Find the area of the region bounded by the graphs of the given equations.
\[
y=7 x-x^{2}, y=2 x
\]

Step 1 ：First, we need to find the points of intersection of the two curves. This is done by setting the two equations equal to each other and solving for \(x\).

Step 2 ：So, we have \(7x - x^2 = 2x\). Simplifying this gives \(x^2 - 5x = 0\).

Step 3 ：Factoring out an \(x\) gives \(x(x - 5) = 0\). Setting each factor equal to zero gives the solutions \(x = 0\) and \(x = 5\).

Step 4 ：These are the \(x\)-coordinates of the points of intersection. We can find the corresponding \(y\)-coordinates by substituting these values into either of the original equations. Using \(y = 2x\), we get \(y = 0\) when \(x = 0\) and \(y = 10\) when \(x = 5\).

Step 5 ：So, the points of intersection are \((0, 0)\) and \((5, 10)\).

Step 6 ：The area of the region bounded by the two curves is given by the integral of the absolute difference of the two functions, from the lower point of intersection to the upper point of intersection.

Step 7 ：So, we need to compute the integral \(\int_{0}^{5} |7x - x^2 - 2x| dx\).

Step 8 ：Simplifying the integrand gives \(\int_{0}^{5} |5x - x^2| dx\).

Step 9 ：Since the integrand is nonnegative on the interval \([0, 5]\), we can remove the absolute value signs, giving \(\int_{0}^{5} (5x - x^2) dx\).

Step 10 ：Computing this integral gives \(\left[\frac{5}{2}x^2 - \frac{1}{3}x^3\right]_{0}^{5}\).

Step 11 ：Evaluating this at \(x = 5\) gives \(\frac{5}{2} \cdot 25 - \frac{1}{3} \cdot 125 = 62.5 - 41.67 = 20.83\).

Step 12 ：Evaluating this at \(x = 0\) gives 0.

Step 13 ：So, the area of the region bounded by the two curves is \(20.83 - 0 = 20.83\).

Step 14 ：Therefore, the area of the region bounded by the graphs of the given equations is \(\boxed{20.83}\) square units.