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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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.

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.

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.

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