The correct option is D 0.02
We 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