Question

Write $\displaystyle \sf \lim_{n \to \infty} \sum \limits_{k = 1}^{n} \bigg( - 1 + k \frac{3}{n}\bigg)^{4} \frac{3}{n}$ as a definite integral.​​

Answer 1

${\huge{\star{\underbrace{\tt{\red{Answer}}}}}}{\huge{\star}}$

∫-12x4dx = x5/5-12 = 32/5 + 1/5 = 33/5.

n → ∞

Answer 2

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

Given expression is

$\rm :\longmapsto\:\displaystyle \sf \lim_{n \to \infty} \sum \limits_{k = 1}^{n} \bigg( - 1 + k \frac{3}{n}\bigg)^{4} \frac{3}{n}$

can be rewritten as

$\rm \:  =  \:3 \displaystyle \sf \lim_{n \to \infty} \sum \limits_{k = 1}^{n} \bigg( - 1 + 3\frac{k}{n}\bigg)^{4} \frac{1}{n}$

Now, using Limit as sum of definite integrals

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

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

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

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

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

So, given expression can be rewritten as

$\rm \:  =  \:3 \displaystyle \sf \int \limits_{0}^{1} \bigg( - 1 + 3x\bigg)^{4} \: dx$

can be rewritten as

$\rm \:  =  \:3 \displaystyle \sf \int \limits_{0}^{1} \bigg(3x - 1\bigg)^{4} \: dx$

We know,

$\boxed{\tt{ \int \: {(ax + b)}^{n}dx = \frac{ {(ax + b)}^{n + 1} }{a(n + 1)} + c}}$

So, using this, we get

$\rm \:  =  \: 3\bigg[\dfrac{ {(3x - 1)}^{4 + 1} }{3(4 + 1)} \bigg]_{0}^{1}$

$\rm \:  =  \: \bigg[\dfrac{ {(3x - 1)}^{5} }{5} \bigg]_{0}^{1}$

$\rm \:  =  \: \dfrac{1}{5}[ {(3 - 1)}^{5} - {(0 - 1)}^{5} ]$

$\rm \:  =  \: \dfrac{1}{5}[ {(2)}^{5} - {( - 1)}^{5} ]$

$\rm \:  =  \: \dfrac{1}{5}[ 32 + 1 ]$

$\rm \:  =  \: \dfrac{33}{5}$

Hence,

$\rm\implies \:\boxed{\tt{ \displaystyle \sf \lim_{n \to \infty} \sum \limits_{k = 1}^{n} \bigg( - 1 + k \frac{3}{n}\bigg)^{4} \frac{3}{n} = \frac{33}{5} \: }}$

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