$$ =-\frac{27}{2}+\frac{72}{2}=\frac{45}{2} $$ b. $$ \begin{array}{l} y=25-x^{2}[1,4] \ A=\lim _{n \rightarrow \infty} \sum_{i=1}^{n}+(c) \Delta x=U \ \Delta x=\frac{4-1}{n}=\frac{3}{n} \quad C i=1+\frac{3}{n} i \ \sum_{i=1}^{n} i=\frac{n(n+1)}{2} \end{array} $$ $$ A=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f(c) \Delta x=\operatorname{un}_{n \rightarrow \infty} \sum_{i=1}^{n}\left(25-\left(1+\frac{3 i}{n}\right)^{2}\right)\left(\frac{3}{n}\right) $$ $$ =\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(25-\left(1+\frac{6 i}{n}+\frac{9 i^{2}}{n^{2}}\right)\right)\left(\frac{3}{n}\right) $$ $$ =\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left(24-\frac{6 i}{n}-\frac{4 i^{2}}{n^{2}}\right)\left(\frac{3}{n}\right) $$ $$ \begin{array}{l} =\lim _{n \rightarrow \infty} \sum_{n=1}^{n}\left(24-\frac{6 i}{n} \cdot \frac{9 i^{2}}{n^{2}}\right)\left(\frac{3}{n}\right) \ A=\lim _{n \rightarrow \infty}\left(\frac{3}{n}\left(24 n-6 \cdot \frac{n(m+n)}{2}-9 \cdot \frac{n(n+1)(2 n+1)}{6 n^{2}}\right)\right) \end{array} $$ this is a review of 4.1-4.3, we

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UpStudy AI · Accepted Answer

It looks like you're working through a problem involving calculating the area under the curve $$ y = 25 - x^2 $$ on the interval $$[1, 4]$$ using Riemann sums. Let's walk through your steps, identify any discrepancies, and ensure the final result is accurate.

### Problem Overview

You're aiming to compute the definite integral:

$$
A = \int_{1}^{4} (25 - x^2) \, dx
$$

This integral represents the area under the curve $$ y = 25 - x^2 $$ from $$ x = 1 $$ to $$ x = 4 $$.

### Your Approach Using Riemann Sums

You've set up the Riemann sum as follows:

1. **Partition the Interval:**
   $$
   \Delta x = \frac{4 - 1}{n} = \frac{3}{n}
   $$
   
2. **Sample Points:**
   $$
   c_i = 1 + \frac{3i}{n}
   $$
   
3. **Function Evaluation at Sample Points:**
   $$
   f(c_i) = 25 - \left(1 + \frac{3i}{n}\right)^2
   $$
   
4. **Form the Riemann Sum:**
   $$
   A = \lim_{n \to \infty} \sum_{i=1}^{n} f(c_i) \Delta x = \lim_{n \to \infty} \sum_{i=1}^{n} \left(25 - \left(1 + \frac{3i}{n}\right)^2\right) \left(\frac{3}{n}\right)
   $$

### Simplifying the Riemann Sum

Let's simplify the expression step by step.

1. **Expand the Squared Term:**
   $$
   \left(1 + \frac{3i}{n}\right)^2 = 1 + \frac{6i}{n} + \frac{9i^2}{n^2}
   $$
   
2. **Substitute Back into the Sum:**
   $$
   25 - \left(1 + \frac{6i}{n} + \frac{9i^2}{n^2}\right) = 24 - \frac{6i}{n} - \frac{9i^2}{n^2}
   $$
   
   **Note:** It seems there was a typographical error in your original work where $$ \frac{9i^2}{n^2} $$ was incorrectly written as $$ \frac{4i^2}{n^2} $$. The correct coefficient should be **9**, not 4.

3. **Updated Riemann Sum Expression:**
   $$
   A = \lim_{n \to \infty} \sum_{i=1}^{n} \left(24 - \frac{6i}{n} - \frac{9i^2}{n^2}\right) \left(\frac{3}{n}\right)
   $$
   
4. **Distribute $$ \frac{3}{n} $$ Across the Terms:**
   $$
   A = \lim_{n \to \infty} \left( \sum_{i=1}^{n} \frac{72}{n} - \sum_{i=1}^{n} \frac{18i}{n^2} - \sum_{i=1}^{n} \frac{27i^2}{n^3} \right)
   $$

### Evaluating Each Sum

Let's evaluate each component of the sum:

1. **First Sum:**
   $$
   \sum_{i=1}^{n} \frac{72}{n} = 72 \cdot \frac{n}{n} = 72
   $$
   
2. **Second Sum:**
   $$
   \sum_{i=1}^{n} \frac{18i}{n^2} = \frac{18}{n^2} \cdot \frac{n(n + 1)}{2} = \frac{18(n + 1)}{2n} = \frac{9(n + 1)}{n}
   $$
   
   As $$ n \to \infty $$, $$ \frac{9(n + 1)}{n} \to 9 $$.
   
3. **Third Sum:**
   $$
   \sum_{i=1}^{n} \frac{27i^2}{n^3} = \frac{27}{n^3} \cdot \frac{n(n + 1)(2n + 1)}{6} = \frac{27(n + 1)(2n + 1)}{6n^2} = \frac{9(n + 1)(2n + 1)}{2n^2}
   $$
   
   As $$ n \to \infty $$, $$ \frac{9(n + 1)(2n + 1)}{2n^2} \to \frac{9 \cdot 2n^2}{2n^2} = 9 $$.
   
### Combining the Results

Putting it all together:

$$
A = \lim_{n \to \infty} \left(72 - 9 - 9\right) = 54
$$

### Verification Using Direct Integration

For confirmation, let's compute the integral directly:

$$
A = \int_{1}^{4} (25 - x^2) \, dx = \left[25x - \frac{x^3}{3}\right]_{1}^{4} 
$$
$$
= \left(25 \cdot 4 - \frac{4^3}{3}\right) - \left(25 \cdot 1 - \frac{1^3}{3}\right) 
$$
$$
= \left(100 - \frac{64}{3}\right) - \left(25 - \frac{1}{3}\right) 
$$
$$
= \left(\frac{300}{3} - \frac{64}{3}\right) - \left(\frac{75}{3} - \frac{1}{3}\right)
$$
$$
= \frac{236}{3} - \frac{74}{3} = \frac{162}{3} = 54
$$

Both methods agree that the area $$ A = 54 $$.

### Summary of Corrections

1. **Typographical Error:** Ensure that $$ \frac{9i^2}{n^2} $$ remains consistent throughout your calculations. It was incorrectly written as $$ \frac{4i^2}{n^2} $$ in one of your steps.

2. **Final Answer Discrepancy:** Your initial calculation concluded with $$ \frac{45}{2} = 22.5 $$, which doesn't align with the correct value of 54. This discrepancy likely arose from the aforementioned typographical error and subsequent miscalculations.

3. **Correct Final Answer:** The accurate area under the curve $$ y = 25 - x^2 $$ from $$ x = 1 $$ to $$ x = 4 $$ is **54**.

If you have any further questions or need clarification on specific steps, feel free to ask!
The area under the curve $$ y = 25 - x^2 $$ from $$ x = 1 $$ to $$ x = 4 $$ is **54**.

b.

this is a review of 4.1-4.3, we

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b. this is a review of 4.1-4.3, we