Chapter 2 | Limits
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s (0.51)− s (0.5) 0.51−0.5
b. v ave =
=−16.16
The instantaneous velocity is somewhere between −15.84 and −16.16 ft/sec. A good guess might be −16 ft/sec.
2.2 An object moves along a coordinate axis so that its position at time t is given by s ( t ) = t 3 . Estimate its instantaneous velocity at time t =2 by computing its average velocity over the time interval [2, 2.001].
The Area Problem and Integral Calculus We now turn our attention to a classic question from calculus. Many quantities in physics—for example, quantities of work—may be interpreted as the area under a curve. This leads us to ask the question: How can we find the area between the graph of a function and the x -axis over an interval ( Figure 2.8 )?
Figure 2.8 The Area Problem: How do we find the area of the shaded region?
As in the answer to our previous questions on velocity, we first try to approximate the solution. We approximate the area by dividing up the interval ⎡ ⎣ a , b ⎤ ⎦ into smaller intervals in the shape of rectangles. The approximation of the area comes from adding up the areas of these rectangles ( Figure 2.9 ).
Figure 2.9 The area of the region under the curve is approximated by summing the areas of thin rectangles.
As the widths of the rectangles become smaller (approach zero), the sums of the areas of the rectangles approach the area between the graph of f ( x ) and the x -axis over the interval ⎡ ⎣ a , b ⎤ ⎦ . Once again, we find ourselves taking a limit. Limits of this type serve as a basis for the definition of the definite integral. Integral calculus is the study of integrals and their applications.
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