Calculus 8th Edition

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

Chapter 4 - Integrals - 4.2 The Definite Integral - 4.2 Exercises - Page 317: 33

Answer

A) 4 B) 10 C) -3 D) 2

Work Step by Step

Using the definition discussed in Section 4.1, we know that the area under the curve of the function and above the x-axis represents the result of the integral, also if there is area below the x-axis, it is subtracted rather than added from the total sum of areas. It follows that, to evaluate the integral it suffices to use geometric approach to find the area. A) We can see that the region under the curve of the function and above the x-axis within the integral limit 0 and 2 can be broken to a rectangle with 2x1 dimensions, and a triangle with a base of 2 and a height of 2 units , Evaluated integral = Total area = 1*2 + 0.5*2*2 = 4. B) Using same argument as before, we break the region under the curve into the same one like part a, in addition to to a 1x3 rectangle and 2x3 right triangle. Evaluated integral = Total area = 4 + 3*1 + 0.5*3*2 = 10 C) Since we have a region below the x-axis, we calculate the normal area but assign a negative sign at the result to indicate the integration limits correspond to a region below x-axis. Evaluated integral = Total area = (-) Area of the right-angle triangle between x= 5 and x= 7 >> 0.5*2*3 = - 3 D) We have the areas till x = 7, we only find the remaining are related to the curve of the integrated function, which is a 2x2 square, and a 2x1 right-angle triangle. Evaluated integral = Total area = 10-3 - (2*2+.05*2*1) = 2
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