Question

Solve the integral

$\int \left[1-\dfrac{x}{1!} + \dfrac{x^2}{2!}-\dfrac{x^3}{3!}+...\infty\right]e^{2x} dx$​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \left[1-\dfrac{x}{1!} + \dfrac{x^2}{2!}-\dfrac{x^3}{3!}+...\infty\right]e^{2x} dx$

We know,

From exponential series,

$\rm :\longmapsto\: {e}^{x} = 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!}+...\infty$

Now, Replace x by - x, we get

$\rm :\longmapsto\: {e}^{ - x} = 1 + \dfrac{( - x)}{1!} + \dfrac{( - x)^2}{2!} + \dfrac{( - x)^3}{3!}+...\infty$

$\rm :\longmapsto\: {e}^{ - x} = 1 - \dfrac{x}{1!} + \dfrac{x^2}{2!} - \dfrac{x^3}{3!}+...\infty$

So, given integral

$\rm :\longmapsto\:\displaystyle\int\rm \left[1-\dfrac{x}{1!} + \dfrac{x^2}{2!}-\dfrac{x^3}{3!}+...\infty\right]e^{2x} dx$

can be rewritten as

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

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

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

$\rm \:  =  \: {e}^{x} \: + \: c$

Hence,

$\boxed{ \tt{ \: \:\displaystyle\int\rm \left[1-\dfrac{x}{1!} + \dfrac{x^2}{2!}-\dfrac{x^3}{3!}+...\infty\right]e^{2x} dx \: = \: {e}^{x} + c \: }}$

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MORE TO EXPLORE

Let prove that,

$\rm :\longmapsto\: {e}^{x} = 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!}+...\infty$

Let assume that

$\rm :\longmapsto\:f(x) = {e}^{x}$

So, differential coefficients of f(x) are as follow :-

$\rm :\longmapsto\:f'(x) = {e}^{x}$

$\rm :\longmapsto\:f''(x) = {e}^{x}$

$\rm :\longmapsto\:f'''(x) = {e}^{x}$

.

.

.

So, differential coefficients at x = 0, are

$\rm :\longmapsto\:f(0) = 1$

$\rm :\longmapsto\:f'(0) = 1$

$\rm :\longmapsto\:f''(0) = 1$

$\rm :\longmapsto\:f'''(0) = 1$

.

.

.

So, Using Mc Lauren's Series Method, we have

$\rm :\longmapsto\:f(x) = f(0) + xf'(0) + \dfrac{ {x}^{2} }{2!}f''(0) + \dfrac{ {x}^{3} }{3!}f'''(0) + - - - -$

So, on substituting above evaluated values, we get

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

Hence, Proved

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