Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.3* The Natural Exponential Function - 6.3* Exercises - Page 454: 99


$\int_{a}^{b} e^{-t^{2}}dt=\frac{1}{2}\sqrt \pi[ erf(b)- erf(a)]$

Work Step by Step

As we are given that $erf(x)=\frac{2}{\sqrt \pi} \int_{0}^{x} e^{-t^{2}} dt$ $\int_{0}^{x} e^{-t^{2}} dt=erf(x)\frac{\sqrt \pi}{2} $ Consider $\int_{a}^{b} e^{-t^{2}}dt=\int_{0}^{b} e^{-t^{2}} dt-\int_{0}^{a} e^{-t^{2}} dt $ Therefore,$\int_{0}^{b} e^{-t^{2}} dt-\int_{0}^{a} e^{-t^{2}} dt =[ \frac{\sqrt \pi}{2}erf(b)-\frac{\sqrt \pi}{2} erf(a)]$ Hence, $\int_{a}^{b} e^{-t^{2}}dt=\frac{1}{2}\sqrt \pi[ erf(b)- erf(a)]$
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