Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - 5.2 Definite Integrals - 5.2 Exercises: 4


This is because every member of the Riemann sum is negative and thus the whole sum is negative as well.

Work Step by Step

Using Riemann sums, the definition of the definite integral says that this integral is equal to the limit when the width of the arbitrary partition $\Delta$ goes to $0$ of the sum $\sum_{i=0}^Nf(x_i^*)\Delta x_i$ where $x_i^*\in(x_{i-1},x_i)$, and $x_i$ are the points defining the partition. Since $\Delta x_i$ are positive and the function $f$ is negative everywhere,which means that every $f(x_i^*)<0$, then every member of the sum $f(x_i^*)\Delta x_i<0$ so the total sum must be negative, and so is the integral. Another way of clearly saying this, is that there is some $M<0$ such that $f(x)\leq M$ for every $x$ so $$\sum_{i=0}^Nf(x_i^*)\Delta x_i\leq\sum_{i=0}^NM\Delta x_i=M\sum_{i=0}^N\Delta x_i=M(b-a)<0,$$ because $M<0$ and $b>a$.
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