Let's address both parts of the question step-by-step.

Part 1: Evaluate the Integral

We are given that $f(x) = k$ and $k = 3.4$. So, $f(x) = 3.4$.

We need to evaluate the definite integral: $\int_{1.3}^{6} f(x) \, dx = \int_{1.3}^{6} 3.4 \, dx$

This is a straightforward integral of a constant function. The formula for integrating a constant $c$ over the interval $[a, b]$ is: $\int_{a}^{b} c \, dx = c \times (b - a)$

Substituting $c = 3.4$, $a = 1.3$, and $b = 6$: $\int_{1.3}^{6} 3.4 \, dx = 3.4 \times (6 - 1.3)$

Calculating the value: $3.4 \times (6 - 1.3) = 3.4 \times 4.7 = 15.98$

So, the value of the integral is: $\int_{1.3}^{6} f(x) \, dx = 15.98$

Part 2: Find $b$ Such That $\int_{1}^{b} f(x) \, dx = 13$

We need to find $b$ such that: $\int_{1}^{b} f(x) \, dx = 13$ Substitute $f(x) = 3.4$: $\int_{1}^{b} 3.4 \, dx = 3.4 \times (b - 1) = 13$

Now solve for $b$: $3.4(b - 1) = 13$ $b - 1 = \frac{13}{3.4}$ $b - 1 = 3.8235$ $b = 3.8235 + 1$ $b = 4.8235$

Rounding to 2 decimal places: $b \approx 4.82$

Summary

1. $\int_{1.3}^{6} f(x) \, dx = 15.98$.
2. $b \approx 4.82$.

Do you have any questions or would you like more details on any part?

Here are 5 related questions for further exploration:

1. How would the integral change if $k$ were different, say 5 instead of 3.4?
2. What does the value of $\int_{1}^{6} f(x) \, dx$ represent geometrically?
3. How would you solve the problem if $f(x)$ were not a constant but a linear function, such as $f(x) = 2x + 1$?
4. What are the properties of the definite integral when the function is negative over the interval?
5. How can we verify our solution using the Fundamental Theorem of Calculus?

Tip: When integrating a constant function over an interval, the result is simply the constant multiplied by the length of the interval.