Question

${\int }_{-1}^1{e}^{|x|}dx$

• A. e
• B. 2e
• C. e-1
• D. 0.02

Answer 1

The correct option is D 0.02

We can see that the limits are (-a, a) so we’ll use property
$\left(i\right){\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$,if f is even,i.e.,
$iff(-x)=f\left(x\right)$
$\left(ii\right){\int }_{-a}^{a}f\left(x\right)dx=0,$ if f is odd,i.e.,$iff(-x)=-f\left(x\right)$.
Now we have to check whether the function is even or odd.
$Letf\left(x\right)be=e^{|x|}f(-x)=e^{|-x|}=e^{|x|}So,f(-x)=f\left(x\right)$
The given function is even.
${\int }_{-1}^1{e}^{|x|}dx=2.{\int }_0^1{e}^{|x|}dx$ (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 $e^x$.
$Asforx>0|x|=x\&forx<0|x|=-x2.{\int }_0^1{e}^{x}dx=2.e^x{|}_0^1=2.(e-1).=2e-2$