Question

Evaluate each improper integral or show that it diverges. \(\int_{1}^{\infty} \frac{d x}{x^{1.00001}}\)

Short answer

The integral converges to 100,000.

Step-by-step solution

Set up the improper integral

The integral given is \( \int_{1}^{\infty} \frac{dx}{x^{1.00001}} \). Since it is improper, we set it up as \( \lim_{b \to \infty} \int_{1}^{b} \frac{dx}{x^{1.00001}} \).

Integrate the function

We need to integrate \( \frac{1}{x^{1.00001}} \). Applying the power rule for integration, we get \( \int x^{-1.00001} dx = \frac{x^{-0.00001}}{-0.00001} + C \).

Evaluate the definite integral

Now substitute the limits of integration into the antiderivative: \[ \lim_{b \to \infty} \left( -100000 x^{-0.00001} \right) \Bigg|_{1}^{b} = \lim_{b \to \infty} \left( -100000 b^{-0.00001} + 100000 \right). \]

Evaluate the limit

Evaluate the limit \( \lim_{b \to \infty} \left( -100000 b^{-0.00001} + 100000 \right) \). As \( b \) approaches infinity, \( b^{-0.00001} \) approaches 0, making the term \( -100000 b^{-0.00001} \) approach 0. Thus, the limit is equal to \( 100000 \).

Conclude the result

Since the limit evaluated to \( 100000 \), the integral converges to \( 100000 \).

Key concepts

Convergence

When dealing with improper integrals, a key concept is convergence. It's like checking whether a process leads to a result or keeps going without limit. If an integral "converges," it adds up to a finite number. Otherwise, it's said to "diverge," implying it doesn't settle to a sum. For instance, the improper integral given in the exercise, \( \int_{1}^{\infty} \frac{dx}{x^{1.00001}} \), must be checked for convergence. We evaluate this by analyzing its limit as the upper boundary approaches infinity. If the result is finite, then the integral converges. In this particular problem, despite the infinite range, the integral converged to a limit of 100,000. This concept helps understand more complicated functions because it shows whether they accumulate to something meaningful or simply blow up to infinity.

Integration Technique

Integration techniques are strategies to find the antiderivative of a function. For this exercise, the power rule is the chosen method due to the function’s form, \( \frac{1}{x^{1.00001}} \). The power rule states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) where \( n eq -1 \). This method simplifies the integration process by transforming the original function into an easier expression to evaluate.
By applying the power rule to \( x^{-1.00001} \), we rewrite the integral as \( \int x^{-1.00001} dx = \frac{x^{-0.00001}}{-0.00001} + C \).
This gives us a new, more manageable expression to evaluate through the fundamental theorem of calculus and our limits of integration. Using established techniques like these not only helps in solving specific problems but also builds deeper understanding of how integrals function in broader contexts.

Definite Integrals

A definite integral computes the accumulated value over an interval. In this problem, the focus is on integrating from 1 to infinity. Definite integrals have boundaries, called limits of integration, which are 1 and infinity here.
The challenge arises because infinity is not a number but rather a concept indicating limitless growth. To deal with this, we modify the integral from \( 1 \) to \( b \) and then take the limit as \( b \) approaches infinity. This transition turns it essentially into a problem about limits and is how definite integrals calculate actual quantities.
Here, the definite integral \( \int_{1}^{b} \frac{1}{x^{1.00001}} dx \) simplifies into \(-100000 x^{-0.00001} \) evaluated at the upper and lower limits, and further evaluated as \( \lim_{b \to \infty} (-100000 b^{-0.00001} + 100000) \). The result was \( 100000 \), indicating that this particular range and function sum up neatly, achieving a precise value, despite spanning towards an infinite limit.