Question

Let $f\left(x\right)$ be a function defined by $f\left(x\right)={\int }_1^{x}x(x^2-3x+2)dx,1\le x\le 3$.Then the range of $f\left(x\right)$is

• A. $[0,2]$
• B. $[-\frac{1}{4},4]$
• C. $[-\frac{1}{4},2]$
• D. none of these

Answer 1

Answer: (C)

$[-\frac{1}{4},2]$

We have $f^{\prime }\left(x\right)=x(x^2-3x+2)=x(x-1)(x-2)$
Clearly $f^{\prime }\left(x\right)\le 0$ in $1\le x\le 2$ and $f^{\prime }\left(x\right)\ge 0$ in $2\le x\le 3$
$\therefore f^{\prime }\left(x\right)$ is monotonic decreasing in $[1,2]$
and monotonic increasing in $[2,3]$
$\therefore minf\left(x\right)=f\left(2\right)={\int }_1^{2}x(x^2-3x+2)dx$
$={[\frac{x^4}{4}-x^3+x^2]}_1^2=\frac{-1}{4}$
$maxf\left(x\right)=$ the greatest among $(f\left(1\right),f\left(3\right))$
Now $f\left(1\right)={\int }_1^{x}x(x^2-3x+2)dx=0$
And $f\left(3\right)={\int }_1^{3}x(x^2-3x+2)dx$
$={[\frac{x^4}{4}-x^3+2]}_1^2=2$
Therefore $maxf\left(x\right)=2$
Hence range $=[\frac{-1}{4},2]$