Question

If $f\left(x\right)={\int }_x^{x^2}(t-1)dt,1\le x\le 2$

then the global maximum value of $f\left(x\right)$ is ?

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is C $4$

$f\left(x\right)={\int }_K^{x^2}(t-1]$ dt
$\Rightarrow f^{\prime }\left(x\right)=(x^2-1\left)\frac{d}{dx}\right(x^2)-(x-1\left)\frac{d}{dx}\right(x⟩$ (by Leibnitz's Rule)
$=2x(x^2-1)-(x-1)$
$=2x^3-3x+1$
$=(x-1)(2x^2+2x-1)\ge 0forx\ge 1$
$\therefore$ the function is increasing in [1,2 ] .
Hence the maximum value $=f\left(2\right)={\int }_2^4(t-1)dt={(\frac{t^2}{2}-t)}_2^4$
$=4$