The following sum \[ \sqrt{36-\left(\frac{6}{n}\right)^{2}} \cdot \frac{6}{n}+\sqrt{36-\left(\frac{12}{n}\right)^{2}} \cdot \frac{6}{n}+\ldots+\sqrt{36-\left(\frac{6 n}{n}\right)^{2}} \cdot \frac{6}{n} \] is a right Riemann sum with $n$ subintervals of equal length for the definite integral \[ \int_{0}^{b} f(x) d x \] where $b=$ and $f(x)=$

Step 1 ：The given sum is a Riemann sum for the definite integral of a function \(f(x)\) over the interval \([0, b]\). The Riemann sum is given by \[\sum_{i=1}^{n} f(x_i) \Delta x\] where \(x_i\) are the points in the interval \([0, b]\) and \(\Delta x\) is the width of each subinterval. In this case, \(x_i = \frac{6i}{n}\) and \(\Delta x = \frac{6}{n}\).

Step 2 ：The function \(f(x)\) is given by the expression inside the square root in the sum, which is \(\sqrt{36 - x^2}\).

Step 3 ：The upper limit of the integral, \(b\), is the maximum value of \(x_i\), which is \(x_n = \frac{6n}{n} = 6\).

Step 4 ：So, \(f(x) = \sqrt{36 - x^2}\) and \(b = 6\).

Step 5 ：Therefore, the definite integral is \[\int_{0}^{6} \sqrt{36 - x^2} dx\]