Express the following definite integrals as limits of left and right Riemann sums. 1. ∫ from 0 to 4 of 3 dx 2. ∫ from 0 to 2 of x dx 3. ∫ from 0 to 1 of √(1 - x²) dx.

Steps to Solve

1. Defining Riemann Sums for Each Integral For each definite integral, define the interval ([a, b]), partition it into (n) equal subintervals, and express the width (w) of each subinterval as:

$$ w = \frac{b - a}{n} $$

2. Integral 1: ( \int_0^4 3 , dx ) The interval is ([0, 4]).

• Left endpoint approach: Set (x_k = 0 + k \cdot w) for (k=0, 1, 2, \ldots, n-1).

The Riemann sum becomes:

$$ \lim_{n \to \infty} \sum_{k=0}^{n-1} 3 \cdot w = \lim_{n \to \infty} 3 \cdot \frac{4-0}{n} \cdot n = 12 $$

3. Integral 2: ( \int_0^2 x , dx ) The interval is ([0, 2]).

• Left endpoint approach: Set (x_k = 0 + k \cdot w) for (k=0, 1, 2, \ldots, n-1).

The Riemann sum becomes:

$$ \lim_{n \to \infty} \sum_{k=0}^{n-1} x_k \cdot w = \lim_{n \to \infty} \sum_{k=0}^{n-1} \left(\frac{2k}{n}\right) \cdot \frac{2}{n} = \lim_{n \to \infty} \frac{2}{n^2} \sum_{k=0}^{n-1} 2k = \frac{2}{n^2} \cdot n(n-1) $$

4. Integral 3: ( \int_0^1 \sqrt{1 - x^2} , dx ) The interval is ([0, 1]).

• Left endpoint approach: Use (x_k = 0 + k \cdot w) for (k=0, 1, \ldots, n-1).

The Riemann sum becomes:

$$ \lim_{n \to \infty} \sum_{k=0}^{n-1} \sqrt{1 - \left(\frac{k}{n}\right)^2} \cdot \frac{1}{n} $$

5. Integral Limits Presentation Present each limit as indicated:

For the first integral:

$$ \lim_{n \to \infty} 3 \cdot 4 = 12 $$

For the second integral:

$$ \lim_{n \to \infty} \frac{1}{2n^2} (n^2 - n) = 1 $$

For the third integral:

$$ \lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n-1} \sqrt{1 - \left(\frac{k}{n}\right)^2} $$