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Part 1

For the function given​ below, find a formula for the Riemann sum obtained by dividing the interval 

​[0,3030​] 

into n equal subintervals and using the​ right-hand endpoint for each 

c Subscript kck. 

Then take a limit of this sum as 

n right arrow infinityn → ∞ 

to calculate the area under the curve over 

​[0,3030​]. f(x)= 900-x^2

To solve this problem, we'll follow these steps:

1. **Divide the interval \([0, 30]\) into \( n \) equal subintervals.**
2. **Find the formula for the Riemann sum using the right-hand endpoints.**
3. **Take the limit of this sum as \( n \) approaches infinity to find the area under the curve.**

### Step 1: Divide the Interval

We divide the interval \([0, 30]\) into \( n \) equal subintervals. The width of each subinterval, \(\Delta x\), is:

\[
\Delta x = \frac{30 - 0}{n} = \frac{30}{n}
\]

### Step 2: Right-Hand Endpoints

The right-hand endpoint of the \( k \)-th subinterval is:

\[
x_k = 0 + k \Delta x = \frac{30k}{n} \quad \text{for} \quad k = 1, 2, \ldots, n
\]

### Step 3: Riemann Sum

The Riemann sum \( S_n \) using the right-hand endpoints is given by:

\[
S_n = \sum_{k=1}^{n} f(x_k) \Delta x
\]

Substitute \( f(x) = 900 - x^2 \) and \( x_k = \frac{30k}{n} \):

\[
S_n = \sum_{k=1}^{n} \left(900 - \left(\frac{30k}{n}\right)^2\right) \cdot \frac{30}{n}
\]

Simplify inside the summation:

\[
S_n = \sum_{k=1}^{n} \left(900 - \frac{900k^2}{n^2}\right) \cdot \frac{30}{n}
\]

Distribute \(\frac{30}{n}\):

\[
S_n = \sum_{k=1}^{n} \left( 900 \cdot \frac{30}{n} - \frac{900k^2}{n^2} \cdot \frac{30}{n} \right)
\]

\[
S_n = \sum_{k=1}^{n} \left( \frac{27000}{n} - \frac{27000k^2}{n^3} \right)
\]

Separate the sums:

\[
S_n = \frac{27000}{n} \sum_{k=1}^{n} 1 - \frac{27000}{n^3} \sum_{k=1}^{n} k^2
\]

### Step 4: Summation Formulas

Use the summation formulas:

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

Substitute these into the Riemann sum:

\[
S_n = \frac{27000}{n} \cdot n - \frac{27000}{n^3} \cdot \frac{n(n+1)(2n+1)}{6}
\]

Simplify:

\[
S_n = 27000 - \frac{27000(n+1)(2n+1)}{6n^2}
\]

### Step 5: Take the Limit as \( n \to \infty \)

Evaluate the limit:

\[
\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( 27000 - \frac{27000(n+1)(2n+1)}{6n^2} \right)
\]

Simplify the fraction:

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

Thus:

\[
\lim_{n \to \infty} S_n = 27000 - \frac{27000 \cdot 2}{6} = 27000 - 9000 = 18000
\]

Therefore, the area under the curve \( f(x) = 900 - x^2 \) over the interval \([0, 30]\) is \( \boxed{18000} \).

Would you like more details or have any questions?

Here are some additional questions you might find interesting:
1. How do you find the area under a curve using the left-hand endpoint for the Riemann sum?
2. Can you explain the difference between Riemann sums and the definite integral?
3. How do you interpret the result of the definite integral in a real-world context?
4. What are the properties of definite integrals that make them useful in calculus?
5. How do you calculate the area between two curves using integration?

**Tip:** When calculating Riemann sums, be careful with the placement of indices and summation limits to ensure accuracy in the final result.

Learn how to calculate the area under the curve using Riemann sums and limits. Solve the problem of finding the area under the curve f(x) = 900 - x^2 over the interval [0, 30] by dividing it into n equal subintervals and using right-hand endpoints. Suitable for advanced high school and college students studying calculus.

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