The value of the definite integral `int_0^1(1+e^-x^2)dx` (b) 2 `1+e^(-1)` (d) none of these

Options

A. `pi`
B. `(3pi)/4`
C. `(pi)/4`
D. `(pi)/2`

Answer

D

Explanation

`I=int_(-1)^(1)(1+x)^(1//2)(1-x)^(3//2)dx`
`:.I=int_(-1)^(1)(1-x)^(1//2)(1+x)^(3//2)dx` (Replacing `x` by` -x`)
Adding we get
`2I=int_(-1)^(1)(1+x)^(1//2)(1-x)^(1//2)[(1-x)+(1+x)]dx`
or `I=int_(-1)^(1)sqrt(1-x^(2))dx`
`:. I=2int_(0)^(1)sqrt(1-x^(2))dx`
`=2[x/2sqrt(1-x^(2)+1/2sin^(-1)x)]_(0)^(1)`
`=2[0+1/2 . (pi)/2-0-0]=(pi)/2`