TAILIEUCHUNG - Calculus and its applications: 4.2

"Calculus and its applications: " - Antiderivatives as Areas have Objective: find the area under a graph to solve real-world problems, use rectangles to approximate the area under a graph. | 2012 Pearson Education, Inc. All rights reserved Slide Antiderivatives as Areas OBJECTIVE Find the area under a graph to solve real-world problems Use rectangles to approximate the area under a graph. 2012 Pearson Education, Inc. All rights reserved Slide Example 1: A vehicle travels at 50 mi/hr for 2 hr. How far has the vehicle traveled? Antiderivatives as Areas The answer is 100 mi. We treat the vehicle’s velocity as a function, We graph this function, sketch a vertical line at and obtain a rectangle. This rectangle measures 2 units horizontally and 50 units vertically. Its area is the distance the vehicle has traveled: 2012 Pearson Education, Inc. All rights reserved Slide Example 2: The velocity of a moving object is given by the function where x is in hours and v is in miles per hour. Use geometry to find the area under the graph, which is the distance the object has traveled: a.) during the first 3 hr b.) between the third hour and the fifth hour Antiderivatives as Areas 2012 Pearson Education, Inc. All rights reserved Slide Example 2 (continued): a.) The graph of the velocity function is shown at the right. We see the region corresponding to the time interval is a triangle with base 3 and height 8 (since ). Therefore, the are of this region is Antiderivatives as Areas The object traveled mi during the first 3 hr. 2012 Pearson Education, Inc. All rights reserved Slide Example 2 (Continued): b.) The region corresponding to the time interval is a trapezoid. It can be decomposed into a rectangle and a triangle as indicated in the figure to the right. The rectangle has abase 2 and height 9, and thus an area Antiderivatives as Areas 2012 Pearson Education, Inc. All rights reserved Slide Example 2 (Concluded): b.) The triangle has base 2 and height 6, for an area Summing the two areas, we get 24. Therefore, the object traveled 24 mi between the third hour and the fifth hour. .

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