At $ x = 2 $: $ I(2) = \frac{1}{4 + 8 + 5} = \frac{1}{17} $

February 22, 2026

$At $ x = 2 $: $ I(2) = \frac{1}{4 + 8 + 5} = \frac{1}{17} $$

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Question: A computer graphics specialist models light intensity $ I(x) $ along a surface as $ I(x) = \frac{1}{x^2 + 4x + 5} $. If the cumulative visual effect over points $ x = 1 $ to $ x = n $ is modeled by $ C(n) = \sum_{x=1}^{n} I(x) $, find $ C(4) $.
Solution: We evaluate $ I(x) = \frac{1}{x^2 + 4x + 5} $ at $ x = 1, 2, 3, 4 $:
At $ x = 1 $: $ I(1) = \frac{1}{1^2 + 4 \cdot 1 + 5} = \frac{1}{10} $
At $ x = 3 $: $ I(3) = \frac{1}{9 + 12 + 5} = \frac{1}{26} $
At $ x = 4 $: $ I(4) = \frac{1}{16 + 16 + 5} = \frac{1}{37} $
C(4) = \frac{1}{10} + \frac{1}{17} + \frac{1}{26} + \frac{1}{37}

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