# Chapter 5: Integrals - Section 5.3 - The Definite Integral - Exercises 5.3 - Page 276: 76

$2 \sqrt 2 \leq \int_0^{1} \sqrt {x+8} dx \leq 3$

#### Work Step by Step

When the function $f(x)$ has maximum and minimum values defined on $[a,b]$ then by Maximum-Minimum inequality , we have $min f(b-a) \leq \int_p^q f(x) dx \leq max f(b-a)$ ....(1) Given: $f(x)= \sqrt {x+8}$ defined on $[0,1]$ and $f(0)= \sqrt {0+8}=\sqrt 8=2\sqrt 2$ and $f(1)= \sqrt {1+8}=\sqrt 9=3$ Equation (1) becomes: $f(0)(1-0) \leq \int_0^{1} f(x) dx \leq f(1) (1-0)$ Hence, $2 \sqrt 2 \leq \int_0^{1} \sqrt {x+8} dx \leq 3$

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