The function f(x)=∫1x[2(t−1)(t−2)3+3(t−1)2(t−2)2] dt  has:

f(x)=∫1x{2(t−1)(t−2)3+3(t−1)2(t−2)2} dt ⇒f1(x)=2(x−1)(x−2)3+3(x−1)2(x−2)2 For maximum (or) minimum f1(x)=0 ⇒2(x−1)(x−2)3+3(x−1)2(x−2)2=0 ⇒(x−1)(x−2)2{2(x−2)+3(x−1)}=0 ⇒(x−1)(x−2)2{5x−7}=0 x=1, 2, 7/5 now  f1(x)=(x−1)(x−2)2(5x−7) ⇒f1(x)=(x−2)2(5x−7)+2(x−1)(x−2)(5x−3)+(x−1)(x−2)25........(2) Putting x=1,x=2,x=7/5  is(2) we get f11(1)<0           f11(2)=0          f11(7/5)>0