What is the area of the region in the first quadrant that is bounded by the line $$y=0.5$$, the curve $$x=y^("5/2")$$, and the y-axis? 

Firstly, let's visualize the bounded region we are being asked to find the area of:

The bounded region is represented by “R” in between the red, blue, and purple lines.

Whenever we are asked to find the area between curves, we know we will need to find an integral. The first step to finding the integral is that we must establish the bounds. 

Since our expression is in terms of y, we want to find the minimum and maximum y values. We are already given one, $y = 0.5$. Since our other bound is given as the y-axis ($x=0$), we must find the value of y where $x = 0$and $x = y^{5/2}$ intersect:

$$\begin{gathered} x = y^{5/2} \\ 0 = y^{5/2} \\ y = 0 \end{gathered}$$

Therefore, the bounds of our integral are y = 0 and y = 0.5. Setting up the rest of our integral according to the basic formula of a definite integral:

${\int }_b^{a}f\left(x\right)dx, where x=a is the upper bound of f\left(x\right) and x=b is the lower bound of f\left(x\right)$

${\int }_0^{0.5}\left(y^{5/2}\right)dy$

Since we know that in order to solve an integral we add +1 to the exponent and divide by the new exponent:

${\int }_0^{0.5}\left(y^{5/2}\right)dy = {\overline{)\frac{\left(y^{7/2}\right)}{(7/2)}}}_0^{0.5}={\overline{)\frac{\left(2y^{7/2}\right)}{7}}}_0^{0.5}$

Now, we plug in our maximum and minimum values of y and subtract:

${\overline{)\frac{\left(2y^{7/2}\right)}{7}}}_0^{0.5}=(\frac{2{(0.5)}^{7/2}}{7}-\frac{2{\left(0\right)}^{7/2}}{7})=\frac{2(.088)}{7}-0=.025$

Therefore, the area of this bounded region is approximately 0.025.