The interval `[0,4]` isdivided into n equal sub-intervals by the points `x_(0),x_(1),x_(2),…,x_(n-1),x_(n)"where" 0=x_(0)ltx_(1)ltx_(2)ltx_(3)lt…ltx_(n)=4`
If`deltax=x_(i)-x_(i-1)"for"i=1,2,3,...,n, "then"lim_(deltaxrarr0)sum_(i-1)^(n)x_(i)deltax` is equal to

Options

A. (a)4
B. (b)8
C. (c)`(32)/(3)`
D. (d)16

Answer

To solve the problem, we need to evaluate the limit of the Riemann sum as \( \Delta x \) approaches 0. The interval \([0, 4]\) is divided into \( n \) equal sub-intervals, and we can express the points as follows:

1. **Define the sub-intervals**:
- The length of each sub-interval is given by:
\[ \Delta x = \frac{b - a}{n} = \frac{4 - 0}{n} = \frac{4}{n} \]
- The points dividing the interval can be expressed as:
\[ x_i = 0 + i \Delta x = \frac{4i}{n} \quad \text{for } i = 0, 1, 2, \ldots, n \]

2. **Set up the Riemann sum**:
- The Riemann sum we need to evaluate is:
\[ \lim_{\Delta x \to 0} \sum_{i=1}^{n} x_i \Delta x \]
- Substituting \( x_i \) and \( \Delta x \):
\[ \sum_{i=1}^{n} x_i \Delta x = \sum_{i=1}^{n} \left(\frac{4i}{n}\right) \left(\frac{4}{n}\right) = \frac{16}{n^2} \sum_{i=1}^{n} i \]

3. **Evaluate the sum of the first \( n \) natural numbers**:
- The formula for the sum of the first \( n \) natural numbers is:
\[ \sum_{i=1}^{n} i = \frac{n(n + 1)}{2} \]
- Substituting this into our expression:
\[ \frac{16}{n^2} \cdot \frac{n(n + 1)}{2} = \frac{16(n + 1)}{2n} = \frac{8(n + 1)}{n} \]

4. **Take the limit as \( n \to \infty \)**:
- As \( n \) approaches infinity, \( \frac{8(n + 1)}{n} \) simplifies to:
\[ \lim_{n \to \infty} \frac{8(n + 1)}{n} = \lim_{n \to \infty} 8 \left(1 + \frac{1}{n}\right) = 8 \cdot 1 = 8 \]

5. **Conclusion**:
- Therefore, the limit is:
\[ \lim_{\Delta x \to 0} \sum_{i=1}^{n} x_i \Delta x = 8 \]

### Final Answer:
The limit is equal to \( 8 \).