Question

If $f\left(x\right)={\int }_0^x(t^2+2t+2)dt,2\le x\le 4,$ then

• A. the maximum value of $f\left(x\right)$ is $\frac{137}{3}$
• B. the minimum value of $f\left(x\right)$ is $10$
• C. the maximum value of $f\left(x\right)$ is $26$
• D. none of these

Answer 1

The correct option is A the maximum value of $f\left(x\right)$ is $\frac{137}{3}$

$f\left(x\right)={\int }_0^x(t^2+2t+2)dt={\int }_0^x\left(\right(t+1{)}^2+1)dt$
$f^{\prime }\left(x\right)=(t+1{)}^2+1>0$
Hence $f\left(x\right)$ is increasing function for all real x
Ergo, $f_{min}=f\left(2\right)={\int }_0^2\left(\right(t+1{)}^2+1)dt=[(t+1{)}^3/3+t{]}_0^2=11$
and $f_{max}=f\left(4\right)={\int }_0^4\left(\right(t+1{)}^2+1)dt=[(t+1{)}^3/3+t{]}_0^4=125/3+4=\frac{137}{3}$