Let f (x) = \int 0 t t (t 2 - 9t + 20) dt, 1 \le x \le 5 . If the range of f is [\alpha, \beta], then 4(\alpha + \beta) equals :
Step-by-Step Solution
Key Concept: Apply the core result for definite integral properties and simplify using the given constraints.
′ f (x) = x (x 2 - 9x + 20) , x \in (1, 5) (4) = (x - 4)x(x - 5) ′ \Rightarrow f (x) > 0\forallx \in (1, 4) ′ \Rightarrow f (x) < 0\forallx \in (4, 5) \Rightarrow f (x) increasing in (1, 4) f (x) decreasing in (4, 5) \Rightarrow critical points to check: x = 1, 4, 5 x 3 2 f (x) = \int (t - 9t + 20t) dt 0 4 4 t x x 3 2 3 2 = - 3t + 10t ∣∣ = - 3x + 10x 0 4 4 1 29 f (1) = - 3 + 10 = 4 4 3 3 2 3 2 f (4) = 4 - 3.4 + 10.4 = -2.4 + 10.4 = 32 4 4 5 5 125 3 f (5) = - 3.5 + 10.25 = - 125 = 4 4 4 29 Range \Rightarrow [ , 32] \Rightarrow 4(\alpha + \beta) = 128 + 29 4 = 157
Correct Answer: 4