f′(x)=x(x2−9x+20)=x(x−4)(x−5) for x∈(1,5).For x∈(1,4), f′(x)>0, so f is increasing on (1,4).For x∈(4,5), f′(x)<0, so f is decreasing on (4,5).Thus the maximum is at x=4, and the minimum is at one of the endpoints x=1 or x=5.
Now compute:f(x)=∫0x(t3−9t2+20t)dt=[4t4−3t3+10t2]0x=4x4−3x3+10x2.f(1)=41−3+10=429.f(4)=444−3(43)+10(42)=64−192+160=32.f(5)=454−3(53)+10(52)=4625−125=4125.Since 4125>429, the minimum is f(1)=429 and the maximum is f(4)=32.Hence the range is [α,β]=[429,32].Therefore 4(α+β)=4(429+32)=29+128=157.So the correct option is 157.
