Setup the Riemann sum = [²³x²³ dx lim n→∞0 [(2+[(3*k)/n])^3]*(3/n) n Entered k=1 [*f(x) dx = f(x) dx = limf(x)Ax for the given integral. k=1 n→∞ (2+3k/n)^3*(3/n) n Answer Preview 3 3k 3 (2 + ³k ) ²2/1/2 n n

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Hey please solve thisc correctly.

(2+3k/n)^3*(3/n) this is not right answer. I want right answer

The image contains a setup for evaluating a definite integral using Riemann sums.

**Entered:**
\[
[(2+[(3*k)/n])^3] \times (3/n)
\]

**Answer Preview:**
\[
\left(2 + \frac{3k}{n}\right)^3 \times \frac{3}{n}
\]

**Instructions:**
(a) Setup the Riemann sum for the given integral:
\[
\int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{k=1}^{n} f(\bar{x}_k) \Delta x
\]

**Example Integral:**
\[
\int_{2}^{5} x^3 \, dx = \lim_{n \to \infty} \sum_{k=1}^{n} \left(2 + 3k/n\right)^3 \times (3/n)
\] 

This explains how to use Riemann sums to approximate the integral from 2 to 5 of the function \(x^3\). The expression inside the sum represents a function value at a sample point within each subinterval, multiplied by the width of the subinterval, \(\Delta x\). This sum approaches the definite integral as the number of subintervals, \(n\), approaches infinity.
Transcribed Image Text:The image contains a setup for evaluating a definite integral using Riemann sums. **Entered:** \[ [(2+[(3*k)/n])^3] \times (3/n) \] **Answer Preview:** \[ \left(2 + \frac{3k}{n}\right)^3 \times \frac{3}{n} \] **Instructions:** (a) Setup the Riemann sum for the given integral: \[ \int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{k=1}^{n} f(\bar{x}_k) \Delta x \] **Example Integral:** \[ \int_{2}^{5} x^3 \, dx = \lim_{n \to \infty} \sum_{k=1}^{n} \left(2 + 3k/n\right)^3 \times (3/n) \] This explains how to use Riemann sums to approximate the integral from 2 to 5 of the function \(x^3\). The expression inside the sum represents a function value at a sample point within each subinterval, multiplied by the width of the subinterval, \(\Delta x\). This sum approaches the definite integral as the number of subintervals, \(n\), approaches infinity.
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