We are given the function $f(x) = k$, where $k = 3.9$, and we need to evaluate two integral problems:

1) Evaluate $\int_{1.9}^{6.4} f(x) \, dx$

Since $f(x) = k = 3.9$, the integral is simply the area of a rectangle with height 3.9 and base equal to the length of the interval $[1.9, 6.4]$.

The formula for this definite integral is: $\int_{1.9}^{6.4} f(x) \, dx = k \times (6.4 - 1.9)$ Substitute $k = 3.9$: $3.9 \times (6.4 - 1.9) = 3.9 \times 4.5 = 17.55$

Answer: $17.55$

2) Find the value of $b$ such that $\int_2^b f(x) \, dx = 23$

For this part, we need to solve for $b$ such that the area under the curve from $x = 2$ to $x = b$ equals 23. Since $f(x) = 3.9$, the integral is:

$\int_2^b f(x) \, dx = 3.9 \times (b - 2)$ We are given that this equals 23, so we set up the equation: $3.9 \times (b - 2) = 23$ Solve for $b$: $b - 2 = \frac{23}{3.9}$ $b - 2 = 5.90$ $b = 5.90 + 2 = 7.90$

Answer: $b = 7.90$

Would you like more details on any of these steps?

Here are 5 related questions to deepen your understanding:

1. How do you calculate the area under a constant function over an interval?
2. How would the integral change if $f(x)$ were not constant but a linear function?
3. What is the geometric interpretation of definite integrals for different types of functions?
4. How do you solve for the upper limit of integration in cases where the integral value is known?
5. How do you generalize this approach to integrals with variable limits?

Tip: When dealing with constant functions, integrals simply compute the area of rectangles, which is the function value (height) multiplied by the width of the interval!