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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
7장, 문제 7.3.43

Evaluate the integrals in Exercises 33–54.
∫ 2t e^(-t²) dt

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Identify the integral to solve: \(\int 2t e^{-t^{2}} \, dt\).
Recognize that the integrand contains a function and its derivative: the exponent \(-t^{2}\) and the factor \$2t\( which is the derivative of \)-t^{2}$ up to a constant.
Use substitution by letting \(u = -t^{2}\). Then compute the differential: \(du = -2t \, dt\), which implies \(-du = 2t \, dt\).
Rewrite the integral in terms of \(u\): \(\int 2t e^{-t^{2}} \, dt = \int e^{u} (-du) = -\int e^{u} \, du\).
Integrate with respect to \(u\): \(-\int e^{u} \, du = -e^{u} + C\), then substitute back \(u = -t^{2}\) to get the final expression in terms of \(t\).

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