a) The approximate area below the curve using five rectangles is 1280 square units.
b) The approximate area below the curve using ten rectangles is 1320 square units.
c) The approximate area below the curve using infinite number of rectangles is 1333.333 square units.
How to find the area below the curve by Riemann's sum
In this problem we must estimate the value of the area below the curve by finite number of rectangles using Riemann sums, whose expression is:
A ≈ [(b - a) / n] · ∑ f[a + i · (b - a) / n], for i = {0, 1, 2, 3, ..., n - 1} (1)
Where:
- n - Number of rectangles
- a - Lower limit of the interval.
- b - Upper limit of the interval.
- i - Index of the rectangle.
The approximate area below the curve using five rectangles is: f(x) = 20 · x - x², a = 0, b = 20, n = 5
A ≈ [(20 - 0) / 5] · ∑ f[0 + i · (20 - 0) / 5]
A ≈ 4 · ∑ f(4 · i)
A ≈ 4 · [f(0) + f(4) + f(8) + f(12) + f(16)]
f(0) = 20 · 0 - 0² = 0
f(4) = 20 · 4 - 4² = 64
f(8) = 20 · 8 - 8² = 96
f(12) = 20 · 12 - 12² = 96
f(16) = 20 · 16 - 16² = 64
A ≈ 4 · (0 + 64 + 96 + 96 + 64)
A ≈ 1280
And using ten rectangles:
A ≈ [(20 - 0) / 10] · ∑ f[0 + i · (20 - 0) / 10]
A ≈ 2 · ∑ f(2 · i)
A ≈ 2 · [f(0) + f(2) + f(4) + f(6) + f(8) + f(10) + f(12) + f(14) + f(16) + f(18)]
f(0) = 20 · 0 - 0² = 0
f(2) = 20 · 2 - 2² = 36
f(4) = 20 · 4 - 4² = 64
f(6) = 20 · 6 - 6² = 84
f(8) = 20 · 8 - 8² = 96
f(10) = 20 · 10 - 10² = 100
f(12) = 20 · 12 - 12² = 96
f(14) = 20 · 14 - 14² = 84
f(16) = 20 · 16 - 16² = 64
f(18) = 20 · 18 - 18² = 36
A ≈ 2 · (0 + 36 + 64 + 84 + 96 + 100 + 96 + 84 + 64 + 36)
A ≈ 1320
And using infinite rectangles:
A ≈ [(b - a) / n] · ∑ f[a + i · (b - a) / n]
A ≈ (20 / n) · ∑ [20 · (20 / n) · i - (20 / n)² · i²]
A ≈ (20 / n) · ∑ [400 · i / n - 400 · i² / n²]
A ≈ ∑ (8000 · i / n² - 8000 · i² / n³)
A ≈ (8000 / n²) · ∑ i - (8000 / n³) · ∑ i²
A ≈ (8000 / n²) · [n · (n + 1) / 2] - (8000 / n³) · [n · (n + 1) · (2 · n + 1) / 6]
A ≈ (8000 / n²) · [(n² + n) / 2] - (8000 / n³) · [(n² + n) · (2 · n + 1) / 6]
A ≈ 4000 · (n² + 2) / n² - (8000 / n³) · [(2 · n³ + 3 · n² + n) / 6]
A ≈ 4000 · (1 + 2 / n²) - (4000 / 3) · [2 + 3 · (1 / n) + (1 / n²)]
As n → + ∞, then:
A ≈ 4000 - 8000 / 3
A ≈ 1333.333
To learn more on Riemann's sums: https://brainly.com/question/21847158
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