    Question

# ∫1−1e|x|dx

A
e
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B
2e
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C
e-1
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D
0.02
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Solution

## The correct option is D 0.02We can see that the limits are (-a, a) so we’ll use property (i)∫a−af(x)dx=2∫a0f(x)dx,if f is even,i.e., if f(−x)=f(x) (ii)∫a−af(x)dx=0, if f is odd,i.e.,if f(−x)=−f(x). Now we have to check whether the function is even or odd. Let f(x) be=e|x|f(−x)=e|−x|=e|x|So,f(−x)=f(x) The given function is even. ∫1−1e|x|dx=2.∫10e|x|dx (Using the above mentioned property) Now we are left with limits 0 to 1. For limits 0 to 1 i.e. for values of x , 0 to 1e|x| will become ex. As for x>0|x|=x& for x<0|x|=−x2.∫10exdx=2.ex|10=2.(e−1).=2e−2  Suggest Corrections  0      Similar questions  Related Videos   Area under the Curve
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