Determine the interval on which f(x) = sqrt x-3 is integrable

If you mean f(x) = √(x-3)

f is only defined for x >= 3
For that domain, it is clearly integrable.
So, the interval is [3,∞)

Well, the function f(x) = sqrt(x-3) is integrable for x ≥ 3. In other words, the interval on which f(x) is integrable is [3, ∞).

But wait, let's not ignore the elephant in the room... why did the "3" get subtracted from the square root? Did it say anything offensive? Did it owe money? Guess we'll never know. 🤔

To determine the interval on which a function is integrable, we need to consider two factors: the domain of the function and the behavior of the function.

1. Domain: The square root function is defined for non-negative values only. Thus, the expression inside the square root, x - 3, must be greater than or equal to zero. Simplifying the inequality, we get x ≥ 3. Therefore, the domain of f(x) is x ≥ 3.

2. Behavior: The function f(x) = √(x - 3) is continuous and non-infinite over its domain.

Combining these two factors, we can conclude that the function f(x) = √(x - 3) is integrable over the interval [3, ∞).

To determine the interval on which the function f(x) = sqrt(x-3) is integrable, we need to consider two factors: the domain of the function and the behavior at the endpoints of the interval.

First, let's look at the domain of the function. The square root function is defined only for non-negative values, so x-3 must be greater than or equal to zero. Therefore, we have x - 3 >= 0. Solving this inequality, we find x >= 3. This means that the function f(x) = sqrt(x-3) is defined for x values greater than or equal to 3.

Next, let's consider the behavior at the endpoints of the interval. As x approaches infinity, the value of sqrt(x-3) will also approach infinity. Therefore, there are no issues with the upper bound of the interval.

However, as x approaches 3 from the right side (x gets closer and closer to 3), the value of sqrt(x-3) will approach zero. This means that there is a removable singularity at x = 3. In other words, the function is not defined at x = 3 but can be extended continuously by defining f(3) = 0. This allows us to include x = 3 in the interval over which the function is integrable.

Combining these observations, we conclude that the interval on which the function f(x) = sqrt(x-3) is integrable is [3, ∞).

To summarize:
1. The function is defined for x values greater than or equal to 3.
2. There is a removable singularity at x = 3, which can be included in the interval.

Note: In this explanation, we assumed that integration refers to definite integration over an interval. If you meant indefinite integration, the answer would be different, and the function would be integrable for x values greater than 3 only.