The value (round off to one decimal place) of \(\mathop \smallint \nolimits_{ - 1}^1 x\;{e^{\left| x \right|}}dx\) ______
Explanation
Given,
Function f(x) = x e|x|
Integral is -1 to 1.
If f(-x) = f(x) then the function is said to be even function
If f(-x) = - f(x) then the function is said to be odd function.
f(-x) = -x e|-x| = -x e|x| = - f(x)
∴ The given function is an odd function.
For an odd function:
\(\mathop \smallint \nolimits_{ - a}^a x\;{f(x)}dx\) = 0
For a even function
\(\mathop \smallint \nolimits_{ - a}^a x\;{f(x)}dx\) = 2 × \(\mathop \smallint \nolimits_{ 0}^a x\;{f(x)}dx\)
Now, as the function is odd
\(\mathop \smallint \nolimits_{ - 1}^1 x\;{e^{\left| x \right|}}dx\) = 0
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