Chapter 8: Problem 21
The integrals converge. Evaluate the integrals without using tables. $$\int_{-\infty}^{0} \theta e^{\theta} d \theta$$
Short Answer
Expert verified
The integral evaluates to \(-1\).
Step by step solution
01
Identify the Integral Type
The given integral is \( \int_{-\infty}^{0} \theta e^{\theta} d\theta \). This is a definite integral of an improper type because it involves integration over an infinite limit.
02
Use the Integration by Parts Formula
Integration by parts formula is given by \( \int u \, dv = uv - \int v \, du \). Here, choose \( u = \theta \) (hence \( du = d\theta \)) and \( dv = e^{\theta} d\theta \) (hence \( v = e^{\theta} \)).
03
Apply the Integration by Parts Formula
Using integration by parts, we have: \[\int \theta e^{\theta} d\theta = \theta e^{\theta} - \int e^{\theta} d\theta = \theta e^{\theta} - e^{\theta} + C.\]
04
Evaluate the Definite Integral
Now substitute the limits of integration from \(-\infty\) to \(0\):\[\left[ \theta e^{\theta} - e^{\theta} \right]_{-\infty}^{0} = \left( 0 \cdot e^{0} - e^{0} \right) - \lim_{a \to -\infty} ( a e^{a} - e^{a} ).\]
05
Evaluate at the Upper Limit
At \(\theta = 0\), the expression becomes:\[0 \cdot e^{0} - e^{0} = -1.\]
06
Evaluate at the Lower Limit
Find the limit as \(\theta\) approaches \(-\infty\):\[\lim_{a \to -\infty} (a e^{a} - e^{a}) = \lim_{a \to -\infty} (a - 1) e^{a}.\] Both \(a e^{a}\) and \(e^{a}\) approach zero as \(a \to -\infty\) when using L'Hospital's Rule or analyzing the exponential decay. Thus, the expression evaluates to 0 for the lower limit.
07
Calculate the Final Result
Combine the two results from handling the limits:\[-1 - 0 = -1.\]Therefore, the integral evaluates to \(-1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by Parts is a useful technique in calculus for finding integrals of products of functions. It transforms a complex integral into simpler parts that are easier to evaluate. The principle is derived from the product rule of differentiation. The formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \]To apply this technique effectively, we must make strategic choices for our functions \( u \) and \( dv \). Typically, we choose \( u \) as the function that becomes simpler when differentiated, and \( dv \) as the function that remains manageable upon integration.
- Choose \( u \) and \( dv \) appropriately. In this exercise, \( u = \theta \) and \( dv = e^{\theta} d\theta \).
- Differentiate \( u \) to find \( du \), and integrate \( dv \) to find \( v \).
- Substitute into the integration by parts formula and simplify.
Exponential Functions
Exponential functions are a category of mathematical functions that grow by rates proportional to their current value. The most common base for exponential functions in calculus is Euler's number \( e \). It provides a constant relative rate of growth, making it a powerful tool in both pure and applied mathematics.
- Euler's number \( e \approx 2.71828 \) and is an irrational number.
- Exponential functions are of the form \( f(x) = e^x \), which grows rapidly as \( x \) increases.
- They exhibit a unique property: the derivative of \( e^x \) is \( e^x \) itself, making it incredibly straightforward to work with in differential and integral calculus.
Definite Integrals
Definite integrals, represented by notation \( \int_a^b f(x) \, dx \), calculate the accumulated area under the curve of \( f(x) \) between the limits \( a \) and \( b \). They provide precise information about the total accumulation or net change of function values over an interval.
- The lower limit of a definite integral is \( a \), and the upper limit is \( b \).
- A definite integral gives a specific numerical value, contrasting with indefinite integrals which include a constant of integration \( C \).
- Definite integrals can involve infinite limits, referred to as improper integrals. These require careful evaluation of limits at infinity.