Question

Kindly solve this? which option is correct?

Answer 1

This is readily seen as not doable by elementary means, so I just solved it numerically and got:

I evaluated the integral for `n = 1, 1.5, 2, . . . , 9.5, 10, 25, 50, 75, 100`. By then it was clearly reaching `0.5`.

Answer 2

See below.

Explanation:

`int_0^1 (n x^(n-1))/(1+x^2) dx le int_0^1n x^(n-1)dx = 1`

`int_0^1 (n x^(n-1))/(1+x^2) dx ge 1/2 int_0^1n x^(n-1)dx = 1/2`

or

`1/2 le int_0^1 (n x^(n-1))/(1+x^2) dx le 1`

Now assuming that one of the answers is true, the most natural seems to be the fourth 4)

NOTE

for `x in [0,1]`
`1/2 le 1/(1+x^2) le 1`

Answer 3

`1/2`

Explanation:

As has already been shown in a previous solution,

`I_n = int_0^1 (nx^(n-1))/(1+x^2)dx`

exists and is bounded :

`1/2 le I_n <1`

Now integration by parts yields

`I_n = ((int nx^(n-1) dx)/(1+x^2))_0^1-int_0^1 x^n times (-(2x)/(1+x^2)^2) dx`
`qquad = (x^n/(1+x^2))_0^1+2int_0^1 x^(n+1)/(1+x^2)^2dx`
`qquad = 1/2 +J_n`

Now , since `0 < (1+x^2)^-1 < 1` in `(0,1)`

`J_n = 2/(n+2) int_0^1 ((n+2)x^(n+1))/(1+x^2)^2 dx`
`qquad <= 2/(n+2) int_0^1 ((n+2)x^(n+1))/(1+x^2) dx = 2/(n+2)I_(n+2)`

Since `lim_(n to oo)I_n` exists, we have

`lim_(n to oo )J_n = lim_(n to oo) 2/(n+2)I_(n+2) = lim_(n to oo)2/(n+2) times lim_(n to oo)I_(n+2) = 0`

Hence

`lim_(n to oo) I_n = 1/2`