Chapter 6, Problem 37RE

To determine

To solve : the given expression.

Answer to Problem 37RE

The answer we get from the expression is: $0\le {{\displaystyle \int }}_0^1\left(\sqrt{1+{\sin }^2(x)}\right)dx\le \sqrt{2}$

Explanation of Solution

Given information:

Given expression is: $0\le {{\displaystyle \int }}_0^1\sqrt{1+{\sin }^{2}x}dx\le \sqrt{2}$

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,

  $f(x)=\sqrt{1+{\sin }^{2}x}$

The graph of the function $f\left(x\right)=\sqrt{1+{\sin }^{2}x}$ is as shown below:

  

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

  $\begin{array}{cc}\min f& =\sqrt{1+{\sin }^{2}0}\\ & =1\\ \max f& =\sqrt{1+{\sin }^{2}1}\\ & \approx 1.307\end{array}$

So,

  $1(1-0)\le {{\displaystyle \int }}_0^1\left(\sqrt{1+{\sin }^2(x)}\right)dx\le 1.307(1-0)$

  $\Rightarrow 1\le {{\displaystyle \int }}_0^1\left(\sqrt{1+{\sin }^2(x)}\right)dx\le 1.307$

Since, 1 is greater than 0 and 1.307 is less than $\sqrt{2}$

Thus,

  $0\le {{\displaystyle \int }}_0^1\left(\sqrt{1+{\sin }^2(x)}\right)dx\le \sqrt{2}$

Hence, the given expression is proved.