Suppose f is continuous on [5,18] and '14 3) [1²₁ /14 (a) 5 (b) 1) fort f(t) dt = 18 (c) √ ¹8: (d) f(t) dt = (e) '18 › [¹⁰/ 3f(t) dt = f(t) + 3 dt = 14 66²4 f(t) dt + 18 [28 FE 14 f(t) dt = 18 6 f(t) dt = -6. Compute each of the following values (f) The average value of f(x) on [5.18] =

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