Let f(x)= ∫ex (x-1)(x-2) dx. Then f decreases in the interval :

(A) (∞,-2) (B) (–2, –1)
(C) (1, 2) (D) (2,+∞)

Hi pallavi,

f(x)= ∫ex (x-1)(x-2) dx

f¹(x)=df(x)/dx =ex (x-1)(x-2) + C

e^x > 0

(x-1)(x-2) ≥0 when x≤1 and x≥2

(x-1)(x-2) < 0 when 1<x<2

therefore      ex (x-1)(x-2) < 0 when 1<x<2

f¹(x) < 0 when 1<x<2

f decreases in the interval (1,2)