Chapter 6.3, Problem 8E

To determine

To show: The value of given expression value $\text{lies between }2\sqrt{2}\approx 2.8\text{and}3$.

Explanation of Solution

Giveninformation :

  ${{\displaystyle \int }}_0^1\sqrt{x+8}dx$

Calculation :

Apply the rule of definite integral-

If max f and min f are the maximum and minimum values of f on [a,b], then

  $\min f(b-a)\le {{\displaystyle \int }}_a^{b}f(x)dx\le \max f(b-a)$

Here,

  $f(x)=\sqrt{x+8}$

  

Observe the graph in the interval [0,1] it is increasing.

  $\begin{array}{cc}\min f& =\sqrt{0+8}\\ & \approx 2.828\\ \max f& =\sqrt{1+8}\\ & =3\end{array}$

So,

  $2.8(1-0)\le {{\displaystyle \int }}_0^1\sqrt{x+8}dx\le 3(1-0)$

  $2.8\le {{\displaystyle \int }}_0^1\sqrt{x+8}dx\le 3$