Let f(x) = int_0^x t(t^2 - 9t + 20) dt,1 leq | vedclass

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(C) Given $f(x) = \int_0^x (t^3 - 9t^2 + 20t) dt$.By the Fundamental Theorem of Calculus,$f'(x) = x^3 - 9x^2 + 20x = x(x - 4)(x - 5)$.For $x \in [1, 5

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$157$

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(C) Given $f(x) = \int_0^x (t^3 - 9t^2 + 20t) dt$.By the Fundamental Theorem of Calculus,$f'(x) = x^3 - 9x^2 + 20x = x(x - 4)(x - 5)$.For $x \in [1, 5]$,$f'(x) = 0$ at $x = 4$.Evaluating $f(x)$ at critical points and endpoints:$f(x) = \int_0^x (t^3 - 9t^2 + 20t) dt = \left[ \frac{t^4}{4} - 3t^3 + 10t^2 \right]_0^x = \frac{x^4}{4} - 3x^3 + 10x^2$.$f(1) = \frac{1}{4} - 3 + 10 = 7.25 = \frac{29}{4}$.$f(4) = \frac{256}{4} - 3(64) + 10(16) = 64 - 192 + 160 = 32$.$f(5) = \frac{625}{4} - 3(125) + 10(25) = 156.25 - 375 + 250 = 31.25 = \frac{125}{4}$.Comparing values,the minimum value $\alpha = f(1) = \frac{29}{4}$ and the maximum value $\beta = f(4) = 32$.Thus,$4(\alpha + \beta) = 4(\frac{29}{4} + 32) = 29 + 128 = 157$.

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

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