Let f(x)=∫_0^t t (t^2-9 t+20 ) d t, 1 ≤ x ≤ 5. If the range of f is [α, β], then 4(α+β) equals :

$f^{\prime}(x)=x\left(x^2-9 x+20\right), x \in(1,5)$f′(x)=x(x2−9x+20),x∈(1,5) $=(x-4) x(x-5)$=(x−4)x(x−5)

$\begin{aligned} & \Rightarrow f^{\prime}(x)>0 \forall x \in(1,4) \\ & \Rightarrow f^{\prime}(x) < 0 \forall x \in(4,5) \\ & \Rightarrow f(x) \text { increasing in }(1,4) \\ & \quad f(x) \text { decreasing in }(4,5) \\ & \Rightarrow \text { critical points to check: }\end{aligned}$​⇒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: ​ $\displaystyle \begin{aligned} & x=1,4,5 \\ & f(x)=\int_0^x\left(t^3-9 t^2+20 t\right) d t \\ & =\frac{t^4}{4}-3 t^3+\left.10 t^2\right|_0 ^x=\frac{x^4}{4}-3 x^3+10 x^2 \\ & f(1)=\frac{1}{4}-3+10=\frac{29}{4} \\ & f(4)=4^3-3.4^3+10.4^2=-2.4^3+10.4^2=32 \\ & f(5)=\frac{5^4}{4}-3.5^3+10.25=\frac{5^4}{4}-125=\frac{125}{4}\end{aligned}$​x=1,4,5f(x)=∫0x​(t3−9t2+20t)dt=4t4​−3t3+10t2​0x​=4x4​−3x3+10x2f(1)=41​−3+10=429​f(4)=43−3.43+10.42=−2.43+10.42=32f(5)=454​−3.53+10.25=454​−125=4125​​ $\begin{aligned} & \text { Range } \Rightarrow\left[\frac{29}{4}, 32\right] \Rightarrow 4(\alpha+\beta)=128+29 \\ & =157\end{aligned}$​ Range ⇒[429​,32]⇒4(α+β)=128+29=157​