9–40. Integration by parts Evaluate the following integrals using integration by parts.
14. ∫ s · e⁻²ˢ ds

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Identify the parts of the integral for integration by parts. Let \(u = s\) and \(dv = e^{-2s} \, ds\).

Compute \(du\) by differentiating \(u\): \(du = ds\).

Compute \(v\) by integrating \(dv\): \(v = \int e^{-2s} \, ds\). Recall that \(\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C\), so here \(v = -\frac{1}{2} e^{-2s}\).

Apply the integration by parts formula: \(\int u \, dv = uv - \int v \, du\). Substitute the expressions for \(u\), \(v\), \(du\) to get \(s \cdot \left(-\frac{1}{2} e^{-2s}\right) - \int \left(-\frac{1}{2} e^{-2s}\right) \, ds\).

Simplify the integral and evaluate \(\int e^{-2s} \, ds\) again to complete the solution.

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주요 개념

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

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It transforms the integral of a product of functions into simpler integrals using the formula ∫u dv = uv - ∫v du. Choosing u and dv wisely simplifies the integration process.

Choosing u and dv

Selecting which part of the integrand to assign as u and which as dv is crucial. Typically, u is chosen as a function that simplifies when differentiated, and dv as a function that is easy to integrate. This choice affects the ease of solving the integral.

Integration of Exponential Functions

Integrating exponential functions like e^(-2s) involves recognizing the integral formula ∫e^(as) ds = (1/a)e^(as) + C. Understanding how to handle constants in the exponent is essential for correctly computing the integral after applying integration by parts.

9–40. Integration by parts Evaluate the following integrals using integration by parts.14. ∫ s · e⁻²ˢ ds