Let f(x)=_0^x t(t^2-9 t+20) dt, 1 x 5.

Question

Let $f(x)=\int\limits_0^x \mathrm{t}\left(\mathrm{t}^2-9 \mathrm{t}+20\right) \mathrm{dt}, 1 \leq x \leq 5$. If the range of $f$ is $[\alpha, \beta]$, then $4(\alpha+\beta)$ equals :

Accepted Answer

$f^{\prime}(x)=x^3-9 x^2+20 x=x(x-4)(x-5)$

$$\begin{aligned}
& \therefore f(x)=\frac{x^4}{4}-\frac{9 x^3}{3}+\frac{20 x^2}{2} \
& f(1)=\frac{1}{4}-3+10=\frac{29}{4}=\alpha \
& \left.f(4)=\frac{256}{4}-3(64) \right\rvert\,+10(16)=32=\beta \
& 4(\alpha+\beta)=4\left(\frac{29}{4}+32\right)=157
\end{aligned}$$

Solutions · 1