# calculate in limit 1/n(f(1/n)+f(2/n)+____+f(1)) if f(x)=arctgx in n --> infinite simbol? use antiderivative

*print*Print*list*Cite

You may replace the limit by Riemann integral, over the interval `[0,1]` , such that:

`lim_(n->oo)(1/n)(f(1/n) + f(2/n) + ... + f(n/n)) = int_0^1 f(x) dx`

The problem provides the equation of the function, such that:

`f(x) = arctan x`

`int_0^1 f(x) dx = int_0^1 arctan x dx`

You need to use integration by parts, such that:

`arctan x = u => 1/(1 + x^2)dx = du`

`1dx = dv => v = x`

`int_0^1 arctan x dx = x*arctan x|_0^1 - int_0^1 x/(1 + x^2)dx`

You should come up with the substitution, such that:

`1 + x^2 = y => 2x dx = dy => x dx = (dy)/2`

`int_0^1 x/(1 + x^2)dx = int_1^2 (dy)/(2y) `

`int_1^2 (dy)/(2y) = (1/2)ln y|_1^2 => int_1^2 (dy)/(2y) = (1/2)(ln 2 - ln 1)`

`int_1^2 (dy)/(2y) = (1/2)*ln 2`

`int_0^1 arctan x dx = 1*arctan 1 - 0*arctan 0 - (1/2)*ln 2`

`int_0^1 arctan x dx = pi/4 - (1/2)*ln 2`

**Hence, evaluating the given limit, using Riemann integral, yields **`lim_(n->oo)(1/n)(f(1/n) + f(2/n) + ... + f(n/n)) = pi/4 - (1/2)*ln 2.`

`lim_(n->oo){(1/n)[f(1/n)+f(2/n)+..........+f(1)]}`

`=lim_(n->oo){(1/n)``[f(0)+f(1/n)+f(2/n)+..........+f(n/n)-f(0)]}`

``

`` `=lim_(n->oo){(1/n)[f(0)+f(1/n)+f(2/n)+..........+f(n/n)]}`

`-lim_(n->oo) (1/n)f(0)`

`=int_0^1tan^(-1)(x)dx-lim_(x->oo)(1/n)*0`

`=(xtan^(-1)x)_0^1-(1/2)(log(1+x^2))_0^1`

`=pi/4-(1/2)log(2)`

``

``

``