Use the definition of the definite integral to evaluate \( \int_{0}^{6}\left(x^{2}-6\right) d x \) Evaluate the right Riemann Sum. Choose the correct answer below. A. \( \int_{a}^{b} f(x) d x=\lim _{\Delta \rightarrow 0} \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x_{k}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\left(\frac{6}{n}\right)^{2}-6\right) \frac{6 k}{n} \) B. \( \int_{a}^{b} f(x) d x=\lim _{\Delta \rightarrow 0} \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x_{k}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\left(\frac{n}{6 k}\right)^{2}-6\right) \frac{n}{6} \) c. \( \int_{a}^{b} f(x) d x=\lim _{\Delta \rightarrow 0} \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x_{k}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\left(\frac{n}{6}\right)^{2}-6\right) \frac{n}{6 k} \) D. \( \int_{a}^{b} f(x) d x=\lim _{\Delta \rightarrow 0} \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x_{k}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\left(\frac{6 k}{n}\right)^{2} \cdot-6\right) \frac{6}{n} \) Evaluate the integral. \[ \int_{0}^{6}\left(x^{2}-6\right) d x=\square \text { (Simplify your answer.) } \]

We wish to evaluate \[ \int_{0}^{6} \left(x^2 - 6\right)\, dx \] using the definition of the definite integral (the limit of a Riemann sum). For an interval \([a,b]\) divided into \( n \) equal subintervals, the width of each subinterval is \[ \Delta x = \frac{b - a}{n} = \frac{6 - 0}{n} = \frac{6}{n}. \] When using the right endpoints, the \( k^{\text{th}} \) sample point is \[ x_{k}^* = a + k\Delta x = 0 + k\frac{6}{n} = \frac{6k}{n}. \] Then the corresponding Riemann sum is \[ \sum_{k=1}^{n} f\left(x_{k}^*\right)\Delta x = \sum_{k=1}^{n} \left[\left(\frac{6k}{n}\right)^2 -6\right]\frac{6}{n}. \] Taking the limit as \( n\to\infty \), we have \[ \int_{0}^{6}\left(x^2-6\right)\,dx = \lim_{n\to\infty} \sum_{k=1}^{n} \left[\left(\frac{6k}{n}\right)^2-6\right]\frac{6}{n}. \] Among the given answer choices, the one that correctly represents the right Riemann sum is choice A with the correction that the term inside the sum should have \(\left(\frac{6k}{n}\right)^2\) (not \(\left(\frac{6}{n}\right)^2\)). Therefore, choice A corresponds to the correct setup. Now we evaluate the integral by finding an antiderivative of \[ f(x)=x^2-6. \] 1. Compute the antiderivative: \[ \int \left(x^2-6\right)\,dx=\frac{x^3}{3}-6x+C. \] 2. Evaluate from \( x=0 \) to \( x=6 \): \[ \left[\frac{x^3}{3}-6x\right]_{0}^{6}=\left(\frac{6^3}{3}-6\cdot 6\right)-\left(\frac{0^3}{3}-6\cdot 0\right). \] Calculate the value at \( x=6 \): \[ \frac{6^3}{3}-6\cdot 6= \frac{216}{3} - 36 = 72 - 36 = 36. \] Thus, the value of the integral is \[ \int_{0}^{6}\left(x^2-6\right) dx = 36. \]