18 Suppose fis continuous on [5,18] and f(t) dt = -6. Compute each of the following values 5 11 (a) f(t) dt = 0 11 5 ( b ) f(t) dt = 6 /18 18 (c) 5f(t) dt = -30 18 (d) f(t) + 5 dt = 59 15 11 18 (e) f(t) dt + f(t) dt = -6 5 11 (f) The average value of f(x) on [5,18] = -0.462 X