f′(x)=x(x2−9x+20),x∈(1,5) =(x−4)x(x−5) 
⇒f′(x)>0∀x∈(1,4)⇒f′(x)<0∀x∈(4,5)⇒f(x) increasing in (1,4)f(x) decreasing in (4,5)⇒ critical points to check: x=1,4,5f(x)=∫0x(t3−9t2+20t)dt=4t4−3t3+10t20x=4x4−3x3+10x2f(1)=41−3+10=429f(4)=43−3.43+10.42=−2.43+10.42=32f(5)=454−3.53+10.25=454−125=4125 Range ⇒[429,32]⇒4(α+β)=128+29=157