Question

plz help to do this jee question​

Answer 1

Answer:

2 answer one is the answer

Explanation:

please mark me Brainlest

Answer 2

Given Question : -

The value of

$\rm \: \: \: \: \:\displaystyle\int_{1}^2\rm {e}^{ - x} \: dx \: is \:$

$\: \: \: \rm \: (1) \: \: \dfrac{1}{e} - \dfrac{1}{ {e}^{2} }$

$\: \: \: \rm \: (2) \: \: \dfrac{1}{e} + \dfrac{1}{ {e}^{2} }$

$\: \: \: \rm \: (3) \: \: \dfrac{1}{e}$

$\: \: \: \rm \: (4) \: \: \dfrac{1}{ {e}^{2} } - \dfrac{1}{e}$

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int_{1}^2\rm {e}^{ - x} \: dx$

We know,

$\red{ \boxed{ \sf{ \:\displaystyle\int\rm {e}^{x} \: dx \: = \: {e}^{ x} + c}}}$

So, using this identity, we get

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

$\rm \:  =  \: - \: \bigg[ {e}^{ - x} \bigg]_{1}^2\rm$

$\rm \:  =  \: - \: \bigg[ {e}^{ - 2} - {e}^{ - 1}\bigg]$

$\rm \:  =  \: - \: \bigg[\dfrac{1}{ {e}^{2} } - \dfrac{1}{e} \bigg]$

$\rm \:  =  \: \dfrac{1}{e} - \dfrac{1}{ {e}^{2} }$

Hence,

$\rm :\longmapsto\:\red{ \boxed{ \sf{ \:\displaystyle\int_{1}^2\rm {e}^{ - x} \: dx \: = \: \dfrac{1}{e} \: - \: \dfrac{1}{ {e}^{2} } \: }}}$

• So, option (1) is correct

Additional Information :-

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