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 )

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 $3888$.

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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.