Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 28: Problem 18
Integrate each of the functions. $$\int 0.8(3+2 \ln u)^{3} \frac{d u}{u}$$
Short Answer
Expert verified
\( 0.1 (3 + 2 \ln u)^4 + C \)
Step by step solution
01
Identify the Type of Function
The function to integrate is \( 0.8(3+2 \ln u)^{3} \frac{1}{u} \). This form suggests that us u-substitution might be suitable, especially given the \( \ln u \) term.
02
Choose an Appropriate Substitution
Set \( v = 3 + 2 \ln u \). Then, differentiate \( v \, \text{with respect to} \, u \) to find \( dv \). We have \( dv = 2 \cdot \frac{1}{u} du \) or \( \frac{dv}{2} = \frac{1}{u} du \). This substitution should simplify the integral.
03
Substitute and Rewrite the Integral
Replace \( 3 + 2 \ln u \) with \( v \), and \( \frac{1}{u} du \) with \( \frac{dv}{2} \). The integral becomes \( 0.8 \int v^3 \cdot \frac{dv}{2} \), simplifying to \( 0.4 \int v^3 dv \).
04
Integrate in Terms of v
The integral \( 0.4 \int v^3 dv \) is a basic polynomial integration. Integrating gives \( 0.4 \cdot \frac{v^4}{4} + C \). This simplifies to \( 0.1 v^4 + C \).
05
Back-Substitute to Original Variable
Substitute back \( v = 3 + 2 \ln u \) to express the solution in terms of \( u \). The result is \( 0.1 (3 + 2 \ln u)^4 + C \).
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!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
u-substitution
U-substitution is a powerful technique in calculus integration that can simplify the integration of composite functions. In this exercise, the function has the form of a polynomial raised to a power, and includes a logarithmic term.
This suggests the use of u-substitution. The idea is to make a substitution that simplifies the integral by replacing a part of the function with a new variable.
This suggests the use of u-substitution. The idea is to make a substitution that simplifies the integral by replacing a part of the function with a new variable.
- We select a substitution that encapsulates the most complex and derivative-friendly part of the function.
- In this case, we define a new variable, say \( v = 3 + 2 \ln u \). This new variable replaces the expression inside the power.
- Differentiating \( v \) with respect to \( u \) yields the differential \( dv = 2 \cdot \frac{1}{u} \, du \).
polynomial integration
Polynomial integration is an integral calculus technique used to find the integral of polynomial functions. In the simplified form of the exercise, the function \( 0.4 \int v^3 \, dv \) is a polynomial. The integration process involves finding the antiderivative of the polynomial term.
- A polynomial \( v^n \) integrates into \( \frac{v^{n+1}}{n+1} \).
- Therefore, \( \int v^3 \, dv \) becomes \( \frac{v^4}{4} \).
- Multiplying by the constant \( 0.4 \) scales the result, leading to \( 0.4 \cdot \frac{v^4}{4} \), which simplifies to \( 0.1 v^4 \).
logarithmic integration
Logarithmic integration often involves integrals that include a logarithmic function or its derivative. In this exercise, \( \ln u \) appears within the expression, necessitating the use of u-substitution.
When substituted correctly, it simplifies into polynomial terms. Hence, in cases where \( \ln u \) is part of an integral:
When substituted correctly, it simplifies into polynomial terms. Hence, in cases where \( \ln u \) is part of an integral:
- The natural logarithm \( \ln u \) differentiates to \( \frac{1}{u} \), simplifying the integration.
- Switching variables helps remove complications introduced by the logarithm.
integral calculus techniques
Integral calculus involves various techniques that simplify the process of finding antiderivatives or integrals. Key techniques include u-substitution, polynomial integration, and recognizing patterns in logarithmic or exponential functions. These strategies allow us to transform complex integrals into simpler, more manageable forms.
- U-substitution helps in converting complicated expressions into basic forms. This is particularly useful when the integral comprises a function and its derivative.
- Polynomial integration simplifies expressions where the function is a polynomial.
- Logarithmic integration occurs when logs are present, suggesting a transformation may lead to easier integration processes.
