How many diagonals does a nanogon have
2 answers:
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Answer:
∫₂⁵ ln(x) dx
Step-by-step explanation:
lim(n→∞) ∑ᵢ₌₁ⁿ (3/n) ln((2n + 3i) / n)
lim(n→∞) ∑ᵢ₌₁ⁿ (3/n) ln(2 + (3/n) i)
The width of the interval is b−a = 3, and there are n rectangles. So the width of each rectangle is 3/n, and the height of each rectangle is ln(2 + (3/n) i).
The ith term is 2 + (3/n) i, so a = x₀ = 2. Therefore, b = 2+3 = 5.
So the region is the area under f(x) = ln(x) between x=2 and x=5.
∫₂⁵ ln(x) dx
