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. $$ \int_{0}^{6}\left(x^{2}-6\right) d x=\square \text { (Simplify your answer.) } $$

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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

$$
\int_{0}^{6}\left(x^2-6\right) dx = 36.
$$
The integral evaluates to 36.

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