/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Calculate (a) \(\int_{1}^{n} k 2... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 21

Calculate (a) \(\int_{1}^{n} k 2^{-k} d k\), (b) \(\lim _{n \rightarrow \infty} \int_{1}^{n} k 2^{-k} d k\), and (c) \(\int_{1}^{\infty} k 2^{-k} d k\).

Short Answer

Expert verified
(a) \( \frac{1}{\ln(2)} \left( -n 2^{-n} + \frac{1}{2} \right) + \frac{1}{\ln^2(2)} \left( -2^{-n} + \frac{1}{2} \right)\), (b) \( \frac{1}{2 \ln(2)} + \frac{1}{2 \ln^2(2)} \), (c) \( \frac{1}{2 \ln(2)} + \frac{1}{2 \ln^2(2)} \).

Step by step solution

01

- Integrate using integration by parts (Part (a))

Use integration by parts where \[u = k\] and \[dv = 2^{-k} dk\]. First, find \(du = dk\) and \(v\) by integrating \(dv\). Let's integrate \(2^{-k}\) with respect to \(k\). \[v = \int 2^{-k} dk = -\frac{2^{-k}}{\ln(2)}.\] Apply the integration by parts formula: \(\int u dv = uv - \int v du\). \[\int_{1}^{n} k 2^{-k} d k = \left[ -k \frac{2^{-k}}{\ln(2)} \right]_1^n + \int_{1}^{n} \frac{2^{-k}}{\ln(2)} d k\].
02

- Evaluate the boundary terms and the second integral (Part (a))

Evaluate the boundary term: \(\left[ -k \frac{2^{-k}}{\ln(2)} \right]_1^n = -\frac{n 2^{-n}}{\ln(2)} + \frac{2^{-1}}{\ln(2)}\). Now, solve the remaining integral \(\int_{1}^{n} \frac{2^{-k}}{\ln(2)} d k\). First take the constant \(\frac{1}{\ln(2)}\) outside: \[\int_{1}^{n} \frac{2^{-k}}{\ln(2)} d k = \frac{1}{\ln(2)} \left[ - \frac{2^{-k}}{\ln(2)} \right]_1^n = \frac{1}{\ln^2(2)}\left( -2^{-n} + 2^{-1} \right).\]
03

- Combine the results (Part (a))

Combine the results from Step 1 and Step 2: \[\int_{1}^{n} k 2^{-k} d k = -\frac{n 2^{-n}}{\ln(2)} + \frac{2^{-1}}{\ln(2)} + \frac{1}{\ln^2(2)} \left( -2^{-n} + 2^{-1} \right).\] Simplify the terms: \[ \int_{1}^{n} k 2^{-k} d k = \frac{1}{\ln(2)} \left( -n 2^{-n} + \frac{1}{2} \right) + \frac{1}{\ln^2(2)} \left(-2^{-n} + \frac{1}{2} \right).\]
04

- Take the limit as n approaches infinity (Part (b))

Consider the limit as \(n\) approaches infinity of the integral: \[\lim_{n \to \infty} \int_{1}^{n} k 2^{-k} d k.\] As \(n\) approaches infinity, \(n 2^{-n} \rightarrow 0 \) and \(2^{-n} \rightarrow 0\). Thus, the integral becomes: \[\lim_{n \to \infty} \left( \frac{1}{\ln(2)} \left( 0 + \frac{1}{2} \right) + \frac{1}{\ln^2(2)} \left( 0 + \frac{1}{2} \right) \right). \] Simplify: \[\lim_{n \to \infty} \int_{1}^{n} k 2^{-k} d k = \frac{1}{2 \ln(2)} + \frac{1}{2 \ln^2(2)}.\]
05

- Evaluate the integral from 1 to infinity (Part (c))

The integral from 1 to infinity is already evaluated in Step 4 when taking the limit. Hence, \[\int_{1}^{\infty} k 2^{-k} d k = \lim_{n \to \infty} \int_{1}^{n} k 2^{-k} d k.\] From Step 4, the answer is: \[ \int_{1}^{\infty} k 2^{-k} d k = \frac{1}{2 \ln(2)} + \frac{1}{2 \ln^2(2)}. \]

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Key Concepts

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

definite integrals
Definite integrals calculate the net area under a curve within a specified interval, usually between two points. In this exercise, we calculate the definite integral \(\int_{1}^{n} k 2^{-k} d k\)\. It鈥檚 important to understand the bounds (1 to \(\)) define where we are evaluating the area. We integrate between these limits using techniques like integration by parts. The definite integral here demonstrates how to calculate a complex integral with finite limits.
limits in calculus
Limits help us understand the behavior of functions as an input approaches a particular value. In part (b) of the exercise, we use the limit \(\lim_{n \rightarrow \infty} \int_{1}^{n} k 2^{-k} d k\)\ to evaluate the definite integral as \(\) approaches infinity. Limits are crucial in determining the convergence of integrals and series, especially to check if a function approaches a particular value or tends to infinity.
improper integrals
Improper integrals extend definite integrals to include infinite limits or integrands with undefined values within the integration range. This exercise looks at \(\int_{1}^{\infty} k 2^{-k} d k,\)\ which is an improper integral because it integrates to infinity. By taking the limit as \(\) approaches infinity, we determine the integral's convergence. The value of the improper integral helps us understand areas under curves that stretch out indefinitely.
logarithmic integration
Logarithmic integration involves integrating functions that result in logarithmic expressions. In solving integrals such as \(\int 2^{-k} dk\),\ we use the properties of logarithms. The antiderivative of \(\2^{-k.}\)\ involves the natural logarithm's base property, specifically \(\frac{-2^{-k}}{\ln(2).}\)\ This is crucial in different parts of the exercise, demonstrating the importance of understanding how logarithms play a role in integration.

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Most popular questions from this chapter

\- Let \(f(x)=m x+b\) and \(\left(x_{0}, y_{0}\right)\) be a point not on the graph of \(f(x)\). Find the point on the graph of \(f(x)\) that is closest to \(\left(x_{0}, y_{0}\right)\).

Evaluate \(\sum_{n=1}^{\infty} \frac{3^{n / 2}}{5^{n}}\). Confirm your result by showing that the series converges and finding its sum by hand.

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