If f(x)=∫limits_0^x≤ft(t^2+2 t+2) d t, where x ∈[2,4], then

We have,

$f(x)=\int\limits_0^x\left(t^2+2 t+2\right) d t$

$\Rightarrow f'(x)=x^2+2 x+2=(x+1)^2+1>0$ for all x

⇒ f(x) is strictly increasing on [2, 4]

∴ Maximum value of f(x)

$=f(4)=\int\limits_0^4\left(t^2+2 t+2\right) d t=\left[\frac{t^3}{3}+t^2+2 t\right]_0^4=\frac{136}{3}$

Minimum value of f(x)

$=f(2)=\int\limits_0^2\left(t^2+2 t+2\right) d t=\left[\frac{t^3}{3}+t^2+2 t\right]_0^2=\frac{32}{3}$