In Problems 17 to 30, for the curve, betweenand, find:

The arc length.

The distance between two places along a portion of a curve is known as arc length. In other words, it can be defined as the measure of the length of a particular portion of some arc.

Draw the bounded region for the curve , between and .

Write the expression for the arc length from the interval to .

$s={\int }_a^b\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}dx}$

Substitute the for $\sqrt{x}fory$.

$$\begin{gathered} \frac{dy}{dx}=\frac{d\left(\sqrt{x}\right)}{dx} \\ =\frac{1}{2\sqrt{x}} \end{gathered}$$

Take square and evaluate.

$$\begin{gathered} {\left(\frac{dy}{dx}\right)}^2={\left(\frac{1}{2\sqrt{x}}\right)}^2 \\ =\frac{1}{4x} \end{gathered}$$

Write the expression for the arc length of the smallest element.

$ds=\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}dx}$

Substitute $\frac{1}{4x}$for ${\left(\frac{dy}{dx}\right)}^2$in the above expression.

$ds=\sqrt{1+\frac{1}{4x}dx}$

Integrate the above expression in the limits 0 to 2.

$$\begin{gathered} {\int }_0^{s}ds={\int }_0^2\sqrt{1+\frac{1}{4x}}dx \\ s={\int }_0^2\sqrt{1+\frac{1}{4x}}dx \\ ={\left(\frac{1}{4}\sqrt{4+\frac{1}{x}}\sqrt{x}\left(\frac{\sin h^{-1}\left(2\sqrt{x}\right)}{\sqrt{4x+1}}+2\sqrt{x}\right)\right)}_0^2 \\ =\left(\frac{1}{4}\sqrt{4+\frac{1}{x}}\sqrt{2}\left(\frac{\sin h^{-1}\left(2\sqrt{2}\right)}{\sqrt{4\left(2\right)+1}}+2\sqrt{2}\right)-\frac{1}{4}\sqrt{4+\frac{1}{0}}\sqrt{10}\left(\frac{\sin h^{-1}\left(2\sqrt{2}\right)}{\sqrt{4\left(2\right)+1}}+2\sqrt{2}\right)\right) \end{gathered}$$

Simplify the above expression.

$$\begin{gathered} s=\frac{1}{2}\left(3\sqrt{2}+\frac{1}{2}\sin h^{-1}\left(2\sqrt{2}\right)\right) \\ =\frac{1}{2}\left(3\sqrt{2}+In\left(1+\sqrt{2}\right)\right) \end{gathered}$$