Question

Evaluate The Following Definite Integrals As Limit Of Sums:-
$\orange{\boxed{ \boxed{ \displaystyle \tt \pink{ \leadsto\int_{1}^{4}\left(3 x^{2}+2 x+5\right) d x}}}}$
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Answer 1

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

Given integral is

$\rm \: \displaystyle\int_1^4\rm ( {3x}^{2} + 2x + 5) \: dx$

So, Here

Lower limit, a = 1

Upper limit, b = 4

$\rm \: f(x) = {3x}^{2} + 2x + 5$

Step :- 1

$\rm \: nh = b - a = 4 - 1 = 3$

Step :- 2

$\rm \: f(x) = {3x}^{2} + 2x + 5$

Put x = a + rh = 1 + rh

So, on substituting the values, we get

$\rm \: f(1 + rh) = {3(1 + rh)}^{2} + 2(1 + rh) + 5$

$\rm \: = \: 3(1 + {r}^{2} {h}^{2} + 2rh) + 2(1 + rh) + 5$

$\rm \: = \: 3 + 3{r}^{2} {h}^{2} + 6rh + 2 + 2rh + 5$

$\rm \: = \:3{r}^{2} {h}^{2} + 8rh + 10$

Step :- 3

By definition of Limit as a Sum,

$\begin{gathered}\rm \: \displaystyle\int_a^b \: f(x) \: dx \: = \lim _{h \to \: 0}\bigg(h \sum_{r = 0}^{n - 1} f(a + rh)\bigg) \\ \end{gathered}$

So, on substituting the values, we get

$\begin{gathered}\rm \: \displaystyle\int_1^4 ( {3x}^{2} + 2x + 5) \: dx \: = \lim _{h \to \: 0}\bigg(h \sum_{r = 0}^{n - 1} f(1 + rh)\bigg) \\ \end{gathered}$

$\rm \: = \lim _{h \to \: 0}\bigg(h \sum_{r = 0}^{n - 1} (3 {r}^{2} {h}^{2} + 8rh + 10) \bigg)$

$\rm \: = \lim _{h \to \: 0}\bigg(3 {h}^{3} \sum_{r = 0}^{n - 1} {r}^{2} + 8{h}^{2}\sum_{r = 0}^{n - 1}r + 10h\sum_{r = 0}^{n - 1}1 \bigg)$

$\rm \: = \lim _{h \to \: 0}\bigg(3 {h}^{3} \frac{n(n - 1)(2n - 1)}{6} + 8 {h}^{2} \frac{n(n - 1)}{2} + 10nh \bigg)$

$\rm \: = \lim _{h \to \: 0}\bigg( \frac{nh(nh - h)(2nh - h)}{2} + 4 nh(nh - h) + 10nh \bigg)$

$\rm \: = \dfrac{3(3 - 0)(6 - 0)}{2} + 4(3)(3) + 10(3)$

$\rm \: = \: 27 + 36 + 30$

$\rm \: = \: 93$

Hence,

$\rm\implies \:\boxed{\tt{ \rm \: \displaystyle\int_1^4\rm ( {3x}^{2} + 2x + 5) \: dx = 93 \: }}$

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Formulae Used

$\boxed{\tt{ \sum_{r = 0}^{n - 1}1 \: = \: n \: }}$

$\boxed{\tt{ \sum_{r = 0}^{n - 1}r \: = \: \frac{n(n - 1)}{2} \: }}$

$\boxed{\tt{ \sum_{r = 0}^{n - 1} {r}^{2} \: = \: \frac{n(n - 1)(2n - 1)}{6} \: }}$

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ADDITIONAL INFORMATION

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

Answer 2

this is the answer

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