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Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 25 maths textbook solution

Answers (1)

Answer: 2\left ( e-1 \right )

Hint: You must know about |x| function.

Given:\int_{-1}^{1} e^{|x|} d x

Solution:  

Let    I=\int_{\mathrm{I}}^{1} e^{|x|} d x 

\therefore I=\int_{-1}^{1} e^{|x|} d x

       =\int_{-1}^{0} e^{-x} d x+\int_{0}^{1} e^{x} d x

       \begin{aligned} &=\left[\frac{e^{0}}{-1}-\frac{e^{-(-1)}}{-1}\right]+\left[\frac{e^{1}}{1}-\frac{e^{0}}{1}\right] \\\\ &=\left[-e^{0}+e^{1}\right]+\left[e^{1}-e^{0}\right] \end{aligned}

       \begin{aligned} &=(-1+e)+(e-1) \\\\ &=2(e-1) \end{aligned}

 

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