Find the formula for the Riemann sum obtained by dividing the interval  
[
0
,
18
]
  into  
n
  equal subintervals and using the right endpoint for each  
c
k
 . Then take the limit of these sums as  
n
→
∞
  to calculate the area under the curve  
f
(
x
)
=
324
−
x
2
  over  
[
0
,
18
]
 .

We want to calculate the area under the curve \( f(x) = 324 - x^2 \) over the interval \([0, 18]\) using Riemann sums, dividing the interval into \(n\) equal subintervals, and using the right endpoints.

### Step 1: Set up the Riemann Sum

First, divide the interval \([0, 18]\) into \(n\) equal subintervals. Each subinterval has width

\[
\Delta x = \frac{18 - 0}{n} = \frac{18}{n}.
\]

The right endpoint of the \(k\)-th subinterval is given by

\[
x_k = 0 + k \Delta x = \frac{18k}{n}.
\]

Thus, the function evaluated at the right endpoint \(x_k\) is:

\[
f(x_k) = 324 - \left( \frac{18k}{n} \right)^2 = 324 - \frac{324k^2}{n^2}.
\]

### Step 2: Write the Riemann Sum

The Riemann sum \(R_n\) using the right endpoints is:

\[
R_n = \sum_{k=1}^{n} f(x_k) \Delta x = \sum_{k=1}^{n} \left( 324 - \frac{324k^2}{n^2} \right) \cdot \frac{18}{n}.
\]

Simplify the expression:

\[
R_n = \frac{18}{n} \sum_{k=1}^{n} \left( 324 - \frac{324k^2}{n^2} \right) = \frac{18}{n} \left( 324n - \frac{324}{n^2} \sum_{k=1}^{n} k^2 \right).
\]

### Step 3: Use the Formula for the Sum of Squares

We can use the known formula for the sum of squares:

\[
\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}.
\]

Substitute this into the Riemann sum expression:

\[
R_n = \frac{18}{n} \left( 324n - \frac{324}{n^2} \cdot \frac{n(n+1)(2n+1)}{6} \right).
\]

Simplify further:

\[
R_n = 18 \left( 324 - \frac{324(n+1)(2n+1)}{6n^2} \right).
\]

### Step 4: Take the Limit as \(n \to \infty\)

To find the exact area under the curve, we take the limit of the Riemann sum as \(n \to \infty\). In the limit, the term \(\frac{(n+1)(2n+1)}{n^2}\) approaches \(2\), since:

\[
\lim_{n \to \infty} \frac{(n+1)(2n+1)}{n^2} = \lim_{n \to \infty} \frac{2n^2 + 3n + 1}{n^2} = 2.
\]

Therefore, the limit of the Riemann sum is:

\[
\lim_{n \to \infty} R_n = 18 \left( 324 - \frac{324 \cdot 2}{6} \right) = 18 \left( 324 - 108 \right) = 18 \cdot 216 = 3888.
\]

### Final Answer:

The area under the curve \( f(x) = 324 - x^2 \) over the interval \([0, 18]\) is \( 3888 \).

---

Would you like more details or have any questions? 

Here are 5 related questions:
1. What is the geometric interpretation of a Riemann sum?
2. How does the right-endpoint approximation compare to other methods like the midpoint or left-endpoint?
3. Can the result change if we divide the interval into non-equal subintervals?
4. How does the function \( f(x) = 324 - x^2 \) behave geometrically?
5. What happens if we change the interval of integration to something like \([0, 20]\)?

**Tip:** Understanding how sums behave in the limit is key to mastering calculus concepts like integration.

Find the formula for the Riemann sum obtained by dividing the interval [0, 18] into n equal subintervals and using the right endpoint for each ck. Then take the limit of these sums as n→∞ to calculate the area under the curve f(x) = 324 − x^2 over [0, 18].

Explore how to calculate the area under the curve of f(x) = 324 - x^2 over [0, 18] using Riemann sums with right endpoints. This step-by-step guide covers key calculus concepts, including Riemann sums, limits, and the fundamental theorem of calculus. Ideal for undergraduate calculus students.

Math Problem Statement

Find the formula for the Riemann sum obtained by dividing the interval
[ 0 , 18 ] into
n equal subintervals and using the right endpoint for each
c k . Then take the limit of these sums as
n → ∞ to calculate the area under the curve
f ( x )

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