\rm \: lower \: limit \: a \: = \: \displaystyle \sf \lim_{n \to \infty}\frac{1}{n} = 0$

$\rm \: \: upper \: limit \: b \: = \: \displaystyle \sf \lim_{n \to \infty}\frac{n}{n} = \: 1$

So, above expression can be rewritten as

$\rm \: =  \: 3\displaystyle\int_0^1\rm (5 + 3x) \: dx$

$\rm \: =  \: 3\bigg[5x + \dfrac{ {3x}^{2} }{2} \bigg]_0^1$

$\rm \: =  \: 3\bigg[5 + \dfrac{3}{2} - 0 - 0 \bigg]$

$\rm \: =  \: 3\bigg[\dfrac{10 + 3}{2} \bigg]$

$\rm \: =  \: 3\bigg[\dfrac{13}{2} \bigg]$

$\rm \: =  \: \dfrac{39}{2}$

Thus,

$\rm\implies \:\boxed{\sf{  \:\displaystyle \rm {\lim_{n \to \infty } \sum_{k = 1}^{n} \bigg( 5 + \frac{3k}{n} \bigg) \frac{3}{n} }\rm \: =  \: \dfrac{39}{2} \: \: }}$

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Formula Used :-

$\boxed{\sf{  \:\displaystyle\int\rm {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: \: }}$

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Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$