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Ahat [919]
2 years ago
5

For the function given below, find a formula for the Riemann sum obtained by dividing the interval (0, 3) into n equal subinterv

als and us right-hand endpoint for each Then take a limit of this sum as c_{k}; n -> ∞ to calculate the area under the curve over [0, 3] . f(x) = 2x ^ 2 Write a formula for a Riemann sum for the function f(x) = 2x ^ 2 over the interval [0, 3]
Mathematics
1 answer:
Viktor [21]2 years ago
8 0

Splitting up [0, 3] into n equally-spaced subintervals of length \Delta x=\frac{3-0}n = \frac3n gives the partition

\left[0, \dfrac3n\right] \cup \left[\dfrac3n, \dfrac6n\right] \cup \left[\dfrac6n, \dfrac9n\right] \cup \cdots \cup \left[\dfrac{3(n-1)}n, 3\right]

where the right endpoint of the i-th subinterval is given by the sequence

r_i = \dfrac{3i}n

for i\in\{1,2,3,\ldots,n\}.

Then the definite integral is given by the infinite Riemann sum

\displaystyle \int_0^3 2x^2 \, dx = \lim_{n\to\infty} \sum_{i=1}^n 2{r_i}^2 \Delta x \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac6n \sum_{i=1}^n \left(\frac{3i}n\right)^2 \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac{54}{n^3} \sum_{i=1}^n i^2 \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac{54}{n^3}\cdot\frac{n(n+1)(2n+1)}6 = \boxed{18}

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Answer:

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Step-by-step explanation:

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Answer:

Question 9

(a) The distance, Jalaj walks in one day is 4.4 km

(b) The amount Jalaj raises after walking for 22 km at the end of the 5 days is $8

Question 10

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Step-by-step explanation:

Question 9

(a) The distance Jalaj walks in 5 days = 22 km

Whereby Jalaj walks equal distance every day, we have;

The distance, Jalaj walks in one day = 22 km/5 days = 4.4 km/day

The distance, Jalaj walks in one day = 4.4 km

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The amount he raises after walking for 22 km at the end of the 5 days = 5 × $1.60  = $8

Question 10

(b) The given side length of the square = 120 meters to the nearest 10 meters

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The maximum dimension for the side length of the square = 120 + 10/2 = 125

The largest possible area of the square, A_l = 125 m × 125 m = 15,625 m²

The minimum dimension for the side length of the square = 120 m - 10 m/2 = 115 m

The smallest possible area of the square, A_s = 115 m × 115 m = 13,225 m².

The difference between the largest and smallest areas, A_l - A_s = 15,625 - 13,225 = 2,400 m².

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