SOLUTION:
Step 1:
In this question, we are given the following:
Step 2:
The details of the solution are as follows:
![\begin{gathered} Given\text{ that:} \\ y\text{ = }√(5x)\text{ --equation 1} \\ and\text{ } \\ y\text{ = 4x - equation 2} \\ Then,\text{ the intersection will be:} \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/mp68ergr9e4b1rcp0fx34u7wpwztkwzbk0.png)
![\begin{gathered} √(5x)\text{ = 4x } \\ square\text{ both sides, we have that:} \\ 5x=16x^2 \\ Then,\text{ this means that:} \\ 5\text{ x - 16 x}^2\text{ = 0} \\ and \\ x\text{ \lparen 5 - 16x \rparen = 0} \\ x\text{ = 0 or 5 - 16 x = 0} \\ x\text{ = 0 or 16x = 5} \\ x\text{ = 0 or x =}(5)/(16) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/xkttke0ctkjifebwmz2jusdmwhc0grlx55.png)
![\begin{gathered} Then\text{ , } \\ a\text{ = 0} \\ b\text{ = }(5)/(16) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7fpo92t23kcomyjedqdfeqn9bx3p8dzkt3.png)
PART TWO:
Between these two points, the curve that takes on smaller values is:
![y_1=√(5x)](https://img.qammunity.org/2023/formulas/mathematics/college/r0ha6fsffoohwvsux8v11dx7bqf78t7puv.png)
and the curve that takes on the larger values is:
![y_2=\text{ 4x}](https://img.qammunity.org/2023/formulas/mathematics/college/cdv4rqr3l9f97zylwlpxdib1vtz44369bv.png)
Thus, to find the area of the region, we must calculate the integral:
![\int_a^by_2-y_1\text{ dx}](https://img.qammunity.org/2023/formulas/mathematics/college/vbu04y1v0kvx35s2gsphz9qlwuth72vt71.png)
So, the area between the two curves is:
![\begin{gathered} The\text{ area between the two curves is:} \\ 0.065\text{ square units \lparen correct to 3 decimal places\rparen} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/rwu37xy7f348ucvjrutjst3ruvl3mlkx3q.png)