Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises - Page 514: 82

Answer

The whole integral $$\int_0^5f(x)dx$$ is negative.

Work Step by Step

We can split the integral into a positive part $$\int_0^2 f(x)dx$$ because the graph is above the $x$ axis and a negative part $$\int_2^5f(x)dx$$ because the graph is below the $x$ axis. We have that $$\int_0^5f(x)dx=\int_0^2 f(x)dx + \int_2^5f(x)dx.$$ If the positive part is larger than the negative then the whole integral is positive and if the negative part is larger then the whole integral is negative. From the graph on the figure we see that the negative part is larger (this is only an estimate) because the area under the graph is there larger, so we can say that the whole integral is negative.
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