/*! 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 45 Evaluate the definite integral. ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 45

Evaluate the definite integral. $$ \int_{0}^{5} \frac{x}{(4+x)^{2}} d x $$

Short Answer

Expert verified
Your short answer is \(5 + 4*ln(\frac{4}{9})\)

Step by step solution

01

Identify the substitution

Identify a suitable substitution. Notice the term \((4+x)^{2}\) in the denominator - it would make sense to use this for a substitute. Let \(u = (4+x)\). Then, calculate \(du\) which would be \(du = dx\)
02

Change the limits of the integration

Substitute the limits of integration from x-values to u-values. When \(x = 0\), \(u = (4+0) = 4\), and when \(x = 5\), \(u = (4+5) = 9\)
03

Substitute x and dx in the integral formula

Express \(x\) in terms of \(u\) and substitute this as well as \(dx\) in the integral. From \(u = 4 + x\), you can express \(x = u - 4\). Substituting this yields, \(\int_{4}^{9} \frac{u-4}{u^{2}} du\)
04

Simplify the Integrals

Break down this integral into simpler parts for easy evaluation. It simplifies to \(\int_{4}^{9} (1 - \frac{4}{u}) du= \int_{4}^{9} du - 4*\int_{4}^{9} \frac{1}{u} du\
05

Evaluate the Integrals

Evaluate each simpler integral separately. The integral \(\int du\) is just \(u\), and the integral of \(\int \frac{1}{u} du\) is \(ln(u)\). Thus this step yields \(u-4*ln(u)|_{4}^{9} = [9-4*ln(9)] - [4-4*ln(4)]\
06

Simplify the Expression

Calculate the expression obtained from step 5: \(9-4*ln(9) - 4 + 4*ln(4) = 5 - 4*ln(9) + 4*ln(4) = 5 + 4*ln(\frac{4}{9})\

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Approximate the integral using (a) the Trapezoidal Rule and (b) Simpson's Rule for the indicated value of \(n\). (Round your answers to three significant digits.) $$ \int_{0}^{1} \sqrt{1-x} d x, n=4 $$

$$ \int_{1 / 2}^{\infty} \frac{1}{\sqrt{2 x-1}} d x $$

Consumer Trends The rate of change \(S\) in the number of subscribers to a newly introduced magazine is modeled by \(d S / d t=1000 t^{2} e^{-t}, 0 \leq t \leq 6\), where \(t\) is the time in years. Use Simpson's Rule with \(n=12\) to estimate the total increase in the number of subscribers during the first 6 years.

Population Growth \(\ln\) Exercises 57 and 58, use a graphing utility to graph the growth function. Use the table of integrals to find the average value of the growth function over the interval, where \(N\) is the size of a population and \(t\) is the time in days. $$ N=\frac{5000}{1+e^{4.8-1.9 t}}, \quad[0,2] $$

Use a spreadsheet to complete the table for the specified values of \(a\) and \(n\) to demonstrate that \(\lim _{x \rightarrow \infty} x^{n} e^{-a x}=0, \quad a>0, n>0\) \begin{tabular}{|l|l|l|l|l|} \hline\(x\) & 1 & 10 & 25 & 50 \\ \hline\(x^{n} e^{-a x}\) & & & & \\ \hline \end{tabular} $$ a=1, n=1 $$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.