Step 3 of 4 The approximate Riemann sum of f(x) = x³ - 9x on [0, 7] using n rectangles with right endpoints X; widths. n Σ f(x)Ax = [ f(²1) ² n i=1 i=1 Using f(x) = x³ - 9x, find f(7i/n). (7) SºF ostly clear 7i 3 Write 72 Step 4 of 4 Thus we have the definite integral is approximately ΣπερΔΥ - Σ1(3), = i = 1 i = 1 Submit 91 = Σ(75 - 6977 n = 1 This approximation will become better, and even exact, as n increases without bound √₁²(x³ - 9x) 7i n L'(x³. (x³ - 9x) dx as the limit of a Riemann sum with intervals of equal width and right endpoints as sample points. - 9x) dx = limf(x)Ax = 1 = lim 310 343;³ 63i 773 = 1 n Skip (you cannot come back) with corresponc Search
Step 3 of 4 The approximate Riemann sum of f(x) = x³ - 9x on [0, 7] using n rectangles with right endpoints X; widths. n Σ f(x)Ax = [ f(²1) ² n i=1 i=1 Using f(x) = x³ - 9x, find f(7i/n). (7) SºF ostly clear 7i 3 Write 72 Step 4 of 4 Thus we have the definite integral is approximately ΣπερΔΥ - Σ1(3), = i = 1 i = 1 Submit 91 = Σ(75 - 6977 n = 1 This approximation will become better, and even exact, as n increases without bound √₁²(x³ - 9x) 7i n L'(x³. (x³ - 9x) dx as the limit of a Riemann sum with intervals of equal width and right endpoints as sample points. - 9x) dx = limf(x)Ax = 1 = lim 310 343;³ 63i 773 = 1 n Skip (you cannot come back) with corresponc Search
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Step 3 of 4**
The approximate Riemann sum of \( f(x) = x^3 - 9x \) on \([0, 7]\) using \( n \) rectangles with right endpoints \( x_i \), with corresponding widths.
\[
\sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} f\left(\frac{7i}{n}\right)\frac{7}{n}
\]
Using \( f(x) = x^3 - 9x \), find \( f\left(\frac{7i}{n}\right) \).
\[
f\left(\frac{7i}{n}\right) = \left(\frac{7i}{n}\right)^3 - 9\left(\frac{7i}{n}\right)
\]
The expression simplifies to:
\[
\frac{343i^3}{n^3} - \frac{63i}{n}
\]
**Step 4 of 4**
Thus we have the definite integral is approximately
\[
\sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} \left(\frac{7i}{n}\right)\frac{7}{n}
\]
\[
= \sum_{i=1}^{n} \left(\frac{(7n)^3}{n^3} - \frac{63i}{n}\right)\frac{7}{n}
\]
This approximation will become better, and even exact, as \( n \rightarrow \infty \).
Write
\[
\int_{0}^{7} (x^3 - 9x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x
\]
\[
= \lim_{n \to \infty} \sum_{i=1}^{n}
\]
The diagram shows a calculation of a Riemann sum being approximated by a series of calculations which result in understanding how the function \( f(x) \) behaves as you increase the number of rectangles, \( n \). As \( n \) increases without bound, the approximation becomes more accurate to the actual integral.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe82d4204-c722-4765-8d45-f66ff9ed2d10%2F29f507bb-4e39-45ef-a061-bee4e47ea659%2Fidniwu5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Step 3 of 4**
The approximate Riemann sum of \( f(x) = x^3 - 9x \) on \([0, 7]\) using \( n \) rectangles with right endpoints \( x_i \), with corresponding widths.
\[
\sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} f\left(\frac{7i}{n}\right)\frac{7}{n}
\]
Using \( f(x) = x^3 - 9x \), find \( f\left(\frac{7i}{n}\right) \).
\[
f\left(\frac{7i}{n}\right) = \left(\frac{7i}{n}\right)^3 - 9\left(\frac{7i}{n}\right)
\]
The expression simplifies to:
\[
\frac{343i^3}{n^3} - \frac{63i}{n}
\]
**Step 4 of 4**
Thus we have the definite integral is approximately
\[
\sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} \left(\frac{7i}{n}\right)\frac{7}{n}
\]
\[
= \sum_{i=1}^{n} \left(\frac{(7n)^3}{n^3} - \frac{63i}{n}\right)\frac{7}{n}
\]
This approximation will become better, and even exact, as \( n \rightarrow \infty \).
Write
\[
\int_{0}^{7} (x^3 - 9x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x
\]
\[
= \lim_{n \to \infty} \sum_{i=1}^{n}
\]
The diagram shows a calculation of a Riemann sum being approximated by a series of calculations which result in understanding how the function \( f(x) \) behaves as you increase the number of rectangles, \( n \). As \( n \) increases without bound, the approximation becomes more accurate to the actual integral.
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