To find the definite integral x³dx by the limit definition, divide the interval [-6, 6] into n subintervals. Then the width of each interval is b-a Ax = = 6 - (-6) 6+6 Note that ||A|| → 0 x³dx = n = = n Step 2 Choose c; as the right endpoint of each subinterval. Then C₁ = a + i(Ax) So the definite integral is given by = n→ 12 0 as n→ ∞0. n lim Σ f(c₁) Ax; ||A||→ 07 i = 1 lim n→ ∞ n n→ lim n→ ∞o = lim n→∞o n→ lim n→∞0 n -£(- i=1 lim i = 1 i = 1 -6 + lim + -216 + -2592 + lim + 12) (1²2) / = 1 n Σf(c) Axi i = 1 + ;2+ 2+ n n -2592 + 20736 n n n n² £³] 7 = 1 ](¹+ ²¹² ) 1 ¹¹) (¹²) n Σ i=1 + 17/1/2) + n³ 1 + n n ++/-)]
To find the definite integral x³dx by the limit definition, divide the interval [-6, 6] into n subintervals. Then the width of each interval is b-a Ax = = 6 - (-6) 6+6 Note that ||A|| → 0 x³dx = n = = n Step 2 Choose c; as the right endpoint of each subinterval. Then C₁ = a + i(Ax) So the definite integral is given by = n→ 12 0 as n→ ∞0. n lim Σ f(c₁) Ax; ||A||→ 07 i = 1 lim n→ ∞ n n→ lim n→ ∞o = lim n→∞o n→ lim n→∞0 n -£(- i=1 lim i = 1 i = 1 -6 + lim + -216 + -2592 + lim + 12) (1²2) / = 1 n Σf(c) Axi i = 1 + ;2+ 2+ n n -2592 + 20736 n n n n² £³] 7 = 1 ](¹+ ²¹² ) 1 ¹¹) (¹²) n Σ i=1 + 17/1/2) + n³ 1 + n n ++/-)]
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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![**Step 1**
To find the definite integral \(\int_{-6}^{6} x^3 \, dx\) by the limit definition, divide the interval \([-6, 6]\) into \(n\) subintervals.
The width of each interval is
\[
\Delta x = \frac{b-a}{n} = \frac{6 - (-6)}{n} = \frac{6 + 6}{n} = \frac{12}{n}
\]
Note that \(\|\Delta\| \to 0\) as \(n \to \infty\).
**Step 2**
Choose \(c_i\) as the right endpoint of each subinterval. Then
\[
c_i = a + i(\Delta x) = -6 + \frac{12i}{n}
\]
So the definite integral is given by
\[
\int_{-6}^{6} x^3 \, dx = \lim_{\|\Delta\| \to 0} \sum_{i=1}^{n} f(c_i) \Delta x_i = \lim_{n \to \infty} \sum_{i=1}^{n} f(c_i) \Delta x_i
\]
\[
= \lim_{n \to \infty} \sum_{i=1}^{n} \left(-6 + \frac{12i}{n}\right)^3 \left(\frac{12}{n}\right)
\]
\[
= \lim_{n \to \infty} \sum_{i=1}^{n} \left(-6 + \frac{12i}{n}\right)^3 \left(\frac{12}{n}\right)
\]
\[
= \lim_{n \to \infty} \sum_{i=1}^{n} \left(-216 + \frac{1296i}{n} - \frac{432i^2}{n^2} + \frac{64i^3}{n^3}\right)\left(\frac{12}{n}\right)
\]
\[
= \lim_{n \to \infty} \left[-2592 + \frac{15552}{n^2}\sum_{i=1}^{n} i - \frac{5184}{n^3}\sum_{i=1}^{n} i](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fab6702de-5579-4c14-bacb-4d8e7e421acd%2F40c0e76f-a7f3-4dc8-9344-3ded577a4df5%2F623dd3i_processed.png&w=3840&q=75)
Transcribed Image Text:**Step 1**
To find the definite integral \(\int_{-6}^{6} x^3 \, dx\) by the limit definition, divide the interval \([-6, 6]\) into \(n\) subintervals.
The width of each interval is
\[
\Delta x = \frac{b-a}{n} = \frac{6 - (-6)}{n} = \frac{6 + 6}{n} = \frac{12}{n}
\]
Note that \(\|\Delta\| \to 0\) as \(n \to \infty\).
**Step 2**
Choose \(c_i\) as the right endpoint of each subinterval. Then
\[
c_i = a + i(\Delta x) = -6 + \frac{12i}{n}
\]
So the definite integral is given by
\[
\int_{-6}^{6} x^3 \, dx = \lim_{\|\Delta\| \to 0} \sum_{i=1}^{n} f(c_i) \Delta x_i = \lim_{n \to \infty} \sum_{i=1}^{n} f(c_i) \Delta x_i
\]
\[
= \lim_{n \to \infty} \sum_{i=1}^{n} \left(-6 + \frac{12i}{n}\right)^3 \left(\frac{12}{n}\right)
\]
\[
= \lim_{n \to \infty} \sum_{i=1}^{n} \left(-6 + \frac{12i}{n}\right)^3 \left(\frac{12}{n}\right)
\]
\[
= \lim_{n \to \infty} \sum_{i=1}^{n} \left(-216 + \frac{1296i}{n} - \frac{432i^2}{n^2} + \frac{64i^3}{n^3}\right)\left(\frac{12}{n}\right)
\]
\[
= \lim_{n \to \infty} \left[-2592 + \frac{15552}{n^2}\sum_{i=1}^{n} i - \frac{5184}{n^3}\sum_{i=1}^{n} i
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