Let $f(x)=\int_0^t t\left(t^2-9 t+20\right) d t, 1 \leq x \leq 5$. If the range of $f$ is $[\alpha, \beta]$, then $4(\alpha+\beta)$ equals :

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Aicharya · Accepted Answer

$f^{\prime}(x)=x\left(x^2-9 x+20ight), x \in(1,5)$ $=(x-4) x(x-5)$ $\begin{aligned} & \Rightarrow f^{\prime}(x)>0 \forall x \in(1,4) \ & \Rightarrow f^{\prime}(x) $\begin{aligned} & x=1,4,5 \ & f(x)=\int_0^x\left(t^3-9 t^2+20 tight) d t \ & =\frac{t^4}{4}-3 t^3+\left.10 t^2ight|_0 ^x=\frac{x^4}{4}-3 x^3+10 x^2 \ & f(1)=\frac{1}{4}-3+10=\frac{29}{4} \ & f(4)=4^3-3.4^3+10.4^2=-2.4^3+10.4^2=32 \ & f(5)=\frac{5^4}{4}-3.5^3+10.25=\frac{5^4}{4}-125=\frac{125}{4}\end{aligned}$ $\begin{aligned} & 	ext { Range } \Rightarrow\left[\frac{29}{4}, 32ight] \Rightarrow 4(\alpha+\beta)=128+29 \ & =157\end{aligned}$

Let $f(x)=\int_0^t t\left(t^2-9 t+20\right) d t, 1 \leq x \leq 5$. If the range of $f$ is $[\alpha, \beta]$,…

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