Problem 9
Question
Evaluate using integration by parts or substitution. (Assume \(u>0\) in \(\ln\) u. Check by differentiating. $$ \int x^{2} \ln x d x $$
Step-by-Step Solution
Verified Answer
The integral \( \int x^2 \ln x \, dx \) is \( \frac{x^3}{3} \ln x - \frac{x^3}{9} + C \). Verify by differentiating.
1Step 1: Choose Substitution for Integration by Parts
In integration by parts, we select parts of the integrand for substitution. For \( \int x^2 \ln x \, dx \), a suitable choice is to let \( u = \ln x \) and \( dv = x^2 \, dx \). This leads us to use the formula for integration by parts: \( \int u \, dv = uv - \int v \, du \).
2Step 2: Find \( du \) and \( v \)
Differentiate \( u = \ln x \) to find \( du = \frac{1}{x} \, dx \). Integrating \( dv = x^2 \, dx \) gives \( v = \frac{x^3}{3} \).
3Step 3: Apply Integration by Parts Formula
Using the integration by parts formula, substitute the values: \( \int x^2 \ln x \, dx = uv - \int v \, du = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx \).
4Step 4: Simplify the Integral
The integral \( \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx \) simplifies to \( \frac{1}{3} \int x^2 \, dx \). Calculate this integral to get \( \frac{1}{3} \cdot \frac{x^3}{3} = \frac{x^3}{9} \).
5Step 5: Write the Final Solution
Combine the results from the previous steps to get the solution for the integral: \( \frac{x^3}{3} \ln x - \frac{x^3}{9} + C \), where \( C \) is the constant of integration.
6Step 6: Verify by Differentiation
To verify, differentiate \( \frac{x^3}{3} \ln x - \frac{x^3}{9} + C \). Use the product rule to differentiate \( \frac{x^3}{3} \ln x \) and check that it simplifies back to the original integrand \( x^2 \ln x \).
Key Concepts
Calculus Problem SolvingDefinite and Indefinite IntegralsDifferentiation Verification
Calculus Problem Solving
Solving calculus problems often involves choosing the right technique based on the structure of the integral. For this example, we can solve \( \int x^2 \ln x \, dx \) using the method of integration by parts. This approach helps us handle integrals involving the product of two functions by breaking down the problem into simpler parts.
**Setting Up the Problem**
To begin, identify the components of the integrand that suit the integration by parts formula, \( \int u \, dv = uv - \int v \, du \). Here, choosing \( u = \ln x \) and \( dv = x^2 \, dx \) is effective because differentiating \( \ln x \) simplifies matters, while integrating \( x^2 \) is straightforward.
**Combining Functions**
After calculation, we get:\[ du = \frac{1}{x} \, dx \quad \text{and} \quad v = \frac{x^3}{3} \]. Now, substitute into the formula to perform the integration by parts. This strategic breakdown reduces a complicated integral into a simpler form, making it easier to evaluate.
**Setting Up the Problem**
To begin, identify the components of the integrand that suit the integration by parts formula, \( \int u \, dv = uv - \int v \, du \). Here, choosing \( u = \ln x \) and \( dv = x^2 \, dx \) is effective because differentiating \( \ln x \) simplifies matters, while integrating \( x^2 \) is straightforward.
**Combining Functions**
After calculation, we get:\[ du = \frac{1}{x} \, dx \quad \text{and} \quad v = \frac{x^3}{3} \]. Now, substitute into the formula to perform the integration by parts. This strategic breakdown reduces a complicated integral into a simpler form, making it easier to evaluate.
Definite and Indefinite Integrals
In calculus, integrals can either be definite or indefinite. An indefinite integral, such as \( \int x^2 \ln x \, dx \), represents a family of functions and includes a constant of integration \( C \). This constant ensures that all possible antiderivatives are accounted for.
**Simplifying the Integral**
Using integration by parts has simplified our original integral to: \[ \int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \frac{x^3}{9} + C \]. This final expression is the indefinite integral. If we were dealing with a definite integral, the process would include evaluating this result within specified bounds, yielding a specific numerical result.
Understanding both types of integrals is key in calculus, providing necessary tools for solving a wide range of real-world and theoretical problems.
**Simplifying the Integral**
Using integration by parts has simplified our original integral to: \[ \int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \frac{x^3}{9} + C \]. This final expression is the indefinite integral. If we were dealing with a definite integral, the process would include evaluating this result within specified bounds, yielding a specific numerical result.
Understanding both types of integrals is key in calculus, providing necessary tools for solving a wide range of real-world and theoretical problems.
Differentiation Verification
Verifying integration results through differentiation is a crucial step to ensure correctness. For the integral \( \int x^2 \ln x \, dx \), we found \( \frac{x^3}{3} \ln x - \frac{x^3}{9} + C \) as the solution. The goal now is to differentiate this expression and confirm it matches the integrand \( x^2 \ln x \).
**Applying the Product Rule**
Differentiating \( \frac{x^3}{3} \ln x \) requires the product rule: \((fg)' = f'g + fg'\). Here, \( f = \frac{x^3}{3} \) and \( g = \ln x \). Calculating these derivatives and simplifying should yield \( x^2 \ln x + \frac{x^2}{3} - \frac{x^2}{3} = x^2 \ln x \). Thus, the differentiation correctly verifies our original integral.
Using differentiation as a verification step is a powerful way to reinforce one's understanding and confidence in solving complex calculus problems.
**Applying the Product Rule**
Differentiating \( \frac{x^3}{3} \ln x \) requires the product rule: \((fg)' = f'g + fg'\). Here, \( f = \frac{x^3}{3} \) and \( g = \ln x \). Calculating these derivatives and simplifying should yield \( x^2 \ln x + \frac{x^2}{3} - \frac{x^2}{3} = x^2 \ln x \). Thus, the differentiation correctly verifies our original integral.
Using differentiation as a verification step is a powerful way to reinforce one's understanding and confidence in solving complex calculus problems.
Other exercises in this chapter
Problem 8
Find the area represented by each definite integral. $$ \int_{-1}^{1}|3 x-2| d x $$
View solution Problem 9
Evaluate. (Be sure to check by differentiating!) $$ \int(\ln x)^{3} \frac{1}{x} d x, x>0 $$
View solution Problem 9
Find each integral. $$ \int\left(2 t^{2}+5 t-3\right) d t $$
View solution Problem 9
Find the area under the given curve over the indicated interval. $$ y=4-x^{2} ; \quad[-2,2] $$
View solution