Let p(x) be a function defined on R such that p'(x)=p' for all `z in[0,1],p(0)=1` and p(1)=41. Then, `int_(0)^(1) p(x)dx` equals

Options

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

Answer

D

Explanation

We have,
p'(x)=p'(1-x)
i.e., `(d)/(dx)p(x)=(d)/(d(1-x)){p(1-x)}`
`rArr p(x)=-p(1-x)+C " " `….(i)
Putting x=0 in (i), we get
`p(0)=-p(1)+C rArr 42`
Putting C=42 in (i), we get
`p(x)=-p(1-x)+42" "`......(ii)
Now, `overset(1)underset(0)int p(x)dx=overset(1)underset(0)int p(1-x)dx`
`rArr overset(1)underset(0)int p(x)=overset(1)underset(0)int {42-p(x)}dx`
`rArr 2overset(1)underset(0)int p(x)dx=42`
`rArr overset(1)underset(0)int p(x)dx=21`