Answer
$2 \sqrt 2 \leq \int_0^{1} \sqrt {x+8} dx \leq 3$
Work Step by Step
When the function $f(x)$ has minimum and maximum on $[a,b]$ then, we have
$min f(b-a) \leq \int_a^b f(x) dx \leq max f(b-a)$
We are given that $f(x)= \sqrt {x+8} $
and $f(0)= \sqrt {0+8}=\sqrt 8=2\sqrt 2 $; $f(1)= \sqrt {1+8}=\sqrt 9=3$
Now, we have $f(0)(1-0) \leq \int_0^{1} f(x) dx \leq f(1) (1-0)$
Thus,
$2 \sqrt 2 \leq \int_0^{1} \sqrt {x+8} dx \leq 3$
