Problem 25
Question
Evaluate the integrals in Exercises \(1-34\) without using tables. $$ \int_{0}^{1} x \ln x d x $$
Step-by-Step Solution
Verified Answer
The integral evaluates to \( -\frac{1}{4} \).
1Step 1: Identify the Integration Technique
Look at the integrand \( x \ln x \). The expression involves a polynomial function \( x \) and a logarithmic function \( \ln x \). The equation suggests using integration by parts, as it involves the product of two functions.
2Step 2: Set Up Integration by Parts Formula
Recall the integration by parts formula:\[ \int u \, dv = uv - \int v \, du \]For our integral \( \int x \ln x \, dx \), we choose: \( u = \ln x \), hence \( du = \frac{1}{x} \, dx \); and \( dv = x \, dx \), hence \( v = \frac{x^2}{2} \).
3Step 3: Apply the Integration by Parts Formula
Substitute \( u, v, du, dv \) into the integration by parts formula:\[ \int x \ln x \, dx = \frac{x^2}{2} \ln x \bigg|_0^1 - \int_0^1 \frac{x^2}{2} \cdot \frac{1}{x} \, dx \]This simplifies to:\[ \int x \ln x \, dx = \frac{x^2}{2} \ln x \bigg|_0^1 - \frac{1}{2} \int_0^1 x \, dx \]
4Step 4: Evaluate the Boundary Terms
First, evaluate \( \frac{x^2}{2} \ln x \bigg|_0^1 \). At \( x=1 \), the value is \( 0 \) since \( \ln 1 = 0 \). At \( x=0 \), \( \lim_{x \to 0^+} x^2 \ln x \) needs L'Hôpital's Rule, which can show that it is \( 0 \) because \( x^2 \) approaches zero faster than \( \ln x \) approaches negative infinity.
5Step 5: Evaluate the Remaining Integral
Compute the remaining integral:\[ \frac{1}{2} \int_0^1 x \, dx = \frac{1}{2} \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} \cdot \left( \frac{1}{2} \right) = \frac{1}{4} \]
6Step 6: Combine the Results
Now combine all the calculations:Since the boundary terms evaluated to \( 0 \), we are left with:\[ 0 - \frac{1}{4} = -\frac{1}{4} \]Thus, the integral\[ \int_{0}^{1} x \, \ln x \, dx = -\frac{1}{4} \].
Key Concepts
Definite IntegralsLogarithmic FunctionsL'Hôpital's Rule
Definite Integrals
Definite integrals are used to calculate the area under a curve within a specified interval. In this case, the integral \( \int_{0}^{1} x \ln x \, dx \) aims to find the area between the function \( x \ln x \) and the x-axis from 0 to 1.
Unlike indefinite integrals, definite integrals have upper and lower limits which make them result in a specific numerical value rather than a function with a constant of integration.
The integral symbol \( \int \) with bounds \( a \) and \( b \) represent the limits \( a \) to \( b \). This calculates the exact accumulation of areas captured between the function and the x-axis over those bounds.
When dealing with definite integrals, it's important to apply appropriate integration techniques such as substitution or by parts, especially when simple algebraic methods are not sufficient. Then you evaluate the resulting antiderivative at its boundary limits, substituting each limit into the antiderivative and subtracting the lower limit result from the upper limit result.
Unlike indefinite integrals, definite integrals have upper and lower limits which make them result in a specific numerical value rather than a function with a constant of integration.
The integral symbol \( \int \) with bounds \( a \) and \( b \) represent the limits \( a \) to \( b \). This calculates the exact accumulation of areas captured between the function and the x-axis over those bounds.
When dealing with definite integrals, it's important to apply appropriate integration techniques such as substitution or by parts, especially when simple algebraic methods are not sufficient. Then you evaluate the resulting antiderivative at its boundary limits, substituting each limit into the antiderivative and subtracting the lower limit result from the upper limit result.
Logarithmic Functions
Logarithmic functions, commonly represented as \( \ln x \) where the base is the natural constant \( e \), are the inverses of exponential functions. In mathematical and real-world problem scenes, they exponentially scale values which are otherwise difficult to handle linearly.
A function involving \( \ln x \) signifies that as \( x \) changes, the rate of change or growth changes logarithmically, typically increasing slowly.\( \ln x \) is undefined for non-positive values which means special attention is needed around zero since \( x \to 0^+ \) encounters unique limits.
In calculations involving \( \ln x \), and especially within integration, applying integration by parts here was crucial. The function \( x \ln x \) in the exercise merges linear growth \( x \) accommodated by \( \ln x \) logarithmic slowing. This interaction forms the core challenge, and such functions frequently necessitate more advanced techniques beyond elementary methods.
A function involving \( \ln x \) signifies that as \( x \) changes, the rate of change or growth changes logarithmically, typically increasing slowly.\( \ln x \) is undefined for non-positive values which means special attention is needed around zero since \( x \to 0^+ \) encounters unique limits.
In calculations involving \( \ln x \), and especially within integration, applying integration by parts here was crucial. The function \( x \ln x \) in the exercise merges linear growth \( x \) accommodated by \( \ln x \) logarithmic slowing. This interaction forms the core challenge, and such functions frequently necessitate more advanced techniques beyond elementary methods.
L'Hôpital's Rule
L'Hôpital's Rule is a mathematical method used to find limits for expressions that present indeterminate forms, such as \( 0/0 \) or \( \infty/\infty \). It uses derivatives of the numerator and the denominator to resolve these forms.
In our exercise, evaluating \( x^2 \ln x \) as \( x \to 0^+ \) poses an indeterminate form, specifically the \( 0 \cdot -\infty \) scenario. By reformulating it into a ratio like \( \frac{\ln x}{1/x^2} \), L'Hôpital's Rule allows for differentiation of both parts independently until a manageable limit appears.
Upon applying L'Hôpital's Rule, repeated differentiation addresses undefined behavior or infinite growth, leading ultimately to finding a clearer limit. Here it was deduced that \( x^2 \ln x \) converges effectively to 0 demonstrating how faster zero approaches in numerator outpaces diverging \( -\infty \) in denominator due to \( x^2 \).
Such rule applications frequently occur in calculus when encountering logarithms paired with polynomials or other exponents, especially near boundaries involving zero and infinity.
In our exercise, evaluating \( x^2 \ln x \) as \( x \to 0^+ \) poses an indeterminate form, specifically the \( 0 \cdot -\infty \) scenario. By reformulating it into a ratio like \( \frac{\ln x}{1/x^2} \), L'Hôpital's Rule allows for differentiation of both parts independently until a manageable limit appears.
Upon applying L'Hôpital's Rule, repeated differentiation addresses undefined behavior or infinite growth, leading ultimately to finding a clearer limit. Here it was deduced that \( x^2 \ln x \) converges effectively to 0 demonstrating how faster zero approaches in numerator outpaces diverging \( -\infty \) in denominator due to \( x^2 \).
Such rule applications frequently occur in calculus when encountering logarithms paired with polynomials or other exponents, especially near boundaries involving zero and infinity.
Other exercises in this chapter
Problem 25
In Exercises \(15-26,\) estimate the minimum number of subintervals needed to approximate the integrals with an error of magnitude less than \(10^{-4}\) by ( a
View solution Problem 25
Use the table of integrals at the back of the book to evaluate the integrals. \(\int \frac{d s}{\left(9-s^{2}\right)^{2}}\)
View solution Problem 25
Evaluate the integrals in Exercises \(23-32\). $$ \int_{0}^{\pi / 4} \sec ^{4} \theta d \theta $$
View solution Problem 25
Evaluate the integrals by using a substitution prior to integration by parts. \(\int e^{\sqrt{3 s+9}} d s\)
View solution