calculate in limit 1nf1nf2nf1 if fxarctgx in n infinite sim

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

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.]