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Question

11e|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 0.02
We can see that the limits are (-a, a) so we’ll use property
(i)aaf(x)dx=2a0f(x)dx,if f is even,i.e.,
if f(x)=f(x)
(ii)aaf(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.
11e|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.(e1).=2e2

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