Example Question - riemann sums
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Approximating the Area Under a Curve Using Riemann Sums
<p>لحل هذا السؤال، سنستخدم مجموع ريمان الأيسر:</p> <p>معادلة الدالة \( y = x^2 + 2 \)</p> <p>المجال هو \([0, 1]\)</p> <p>عدد المستطيلات \( n = 12 \)</p> <p>عرض كل مستطيل \( \Delta x = \frac{1 - 0}{12} = \frac{1}{12} \)</p> <p>نقاط التقييم اليسرى هي \( x_i = 0, \frac{1}{12}, \frac{2}{12}, \ldots, \frac{11}{12} \)</p> <p>الآن نقوم بحساب قيمة المجموع:</p> <p>\( \sum_{i=0}^{11} f(x_i) \Delta x \)</p> <p>\( = \sum_{i=0}^{11} \left( ( \frac{i}{12} )^2 + 2 \right) \frac{1}{12} \)</p> <p>\( = \frac{1}{12} \sum_{i=0}^{11} \left( \frac{i^2}{144} + 2 \right) \)</p> <p>\( = \frac{1}{12} \left( \sum_{i=0}^{11} \frac{i^2}{144} + \sum_{i=0}^{11} 2 \right) \)</p> <p>\( = \frac{1}{12} \left( \frac{1}{144} \sum_{i=0}^{11} i^2 + 2 \cdot 12 \right) \)</p> <p>حيث أن \( \sum_{i=0}^{11} i^2 \) هو مجموع مربعات العداد الأول 11 عدد طبيعي \( = 0^2 + 1^2 + 2^2 + \ldots + 11^2 \)</p> <p>\( = 0 + 1 + 4 + \ldots + 121 \)</p> <p>\( = 506 \)</p> <p>إذاً:</p> <p>\( \frac{1}{12} \left( \frac{506}{144} + 24 \right) \)</p> <p>\( = \frac{1}{12} \left( \frac{506}{144} + \frac{3456}{144} \right) \)</p> <p>\( = \frac{1}{12} \cdot \frac{3962}{144} \)</p> <p>\( = \frac{3962}{1728} \)</p> <p>\( \approx 2.292 \)</p> <p>النتيجة التقريبية لمساحة المنطقة تحت المنحنى هي \( \approx 2.292 \) وحدة مربعة.</p>
Analyzing Riemann Sums for Integral Approximation
To find the Riemann sum that gives the value of the definite integral \(\int_{-2}^{3} (x^2 + 4x) dx\), we should identify which sum corresponds to calculating the integral using left-hand, right-hand, or midpoint Riemann sums. The integral given is from -2 to 3. Let's start by defining a partition of the interval [-2, 3] into n subintervals; this means each subinterval has width Δx. Given the boundaries a = -2 and b = 3, the width of each subinterval is: \[ Δx = \frac{{b - a}}{n} = \frac{{3 - (-2)}}{n} = \frac{5}{n} \] The general x-value for the k-th subinterval can be expressed in three different forms depending on the Riemann sum method: - Left-hand: \( x_k = a + kΔx \) - Right-hand: \( x_k = a + (k+1)Δx \) - Midpoint: \( x_k = a + \left(k + \frac{1}{2}\right)Δx \) We need to find out which option corresponds to using one of these forms. (A) The term inside the summation \( -2 + \left(\frac{3}{2}\frac{k}{n} + \frac{3}{2}\frac{k}{n} \right) \) doesn't seem to fit any of the standard forms for left, right, or midpoint Riemann sums. (B) \( -2 + \frac{5k}{n} \) fits the left-hand rule where \( x_k = -2 + k Δx \) because cooresponds \( Δx = \frac{5}{n} \). (C) \( -2 + \frac{5}{2} + \left( \frac{5k}{n} \right) \) corresponds to the midpoint Riemann sum, where we start at the midpoint of the first interval and add \( k Δx \). (D) \( -2 + \frac{5}{2} \left( \frac{4}{n} + \frac{2}{n}k \right) \) doesn't match any of the standard forms either. Now using the function \( f(x) = x^2 + 4x \), we can express the Riemann sum based on the left-hand rule (which option B represents). The terms inside the summation should be \( f(x_k)Δx \), which would be \( f(-2 + \frac{5k}{n})Δx \). Substitute \( f(x) \) and \( Δx \): \[ f(x_k) = (-2 + \frac{5k}{n})^2 + 4(-2 + \frac{5k}{n}) \] \[ Δx = \frac{5}{n} \] Hence, the sum is \( Σ [(-2 + \frac{5k}{n})^2 + 4(-2 + \frac{5k}{n})]·\frac{5}{n} \), which simplifies to: \[ Σ \frac{5}{n}[(-2 + \frac{5k}{n})^2 + 4(-2 + \frac{5k}{n})] \] \[ Σ \frac{5}{n}[4 + \frac{25k^2}{n^2} - 20\frac{k}{n} - 8 + 20\frac{k}{n} - \frac{20k}{n}] \] \[ Σ \frac{5}{n}[-4 + \frac{25k^2}{n^2}] \] This expression doesn't match any of the options exactly, but it is closest to option (B), which suggests that there may be an error in the transcription of the options or a simplification error. Thus, option (B) most closely resembles the expression we derived for the Riemann sum based on the left-hand rule: \[ \lim_{n \to \infty} \sum_{k=1}^{n} f(-2 + \frac{5k}{n})·\frac{5}{n} \] However, the terms for \( f(-2 + \frac{5k}{n}) \) must be \( (-2 + \frac{5k}{n})^2 + 4(-2 + \frac{5k}{n}) \) to properly represent the left-hand Riemann sum for this function \( (x^2 + 4x) \) over the interval [-2, 3], and these terms should be inside the summation, not outside as shown in option (B). If there is an error in the transcription or presentation of the options, it might be wise to revisit the original problem to ensure that option (B) was copied correctly. Given the provided choices, none match the correct Riemann sum for the integral exactly, but the form of (B) suggests that it's meant to be the left endpoint approximation.
Analyzing Riemann Sums for Integral Calculation
The question asks which of the given limits of a Riemann sum represents the value of the integral ∫ from -2 to 2 of (x^3 - 4x) dx. For a Riemann sum to represent the integral of a function, it must be in the form of the limit as n approaches infinity of the sum from k=1 to n of f(x_k)Δx, where Δx is the width of each sub-interval, and x_k is a sample point in the k-th sub-interval. Now, because the interval of integration is from -2 to 2, the total width of the interval is 4. Therefore, Δx = (b-a)/n = (2 - (-2))/n = 4/n. Next, we need to pick the sample points. For problems like this, we often use either left endpoints, right endpoints, or midpoints. Since the problem does not specify, we'll examine each of the options provided to see which one matches up. Looking at the options, the sample points x_k should be something like -2 + k(Δx) or -2 + (k - 1/2)(Δx) or -2 + (k-1)(Δx) for the left, midpoint, and right Riemann sums respectively. Let's examine the structure of the Riemann sums in the options: (A) x_k appears to be -2 + (3/2)(k/n) which is not of the correct form for any of the typical Riemann sum sample points. (B) x_k appears to be -2 + (k/n) which is not correct because it lacks the width of the sub-interval, Δx, in the multiplication with k. (C) x_k appears to be -2 + (5/2)(k/n) which is again not of the correct form for the Riemann sum sample points. (D) The sum in this option has not boxed or circled any term within it to identify x_k. Since options A, B, and C have clear structural issues with their x_k values, none of these correspond to the correct form for a Riemann sum for the integral given. Now, let's re-evaluate option D. The width of each sub-interval is (4/n) and when we plug in k as our index we should get: x_k = -2 + (k-1)(4/n) for the left Riemann sum or x_k = -2 + k(4/n) for the right Riemann sum or x_k = -2 + (4/n)(k - 1/2) for the midpoint Riemann sum. The expression in option D suggests that it's probably intending the midpoint Riemann sum, because it has an additional 1/2 term in the representation of x_k (since it doesn't fit the left or right endpoint patterns). Therefore, let's confirm the structure of x_k for the midpoint Riemann sum by expanding the expression used for x_k in option D: x_k = -2 + (2 + 5/2k/n) = -2 + (2 + (5k)/(2n)) This is not immediately recognizable as a midpoint sum expression. To match the desired structure, we should see: x_k = -2 + (4/n)(k - 1/2) = -2 + (4k/n) - (2/n) However, it's evident that the expressions in option D do not match the form of the midpoint Riemann sum after manipulating the terms. Therefore, we need to correct the expression provided: The midpoint for each sub-interval would be at x_k = -2 + (4/n)(k - 1/2) which simplifies to x_k = -2 + (4k/n) - (2/n) And we would then construct the sum as: Σ_{k=1}^n f(x_k)Δx It seems that none of the options A-D perfectly match the expected structure for a Riemann sum that would calculate the integral ∫ from -2 to 2 of (x^3 - 4x) dx. Option D is the closest, but the form of x_k provided is not correct for a typical midpoint Riemann sum, assuming that was the intention. If none of the options matches, an error may be present in the formulation of the answers, or additional context may be required to determine the correct response.
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