University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 401: 17


$-e^{-t^2} +c$

Work Step by Step

Solve $\int 2te^{-t^2}dt$ Let us $p=-t^2$ and $dp=-2t dt$ This implies $\int 2te^{-t^2}dt=-\int e^p dp $ $=-e^{p} +c$ $=-e^{-t^2} +c$ Hence,$\int 2te^{-t^2}dt=-e^{-t^2} +c$
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