Question

integrate the example given in image.​

Answer 1

Given Question

Evaluate the following integral

$\displaystyle\int\rm\bigg[1 - \dfrac{x}{1!} + \dfrac{ {x}^{2} }{2!} + - - - \infty \bigg] {e}^{2x} \: dx$

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

Given integral is

$\displaystyle\int\rm\bigg[1 - \dfrac{x}{1!} + \dfrac{ {x}^{2} }{2!} + - - - \infty \bigg] {e}^{2x} \: dx$

We know,

From exponential series

$\boxed{ \tt{ \: {e}^{x} = 1 + x + \dfrac{ {x}^{2} }{2!} + \dfrac{ {x}^{3} }{3!} + - - - \infty \: }}$

So,

On replacing x by (- x), we get

$\boxed{ \tt{ \: {e}^{ - x} = 1 - x + \dfrac{ {x}^{2} }{2!} - \dfrac{ {x}^{3} }{3!} + - - - \infty \: }}$

So, given integral can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \: {e}^{ - x} \times {e}^{2x} \: dx$

$\rm \:  =  \: \displaystyle\int\rm \: {e}^{ - x + 2x} \: dx$

$\rm \:  =  \: \displaystyle\int\rm \: {e}^{x} \: dx$

$\rm \:  =  \: {e}^{x} + c$

Therefore

$\boxed{ \tt{ \: \displaystyle\int\rm\bigg[1 - \dfrac{x}{1!} + \dfrac{ {x}^{2} }{2!} + - - - \infty \bigg] {e}^{2x} \: dx \: = \: {e}^{x} \: + \: c \: }}$

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More to Know

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