To solve the given integral using Riemann sums as specified in the problem, let’s carefully break down the task:

Problem Recap:

We need to evaluate the definite integral $\int_{-1}^3 (3 - 2x) \, dx$ using the limit definition of the integral through Riemann sums. This involves dividing the interval into $n$ subintervals and applying the summation formula. Here's the step-by-step approach:

Step 1: Riemann Sum Formula

The integral is defined as the limit of Riemann sums: $\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x$ where:

• $a = -1$, $b = 3$ (the limits of integration),
• $f(x) = 3 - 2x$,
• $\Delta x = \frac{b-a}{n} = \frac{3 - (-1)}{n} = \frac{4}{n}$,
• $x_i^*$ is the representative point in the $i$-th subinterval.

Step 2: Choose $x_i^*$ (Right Endpoint)

For simplicity, let $x_i^*$ be the right endpoint of each subinterval: $x_i^* = a + i \Delta x = -1 + i \frac{4}{n}.$

Step 3: Substitute into the Riemann Sum

The Riemann sum becomes: $\sum_{i=1}^n f(x_i^*) \Delta x = \sum_{i=1}^n \left[3 - 2\left(-1 + i \frac{4}{n}\right)\right] \cdot \frac{4}{n}.$ Simplify $f(x_i^*)$: $f(x_i^*) = 3 - 2\left(-1 + i \frac{4}{n}\right) = 3 + 2 - \frac{8i}{n} = 5 - \frac{8i}{n}.$ Thus, the Riemann sum becomes: $\sum_{i=1}^n \left(5 - \frac{8i}{n}\right) \cdot \frac{4}{n}.$ Distribute $\frac{4}{n}$: $\sum_{i=1}^n \left(20 - \frac{32i}{n}\right) \cdot \frac{1}{n}.$

Step 4: Split the Sum

Separate the terms in the summation: $\sum_{i=1}^n \left(20 - \frac{32i}{n}\right) \cdot \frac{1}{n} = \frac{4}{n} \sum_{i=1}^n 5 - \frac{32}{n^2} \sum_{i=1}^n i.$

1. The sum of a constant $c$: $\sum_{i=1}^n c = c \cdot n.$ So: $\frac{4}{n} \sum_{i=1}^n 5 = 20.$

2. The sum of $i$: $\sum_{i=1}^n i = \frac{n(n+1)}{2}.$ So: $\frac{32}{n^2} \sum_{i=1}^n i = \frac{32}{n^2} \cdot \frac{n(n+1)}{2} = \frac{16(n+1)}{n}.$

Step 5: Take the Limit as $n \to \infty$

Combine the results: $\sum_{i=1}^n \left(5 - \frac{8i}{n}\right) \cdot \frac{4}{n} = 20 - \frac{16(n+1)}{n}.$ As $n \to \infty$: $\frac{16(n+1)}{n} = \frac{16n + 16}{n} = 16 + \frac{16}{n} \to 16.$ Thus: $20 - 16 = 4.$

Final Answer:

$\int_{-1}^3 (3 - 2x) \, dx = 4.$

Would you like further clarification on any step or additional details?

Related Questions:

1. How do Riemann sums differ when using left endpoints versus right endpoints?
2. How would the summation change if we used midpoint Riemann sums?
3. Can we derive the same result using the trapezoidal rule for this integral?
4. How does the definite integral geometrically represent the area under the curve?
5. How do we evaluate improper integrals using similar limit techniques?

Tip: When working with Riemann sums, always simplify $f(x_i^*)$ before substituting it into the summation—it saves time and avoids algebraic errors!