Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Problem Statement
Use integration by parts to evaluate the definite integral:
\[ \int_{1}^{e} t^2 \ln(t) \, dt \]
Give an exact answer:
[Input Box]
---
### Detailed Explanation
**Integration by Parts:**
Integration by parts is a technique that uses the formula:
\[ \int u \, dv = uv - \int v \, du \]
To apply this method, choose parts of the integrand \( t^2 \ln(t) \) as follows:
- Let \( u = \ln(t) \) and thus \( du = \frac{1}{t} \, dt \)
- Let \( dv = t^2 \, dt \) and thus \( v = \frac{t^3}{3} \)
**Steps:**
1. Substitute \( u = \ln(t) \) and \( dv = t^2 \, dt \):
- \( u = \ln(t) \) ⟹ \( du = \frac{1}{t} \, dt \)
- \( dv = t^2 \, dt \) ⟹ \( v = \frac{t^3}{3} \)
2. Apply the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
So,
\[ \int_{1}^{e} t^2 \ln(t) \, dt = \left. \frac{t^3}{3} \ln(t) \right|_{1}^{e} - \int_{1}^{e} \frac{t^3}{3} \cdot \frac{1}{t} \, dt \]
3. Simplify the remaining integral:
\[ \int_{1}^{e} \frac{t^3}{3} \cdot \frac{1}{t} \, dt = \frac{1}{3} \int_{1}^{e} t^2 \, dt \]
4. Evaluate the known integral:
\[ \int_{1}^{e} t^2 \, dt = \left. \frac{t^3}{3} \right|_{1}^{e} = \frac{e^3}{3} - \frac{1^3}{3} = \frac{e^3}{3} - \frac{1}{3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9dc9f679-6754-47d7-a5b7-cef477361a1e%2Fb23599f3-3387-4335-8ed1-a8c94a61f6ec%2F9x2chrb_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Use integration by parts to evaluate the definite integral:
\[ \int_{1}^{e} t^2 \ln(t) \, dt \]
Give an exact answer:
[Input Box]
---
### Detailed Explanation
**Integration by Parts:**
Integration by parts is a technique that uses the formula:
\[ \int u \, dv = uv - \int v \, du \]
To apply this method, choose parts of the integrand \( t^2 \ln(t) \) as follows:
- Let \( u = \ln(t) \) and thus \( du = \frac{1}{t} \, dt \)
- Let \( dv = t^2 \, dt \) and thus \( v = \frac{t^3}{3} \)
**Steps:**
1. Substitute \( u = \ln(t) \) and \( dv = t^2 \, dt \):
- \( u = \ln(t) \) ⟹ \( du = \frac{1}{t} \, dt \)
- \( dv = t^2 \, dt \) ⟹ \( v = \frac{t^3}{3} \)
2. Apply the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
So,
\[ \int_{1}^{e} t^2 \ln(t) \, dt = \left. \frac{t^3}{3} \ln(t) \right|_{1}^{e} - \int_{1}^{e} \frac{t^3}{3} \cdot \frac{1}{t} \, dt \]
3. Simplify the remaining integral:
\[ \int_{1}^{e} \frac{t^3}{3} \cdot \frac{1}{t} \, dt = \frac{1}{3} \int_{1}^{e} t^2 \, dt \]
4. Evaluate the known integral:
\[ \int_{1}^{e} t^2 \, dt = \left. \frac{t^3}{3} \right|_{1}^{e} = \frac{e^3}{3} - \frac{1^3}{3} = \frac{e^3}{3} - \frac{1}{3
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