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Inequalities of Integrals

28. Given f(x...

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28. Given f(x)=[1/3+x/66], find sum of f(x) for x=1 to 66

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f (x) = [\frac{1}{3} + \frac{x}{66}]

66 \sum x = 1 f (x)

f (x) = [\frac{1}{3} + \frac{x}{66}]

66 \sum x = 1 f (x)

f (x) = \frac{1}{x^{2} - 17 x + 66}

f (\frac{2}{x - 2})

is discontinuous at

x =

f (x) = \frac{1}{x^{2} - 17 x + 66}

f (\frac{2}{x - 2})

is continuous at R except

x =

f (x) = \frac{1}{x^{2} - 17 x + 66}

then points of discontinuity of

f (\frac{2}{x - 2})

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