find the area of the region bounded by the graphs of the given equations. enter your answer in exact form or…

Explanation:

Step1: Determine the upper - lower functions

We need to find which function is on top and which is on the bottom in the interval $[1,7]$. Let's find the difference $f(x)=(6 + 2x)-\sqrt{4x}=6 + 2x-2\sqrt{x}$. We can check the value of $f(x)$ at a test - point in the interval $[1,7]$, say $x = 1$. $f(1)=6+2\times1 - 2\sqrt{1}=6 + 2-2=6>0$. So, $y = 6 + 2x$ is the upper function and $y=\sqrt{4x}=2\sqrt{x}$ is the lower function on the interval $[1,7]$.

Step2: Use the area formula

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}(f(x)-g(x))dx$. Here, $a = 1$, $b = 7$, $f(x)=6 + 2x$, and $g(x)=2\sqrt{x}$. So, $A=\int_{1}^{7}((6 + 2x)-2\sqrt{x})dx=\int_{1}^{7}(6 + 2x-2x^{\frac{1}{2}})dx$.

Step3: Integrate term - by - term

Using the power rule of integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have: $\int(6 + 2x-2x^{\frac{1}{2}})dx=6x+x^{2}-\frac{4}{3}x^{\frac{3}{2}}+C$.

Step4: Evaluate the definite integral

$A=\left[6x+x^{2}-\frac{4}{3}x^{\frac{3}{2}}\right]_{1}^{7}$ $=(6\times7 + 7^{2}-\frac{4}{3}\times7^{\frac{3}{2}})-(6\times1+1^{2}-\frac{4}{3}\times1^{\frac{3}{2}})$ $=(42 + 49-\frac{4}{3}\times7\sqrt{7})-(6 + 1-\frac{4}{3})$ $=(91-\frac{28\sqrt{7}}{3})-(7-\frac{4}{3})$ $=91-\frac{28\sqrt{7}}{3}-7+\frac{4}{3}$ $=84+\frac{4 - 28\sqrt{7}}{3}$ $\approx84+\frac{4-28\times2.646}{3}$ $\approx84+\frac{4 - 74.088}{3}$ $\approx84+\frac{-70.088}{3}$ $\approx84 - 23.36$ $\approx60.64$

Answer:

$84+\frac{4 - 28\sqrt{7}}{3}\approx60.64$