Let $f(x) = \displaystyle\int_0^x t(t^2-9t+20)\,dt$, $1\leq x\leq 5$. If the range of $f$ is $[\alpha,\beta]$, then $4(\alpha+\beta)$ equals:
Step-by-Step Solution
Key Concept: Compute $f'(x) = x(x^2-9x+20) = x(x-4)(x-5)$; on $[1,5]$ the critical points are $x=4$ and $x=5$. Evaluate $f$ at the endpoints $x=1,5$ and the interior critical point $x=4$ to find the range.
$f'(x) = x(x-4)(x-5)$. On $[1,5]$: $f'>0$ on $(1,4)$, $f'<0$ on $(4,5)$.
$$f(x) = \frac{x^4}{4}-3x^3+10x^2.$$
$$f(1) = \frac{1}{4}-3+10 = \frac{29}{4},\quad f(4) = 64-192+160 = 32,\quad f(5) = \frac{625}{4}-375+250 = \frac{125}{4}.$$
Range: $\left[\dfrac{29}{4},32\right]$, so $\alpha=\dfrac{29}{4}$, $\beta=32$.
$$4(\alpha+\beta) = 4\cdot\frac{29}{4}+4\cdot 32 = 29+128 = 157.$$
Correct Answer: 4