Let `p(x)` be a function defined on `R` such that `p'(x)=p'(1-x)` for all `x epsilon[0,1],p(0)=1,` and `p(1)=41`.
Then `int_(0)^(1)p(x)dx` is equal to

Options

A. `42`
B. `sqrt(41)`
C. `21`
D. `41`

Answer

C

Explanation

`p'(x)=p'(1-x)`
`impliesp(x)=-p(1-x)+c`
At `x=0`
`p(0)=-p(1)+cimplies42=c`
`:.p(x)=-p(1-x)+42`
`impliesp(x)+p(1-x)=42`
`I=int_(0)^(1)p(x)dx=int_(0)^(1)p(1-x)dx`
`:. 2I=int_(0)^(1)(p(x)+p(1-x))dx=int_(0)^(1)(42)dx`
`impliesI=21`