To express the integral $\int_3^5 \sqrt{x} \, dx$ as a limit, we can utilize the definition of a definite integral in terms of Riemann sums.

The integral can be written as a limit of a Riemann sum like this:

$\int_3^5 \sqrt{x} \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} \sqrt{x_i} \Delta x$

Where:

• $n$ is the number of subintervals (divisions) of the interval $[3, 5]$,
• $\Delta x = \frac{5 - 3}{n} = \frac{2}{n}$ is the width of each subinterval,
• $x_i$ is a sample point in the $i$-th subinterval, which could be chosen as the right endpoint, left endpoint, or midpoint. For simplicity, we'll choose the right endpoint: $x_i = 3 + i \Delta x = 3 + i \frac{2}{n}$

Thus, the Riemann sum becomes:

$\sum_{i=1}^{n} \sqrt{3 + i \frac{2}{n}} \cdot \frac{2}{n}$

Now, the definite integral is given by the limit as $n$ approaches infinity:

$\int_3^5 \sqrt{x} \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} \sqrt{3 + i \frac{2}{n}} \cdot \frac{2}{n}$

This is the desired expression of the integral as a limit.

Would you like more details or have any questions?

Here are 5 related questions:

1. How can the choice of sample points affect the Riemann sum?
2. How do you compute the definite integral of $\sqrt{x}$ directly?
3. How does the definition of a definite integral relate to the concept of area?
4. What is the difference between left Riemann sums, right Riemann sums, and midpoint Riemann sums?
5. How can numerical integration techniques approximate definite integrals?

Tip: When using Riemann sums, the more subintervals you use (larger $n$), the more accurate your approximation of the integral becomes.