Question

Determine whether the function \(f(x)=\frac{1}{x-4}\) is integrable on the interval [3,5] . Explain.

Short answer

The function \(f(x)=\frac{1}{x-4}\) is not integrable on the interval [3,5] because it has a vertical asymptote at x = 4, which makes the function unbounded within the said interval.

Step-by-step solution

Identify the Function and the Interval

The function given is \(f(x)=\frac{1}{x-4}\) and it is to be analyzed on the interval [3, 5].

Check for Boundness

The function has a vertical asymptote at x = 4. Therefore, the function will not be bounded on the interval [3, 5]. An infinite discontinuity, as the one we have at x = 4, does not allow for the function to be bounded.

Determine Integrability

For a function to be integrable on interval [a, b], it must be bounded on [a, b]. The function, \(f(x)=\frac{1}{x-4}\), is not bounded on the interval [3, 5] due to the asymptote at x = 4. Therefore, the function is not integrable on the interval [3, 5].

Key concepts

Vertical Asymptote

In calculus, a vertical asymptote refers to a line that a function's graph approaches but never actually touches or crosses. It essentially represents a boundary where the function's value becomes infinitely large or decreases infinitely without bound. For the function in question, let's explore the vertical asymptote of \( f(x)=\frac{1}{x-4} \).

As we observe the function, it is clear that as the variable \(x\) approaches 4, the denominator \(x-4\) approaches zero, which makes the value of the function grow without limit. This not only affects the shape of the graph but also has implications for the function’s integrability. A common misconception is that a vertical asymptote always exists at a certain \(x\)-value; however, it is the behavior of the function near that \(x\)-value that determines its presence.

Bounded Function

A bounded function is one that has both an upper and a lower limit within which all its values lie over a certain interval. Think of it as a function that remains contained within an invisible box, never stepping outside its limits as it moves along the \(x\)-axis. For the function \(f(x)=\frac{1}{x-4}\), the situation becomes interesting.

When considering boundedness on the interval [3, 5], we must check if there are any values for \(x\) that would cause the function to exceed any limits. In our case, we find that at \(x = 4\), the denominator of the function becomes zero, and hence the function's value surges towards infinity. This singular point prevents the function from being considered bounded, since the 'invisible box' would have to be infinitely tall to contain the function's value at \(x = 4\).

Infinite Discontinuity

An infinite discontinuity occurs at a point on a graph where the function literally 'jumps' to infinity. With the function we are considering, \(f(x)=\frac{1}{x-4}\), we see exactly this happening as \(x\) nears 4. The 'jump' to infinity is what creates the vertical asymptote we discussed earlier.

Discontinuities, especially infinite ones, are critical when we discuss the integrability of a function. They signify that the function is not defined at that particular point and thus, the area under the curve cannot be calculated in the standard way using definite integrals. This is because calculating area involves summation of infinitely many infinitesimally small values, and when the function is unbounded, this process does not converge to a finite number. As such, the presence of an infinite discontinuity in the interval of integration is a key factor to consider, and in this case, is the reason why our function is not integrable on the interval [3, 5].