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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 8, Problem 8.2.11

9–40. Integration by parts Evaluate the following integrals using integration by parts.
11. ∫ t · e⁶ᵗ dt

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Identify the parts of the integral for integration by parts. Let \(u = t\) and \(dv = e^{6t} dt\).
Compute the derivatives and integrals needed: find \(du = dt\) and integrate \(dv\) to get \(v = \frac{1}{6} e^{6t}\).
Apply the integration by parts formula: \(\int u \, dv = uv - \int v \, du\).
Substitute the expressions for \(u\), \(v\), \(du\) into the formula: \(\int t e^{6t} dt = t \cdot \frac{1}{6} e^{6t} - \int \frac{1}{6} e^{6t} dt\).
Evaluate the remaining integral \(\int e^{6t} dt\) and simplify the expression to write the final integral in terms of \(t\) and \(e^{6t}\) plus the constant of integration.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Integration by Parts Formula

Integration by parts is a technique derived from the product rule of differentiation. It states that ∫u dv = uv - ∫v du, where u and dv are parts of the original integral chosen to simplify the problem. Selecting u and dv wisely is crucial for easier integration.
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Introduction to Integration by Parts

Choosing u and dv

In integration by parts, u is typically chosen as a function that simplifies when differentiated, while dv is chosen as a function that is easy to integrate. For example, in ∫ t · e^(6t) dt, choosing u = t and dv = e^(6t) dt helps reduce the integral complexity.
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Choosing a Convergence Test

Integrating Exponential Functions

Integrating exponential functions like e^(kt) involves dividing by the constant k after integration, resulting in (1/k) e^(kt). This property is essential when integrating dv in problems involving exponentials, ensuring correct evaluation of the integral.
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Integrals of General Exponential Functions