Question
By using properties of definite integral, evaluate: $\int_{-1}^1 e^{|x|} d x$
✓
Answer
Let $f ( x )= e ^{ IxI } \Rightarrow f (- x )= e ^{|- x |}= e ^{| x |}= f ( x )$
$\Rightarrow f(x)$ is an even function; therefore,
$
\begin{array}{l}
\therefore \int_{-1}^1 e^{|x|} d x=2 \int_0^1 e^{|x|} d x=2 \int_0^1 e^x d x(\because 0 \leq x \leq 1 \Rightarrow|x|=x) \\
=2\left[e^x\right]_0^1=2\left(e^1-e^0\right)=2(e-1)
\end{array}
$
$\Rightarrow f(x)$ is an even function; therefore,
$
\begin{array}{l}
\therefore \int_{-1}^1 e^{|x|} d x=2 \int_0^1 e^{|x|} d x=2 \int_0^1 e^x d x(\because 0 \leq x \leq 1 \Rightarrow|x|=x) \\
=2\left[e^x\right]_0^1=2\left(e^1-e^0\right)=2(e-1)
\end{array}
$
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