Let f: (0, 2) → R be defined as f(x) = log?(1 + tan(πx/4)). Then, lim (n→∞) (2/n) [f(1/n) + f(2/n) + .... + f(1)] is equal to.......

<p>A = lim (n→∞) (2/n) ∑ (r=1 to n) f (r/n + n/ (n²)<br> (The term n/n² seems intended to be part of the function argument, not simply added. The solution proceeds as if it's f (r/n)<br>A = lim (n→∞) (2/n) ∑ (r=1 to n) [ f (r/n) + f (1/n) + . + f (n-1)/n) ]<br>The expression in the image seems to be: A = lim (n→∞) (2/n) [ f (1/n) + f (2/n) + . + f (n-1)/n) ]<br>A = 2 ∫? ¹ f (x) dx = 2 ∫? ¹ log? (1 + tan (πx/4) dx<br>put πx/4 = t ⇒ dx = 4/π dt<br>A = 2 ∫? ^ (π/4) log? (1 + tan (t) * (4/π) dt = (8/π) ∫? ^ (π/4) log? (1 + tan (t) dt<br>Using the property ∫? f (x)dx = ∫? f (a-x)dx, the integral ∫? ^ (π/4) log (1 + tan (t)dt evaluates to (π/8)ln (2).<br>A = (8/π) * (π/8)ln (2) / ln (2) = 1.</p>

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