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: 29



Work Step by Step

$$A=\int^{\sqrt2}_{1}x2^{(x^2)}dx$$ We set $x^2=u$, which means $$2xdx=du$$ $$xdx=\frac{1}{2}du$$ - For $x=\sqrt2$, we have $u=2$ and for $x=1$, we have $u=1$ Therefore, $$A=\frac{1}{2}\int^{2}_{1}2^udu$$ $$A=\frac{1}{2}\times\frac{2^u}{\ln2}\Big]^{2}_{1}$$ $$A=\frac{1}{2\ln2}(2^2-2^1)$$ $$A=\frac{2}{2\ln2}=\frac{1}{\ln2}$$
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