Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 8: Problem 18
Evaluate the integrals using integration by parts. $$\int\left(r^{2}+r+1\right) e^{r} d r$$
Short Answer
Expert verified
\( \int (r^2 + r + 1)e^r \, dr = e^r(r^2 + r - 1) + C \)
Step by step solution
01
Identify Parts for Integration by Parts
In integration by parts, we use the formula \( \int u \, dv = uv - \int v \, du \). For the integral \( \int(r^2 + r + 1)e^r \, dr \), let's choose \( u = r^2 + r + 1 \) and \( dv = e^r \, dr \).
02
Differentiate and Integrate Parts
Calculate \( du \) and \( v \):- Differentiate \( u \, \rightarrow \, du = (2r + 1) \cdot dr \).- Integrate \( dv \, \rightarrow \, v = e^r \).
03
Apply Integration by Parts Formula
Apply the formula \( \int u \, dv = uv - \int v \, du \):\[ \int (r^2 + r + 1) e^r \, dr = (r^2 + r + 1) e^r - \int e^r (2r + 1) \, dr \]
04
Simplify the New Integral
This step simplifies the integral \( \int e^r (2r + 1) \, dr \). We perform integration by parts again, choosing \( u = 2r + 1 \) and \( dv = e^r \, dr \).
05
Differentiate and Integrate for New Terms
Calculate for the new \( u \) and \( dv \):- Differentiate \( u \, \rightarrow \, du = 2 \, dr \).- \( v = e^r \) remains the same as before.
06
Apply Integration by Parts Again
Apply the formula again:\[ \int e^r (2r + 1) \, dr = (2r + 1)e^r - \int 2e^r \, dr \]
07
Evaluate the Remaining Integral
The remaining integral is \( \int 2e^r \, dr \) which is straightforward:\[ \int 2e^r \, dr = 2e^r \]
08
Combine All Parts Together
Combine all parts:\[ \int (r^2 + r + 1) e^r \, dr = (r^2 + r + 1)e^r - ((2r + 1)e^r - 2e^r) + C \]Simplify:\[ \int (r^2 + r + 1) e^r \, dr = (r^2 + r + 1)e^r - (2re^r + C) \]
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.
Definite Integrals
Definite integrals are a key concept in calculus and serve an important role in calculating the area under a curve. Unlike indefinite integrals, which result in a family of functions, definite integrals evaluate the net area between the curve of a function and the x-axis, between two points. This means a definite integral has specific limits that define where the area begins and ends.
- For definite integrals, the notation is \( \int_{a}^{b} f(x) \, dx \), where \(a\) and \(b\) are the limits of integration.
- The result of a definite integral is a scalar value, giving a precise measurement of the enclosed area.
- It involves applying the fundamental theorem of calculus, which relates differentiation to integration.
Exponential Functions
Exponential functions are functions where the variable appears in the exponent. They are represented by \( a^x \) or in our more common calculus practice, \( e^x \), where \( e \) is the base of natural logarithms, approximately equal to 2.718.
- Exponential functions are important due to their unique properties, such as growth and decay models.
- The derivative and the integral of \( e^x \) remain the same, which simplifies calculations in many calculus problems.
- They exhibit continuous growth or decay, making them useful for modeling phenomena like population growth, radioactive decay, and interest calculations.
Differentiation
Differentiation is a fundamental concept in calculus that involves finding the derivative of a function, which represents the rate of change of the function with respect to a variable. It's what allows us to determine how a function behaves at any given point.
- The derivative of a function \( f(x) \) is denoted as \( f'(x) \) or \( \frac{df}{dx} \).
- To differentiate \( u \) in the original problem, you simply apply the power rule. For example, if \( u = r^2 + r + 1 \), then \( du = (2r + 1) \, dr \).
- Differentiation is used not only in calculus problems but also in real-life applications like finding velocity, acceleration, and in optimizing certain outcomes.
