Find $L_{n}, R_{n}$, and their average for the definite integral below using the indicated value of $n$.
\[
\int_{1}^{4}\left(5 x^{3}-x\right) d x, n=6
\]
The left Riemann sum, $L_{n}$, is
(Do not round until the final answer. Then round to three decimal places as needed.)

Step 1 ：First, we need to find the width of each subinterval. The width, \(\Delta x\), is given by \(\frac{b-a}{n}\), where \(a\) and \(b\) are the limits of integration and \(n\) is the number of subintervals. In this case, \(a = 1\), \(b = 4\), and \(n = 6\), so \(\Delta x = \frac{4-1}{6} = \frac{1}{2}\).

Step 2 ：Next, we find the left endpoints of each subinterval. The \(i\)th left endpoint, \(x_i\), is given by \(a + i\Delta x\). For \(i = 0, 1, 2, ..., 5\), the left endpoints are \(1, 1.5, 2, 2.5, 3, 3.5\).

Step 3 ：We then evaluate the function at each left endpoint. The function is \(f(x) = 5x^3 - x\), so the values of the function at the left endpoints are \(f(1) = 4, f(1.5) = 9.875, f(2) = 36, f(2.5) = 76.375, f(3) = 132, f(3.5) = 204.625\).

Step 4 ：The left Riemann sum, \(L_n\), is the sum of the products of the function values at the left endpoints and the width of the subintervals. So, \(L_n = \Delta x \sum_{i=0}^{n-1} f(x_i) = 0.5(4 + 9.875 + 36 + 76.375 + 132 + 204.625) = 231.4375\).

Step 5 ：Next, we find the right endpoints of each subinterval. The \(i\)th right endpoint, \(x_i\), is given by \(a + (i+1)\Delta x\). For \(i = 0, 1, 2, ..., 5\), the right endpoints are \(1.5, 2, 2.5, 3, 3.5, 4\).

Step 6 ：We then evaluate the function at each right endpoint. The function is \(f(x) = 5x^3 - x\), so the values of the function at the right endpoints are \(f(1.5) = 9.875, f(2) = 36, f(2.5) = 76.375, f(3) = 132, f(3.5) = 204.625, f(4) = 300\).

Step 7 ：The right Riemann sum, \(R_n\), is the sum of the products of the function values at the right endpoints and the width of the subintervals. So, \(R_n = \Delta x \sum_{i=0}^{n-1} f(x_i) = 0.5(9.875 + 36 + 76.375 + 132 + 204.625 + 300) = 379.4375\).

Step 8 ：Finally, the average of the left and right Riemann sums is \(\frac{L_n + R_n}{2} = \frac{231.4375 + 379.4375}{2} = 305.4375\).

Step 9 ：So, the left Riemann sum, \(L_n\), is \(\boxed{231.4375}\), the right Riemann sum, \(R_n\), is \(\boxed{379.4375}\), and their average is \(\boxed{305.4375}\).