Question

Find the following integrals. ​

Answer 1

$\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}$

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Given,

$\displaystyle\rm \int \frac{(2x - 1)(x + 3)}{6} \: dx$

Use Constant Factor Rule:

$\displaystyle \rm\int cf(x) \, dx=c\int f(x) \, dx$

$\\ \rm \implies \: \frac{1}{6}\int (2x-1)(x+3) \, dx$

$\\ \rm\implies \frac{1}{6} \int {2x}^{2} + 5x - 3 \: dx$

Use Power Rule:

$\displaystyle \rm\int {x}^{n} \, dx=\frac{{x}^{n+1}}{n+1}+C$

$\\ \rm\frac{{x}^{3}}{9}+\frac{5{x}^{2}}{12}-\frac{x}{2}+C$

$\pmb{ \red{ \boxed{\displaystyle\rm \int \frac{(2x - 1)(x + 3)}{6} \: dx = \frac{{x}^{3}}{9}+\frac{5{x}^{2}}{12}-\frac{x}{2}+C}}}$

Answer 2

$\rm Question$ :- Find the following integrals

$\rm \: a. \: \: \: \: \displaystyle\int\rm \dfrac{(2x - 1)(x + 3)}{6} \: dx$

$\rm \: b. \: \: \: \: \displaystyle\int_{4}^{9}\rm \bigg(\dfrac{1}{ \sqrt{x} } - 2\bigg) \: dx$

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

Given integral is

$\rm \: \displaystyle\int\rm \frac{(2x - 1)(x + 3)}{6} \: dx$

can be rewritten as

$\rm \: = \: \dfrac{1}{6}\displaystyle\int\rm (2x - 1)(x + 3)dx$

$\rm \: = \: \dfrac{1}{6}\displaystyle\int\rm (2 {x}^{2} - x + 6x - 3)dx$

$\rm \: = \: \dfrac{1}{6}\displaystyle\int\rm (2 {x}^{2}+ 5x - 3)dx$

$\rm \: = \: \dfrac{1}{6}\bigg[2 \times \dfrac{ {x}^{3} }{3} + 5 \times \frac{ {x}^{2} }{2} - 3x\bigg] + c$

$\rm \: = \: \dfrac{1}{6}\bigg[\dfrac{ 2{x}^{3} }{3} + \frac{5 {x}^{2} }{2} - 3x\bigg] + c$

$\rm \: = \: \dfrac{{x}^{3} }{9} + \frac{5 {x}^{2} }{12} - \frac{x}{2} + c$

Hence,

$\boxed{ \rm{ \:\rm \displaystyle\int\rm \frac{(2x - 1)(x + 3)}{6}dx = \dfrac{{x}^{3} }{9} + \frac{5 {x}^{2} }{12} - \frac{x}{2} + c \: }}$

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

Given integral is

$\rm \: \displaystyle\int_{4}^{9}\rm \bigg(\dfrac{1}{ \sqrt{x} } - 2\bigg) \: dx$

can be rewritten as

$\rm \: = \: \displaystyle\int_{4}^{9}\rm \bigg( {(x)}^{ - \frac{1}{2} } - 2\bigg) \: dx$

$\rm \: = \: \bigg[\dfrac{ {x}^{ - \frac{1}{2} + 1} }{ - \frac{1}{2} + 1} - 2x \bigg]_{4}^{9}$

$\rm \: = \: \bigg[\dfrac{ {x}^{\frac{1}{2}} }{\frac{1}{2}} - 2x \bigg]_{4}^{9}$

$\rm \: = \: \bigg[2 \sqrt{x} - 2x \bigg]_{4}^{9}$

$\rm \: = \:2 \bigg[\sqrt{x} - x \bigg]_{4}^{9}$

$\rm \: = \:2 \bigg[(\sqrt{9} - \sqrt{4}) - (9 - 4) \bigg]$

$\rm \: = \:2 \bigg[3 - 2 -5 \bigg]$

$\rm \: = \:2 \bigg[3 - 7 \bigg]$

$\rm \: = \:2 \times ( - 4)$

$\rm \: = \: - \: 8$

Hence,

$\rm\implies \:\rm \boxed{ \rm{ \:\: \displaystyle\int_{4}^{9}\rm \bigg(\dfrac{1}{ \sqrt{x} } - 2\bigg) \: dx \: = - \: 8 \: }}$

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Formula Used :-

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

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Additional Information :-

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