Question

If $\varphi \left(x\right)={\int }_x^{x^2}(t-1)dt,1\le x\le 2$,then the greatest value of$\varphi \left(x\right)$ is

• A. $2$
• B. $4$
• C. $8$
• D. none of these

Answer 1

Answer: (B)

$4$

$\varphi \left(x\right)={\int }_x^{x^2}(t-1)dt={[\frac{t^2}{2}-t]}_x^{x^2}$
$=\frac{x^4}{2}-x^2-\frac{x^2}{2}+x=\frac{x^4-3x^2+2x}{2}$
Hence greatest value is $4$