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Chapter 5: Problem 21

Determine whether each improper integral is convergent or divergent, and calculate its value if it is convergent. $$\int_{\pi}^{\infty} \frac{d t}{t^{1.001}}$$

Short Answer

Expert verified
The integral diverges.

Step by step solution

01

- Understand the improper integral

Recognize that the integral \( \int_{\pi}^{\infty} \frac{d t}{t^{1.001}} \) is an improper integral because it has an infinite upper limit.
02

- Set up the improper integral

Set up the integral as a limit to manage the infinity properly: \[ \int_{\pi}^{\infty} \frac{d t}{t^{1.001}} = \lim_{{b \to \infty}} \int_{\pi}^{b} \frac{d t}{t^{1.001}}. \]
03

- Integrate the function

Find the antiderivative of \( \frac{1}{t^{1.001}} \). This can be rewritten as \( t^{-1.001} \). The antiderivative is \[ \int t^{-1.001} d t = \frac{t^{-0.001}}{-0.001} = -1000 t^{-0.001} = -1000 t^{0.001}. \]
04

- Evaluate the definite integral

Evaluate the definite integral from \( \pi \) to \( b \): \[ \int_{\pi}^{b} \frac{d t}{t^{1.001}} = \left[ -1000 t^{0.001} \right]_{\pi}^{b} = -1000 b^{0.001} + 1000 \pi^{0.001}. \]
05

- Take the limit

Take the limit as \( b \) approaches infinity: \[ \lim_{{b \to \infty}} \left( -1000 b^{0.001} + 1000 \pi^{0.001} \right). \]
06

- Determine convergence or divergence

Evaluate the limit. As \( b \to \infty \, b^{0.001} \) approaches infinity (since \( 0.001 > 0 \)), hence \( -1000 b^{0.001} \to -\infty \). Therefore, the integral diverges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convergent Integrals
A convergent integral is an integral that approaches a finite value as the limit of integration extends to infinity or a singularity. When dealing with improper integrals, determining whether an integral is convergent is crucial. For instance, if the integral of a function over a specified range converges, it equals a real number:
  • Example: \( \int_{a}^{b} f(x) dx \) where \( a \) and \( b \) are finite
  • If extending to infinity: \( \lim_{b\to\infty} \int_{a}^{b} f(x) dx \)
Divergent Integrals
A divergent integral does not approach a finite value as the limit of integration extends to infinity or a singularity; instead, it approaches infinity. For improper integrals, recognizing divergence is important to avoid erroneous conclusions:
  • Example: \$ \[ \int_{1}^{\infty} \frac{1}{x} dx \rightarrow \infty \]
  • If the integral grows infinitely, it diverges: \$ \[ \lim_{b\to\infty} \int_{a}^{b} f(x) dx \rightarrow \infty \]
Limits in Calculus
Limits are fundamental in calculus to handle expressions involving infinity or approaching a specific value. They are crucial for solving improper integrals:
  • To evaluate, replace infinity with a variable approaching infinity, such as: \$ \[ \lim_{b\to\infty} \int_{a}^{b} f(x) dx \]
  • Use algebraic manipulation to find the limit as \( b \) approaches infinity
Integrals are often solved by breaking down the problem with limits, especially improper integrals.
Antiderivatives
Finding the antiderivative, or primitive, is the process of determining a function whose derivative is the given function. This is essential in evaluating integrals:
  • For \( f(x) = \frac{1}{x^{1.001}} \), rewriting it as \( x^{-1.001} \) simplifies finding the antiderivative
  • The antiderivative of \( x^{-1.001} \) is \[ \int x^{-1.001} dx = \-1000 x^{0.001} \]
The antiderivative helps in applying the Fundamental Theorem of Calculus to evaluate definite integrals.
Definite Integrals
Definite integrals compute the area under the curve within a specified range of integration. They are evaluated using limits and antiderivatives:
  • Example: \$ \[ \int_{a}^{b} f(x) dx \]
  • For \(\frac{1}{t^{1.001}} \), using the antiderivative \-1000 x^{0.001}, evaluate from \( \pi \) to \( b \)
  • Apply the Fundamental Theorem of Calculus:
  • \

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