Answer
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Work Step by Step
Letting $u=1-x$, we have $du=-dx$ and the endpoints of the integration are $x=0\to u=1$ and $x=1\to u=0$. Thus, we have $RHS=\int_0^1 f(1-x)dx=\int_1^0 f(u)(-du)=-\int_1^0 f(u)(du)=\int_0^1 f(u)(du)=LHS$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 305: 84
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