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To solve this problem, we need to approximate the area under the curve of \( f(x) = 5x^2 \) from \( x_0 = 0 \) to \( x_n = 10 \) using finite approximations. We will calculate both the lower and upper sums with different numbers of rectangles as specified.

Given \( f(x) = 5x^2 \), here’s how we proceed with each part:

### Part i: Lower sum with two rectangles of equal width
1. Divide the interval \([0, 10]\) into two subintervals: \([0, 5]\) and \([5, 10]\).
2. The width of each rectangle is \( \Delta x = 5 \).
3. For a lower sum, we evaluate \( f(x) \) at the left endpoints:
   - For the first rectangle, \( x = 0 \): \( f(0) = 5 \times 0^2 = 0 \)
   - For the second rectangle, \( x = 5 \): \( f(5) = 5 \times 5^2 = 125 \)
4. The lower sum \( L_2 \) is:
   \[
   L_2 = f(0) \cdot \Delta x + f(5) \cdot \Delta x = 0 \cdot 5 + 125 \cdot 5 = 625
   \]

### Part ii: Lower sum with four rectangles of equal width
1. Divide the interval \([0, 10]\) into four subintervals: \([0, 2.5]\), \([2.5, 5]\), \([5, 7.5]\), and \([7.5, 10]\).
2. The width of each rectangle is \( \Delta x = 2.5 \).
3. For a lower sum, we evaluate \( f(x) \) at the left endpoints:
   - \( f(0) = 0 \)
   - \( f(2.5) = 5 \times (2.5)^2 = 31.25 \)
   - \( f(5) = 125 \)
   - \( f(7.5) = 5 \times (7.5)^2 = 281.25 \)
4. The lower sum \( L_4 \) is:
   \[
   L_4 = f(0) \cdot 2.5 + f(2.5) \cdot 2.5 + f(5) \cdot 2.5 + f(7.5) \cdot 2.5
   = 0 \cdot 2.5 + 31.25 \cdot 2.5 + 125 \cdot 2.5 + 281.25 \cdot 2.5 = 1093.75
   \]

### Part iii: Upper sum with two rectangles of equal width
1. Divide the interval \([0, 10]\) into two subintervals: \([0, 5]\) and \([5, 10]\).
2. The width of each rectangle is \( \Delta x = 5 \).
3. For an upper sum, we evaluate \( f(x) \) at the right endpoints:
   - For the first rectangle, \( x = 5 \): \( f(5) = 125 \)
   - For the second rectangle, \( x = 10 \): \( f(10) = 5 \times 10^2 = 500 \)
4. The upper sum \( U_2 \) is:
   \[
   U_2 = f(5) \cdot \Delta x + f(10) \cdot \Delta x = 125 \cdot 5 + 500 \cdot 5 = 3125
   \]

### Part iv: Upper sum with four rectangles of equal width
1. Divide the interval \([0, 10]\) into four subintervals: \([0, 2.5]\), \([2.5, 5]\), \([5, 7.5]\), and \([7.5, 10]\).
2. The width of each rectangle is \( \Delta x = 2.5 \).
3. For an upper sum, we evaluate \( f(x) \) at the right endpoints:
   - \( f(2.5) = 31.25 \)
   - \( f(5) = 125 \)
   - \( f(7.5) = 281.25 \)
   - \( f(10) = 500 \)
4. The upper sum \( U_4 \) is:
   \[
   U_4 = f(2.5) \cdot 2.5 + f(5) \cdot 2.5 + f(7.5) \cdot 2.5 + f(10) \cdot 2.5
   = 31.25 \cdot 2.5 + 125 \cdot 2.5 + 281.25 \cdot 2.5 + 500 \cdot 2.5 = 2343.75
   \]

Would you like more details or have any questions?

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Here are some related questions for further exploration:

1. How does increasing the number of rectangles affect the accuracy of the approximation?
2. How would the calculation change if \( f(x) \) was a different function?
3. What is the actual integral of \( f(x) = 5x^2 \) from 0 to 10, and how do the approximations compare?
4. How could you set up a Riemann sum for this function and interval?
5. What are the implications of using the midpoint for each interval instead?

**Tip:** As the number of rectangles increases in finite approximations, the approximation generally becomes closer to the true area (the definite integral), especially if the function is continuous and well-behaved over the interval.

Use finite approximation to estimate the area under the graph of f(x) = 5x^2 and above the graph of f(x) = 0 from x_0 = 0 to x_n = 10 using:

i) a lower sum with two rectangles of equal width.
ii) a lower sum with four rectangles of equal width.
iii) an upper sum with two rectangles of equal width.
iv) an upper sum with four rectangles of equal width.

Explore how to use finite approximation to estimate the area under the curve of f(x) = 5x^2 between x = 0 and x = 10. This problem includes calculations for lower and upper sums with both two and four rectangles. A great problem for understanding Riemann sums and approximating areas under quadratic functions.

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