Setup the Riemann sum n ['so k=1 f(x) dx : = •5 x³ dx = lim ((2+3/nk)^3)*(3/n) n→∞ n lim f(x)Ax for the given integral. n→∞ k=1

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**Setup the Riemann Sum for the Given Integral:**

Given the integral:

\[
\int_{a}^{b} f(x) \, dx = \lim_{{n \to \infty}} \sum_{{k=1}}^{n} f(\bar{x}_k) \Delta x
\]

This is the setup for the given integral.

**Evaluate the Integral:**

\[
\int_{2}^{5} x^3 \, dx = \lim_{{n \to \infty}} \sum_{{k=1}}^{n} \left(\left(2 + \frac{3}{n}k\right)^3 \cdot \frac{3}{n}\right)
\]

The expression inside the sum represents the Riemann sum approximation for the integral of \(x^3\) from 2 to 5. Here, \(\Delta x = \frac{3}{n}\), and the sample point \(\bar{x}_k = 2 + \frac{3}{n}k\).
Transcribed Image Text:**Setup the Riemann Sum for the Given Integral:** Given the integral: \[ \int_{a}^{b} f(x) \, dx = \lim_{{n \to \infty}} \sum_{{k=1}}^{n} f(\bar{x}_k) \Delta x \] This is the setup for the given integral. **Evaluate the Integral:** \[ \int_{2}^{5} x^3 \, dx = \lim_{{n \to \infty}} \sum_{{k=1}}^{n} \left(\left(2 + \frac{3}{n}k\right)^3 \cdot \frac{3}{n}\right) \] The expression inside the sum represents the Riemann sum approximation for the integral of \(x^3\) from 2 to 5. Here, \(\Delta x = \frac{3}{n}\), and the sample point \(\bar{x}_k = 2 + \frac{3}{n}k\).
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