Calculus Volume 1

572

Chapter 5 | Integration

1/2

1 2 v ( t ) dt = ⌠ ⌡ 0

1/2 1

2 v ( t ) dt + ∫

2 v ( t ) dt

0

1/2

= ⌠ = ⌠

1/3 1 1/2 1

2(20 t +5) dt + ∫

2(15) dt

⌡ 0

1/2

(40 t +10) dt + ∫

30 dt

⌡ 0

⎡ ⎣ 20 t 2 +10 t ⎤

1/2 +[30 t ] |

1/2 1

⎦ | 0

= =

⎛ ⎝ 20

⎞ ⎠ −0+(30−15)

4 +5

=25. Andrew is 25 mi from his starting point after 1 hour.

5.23 Suppose that, instead of remaining steady during the second half hour of Andrew’s outing, the wind starts to die down according to the function v ( t ) =−10 t +15. In other words, the wind speed is given by

v ( t ) = ⎧ ⎩ ⎨ 20 t +5 for 0≤ t ≤ 1 2 −10 t +15 for 1 2 ≤ t ≤1. Under these conditions, how far from his starting point is Andrew after 1 hour?

Integrating Even and Odd Functions We saw in Functions and Graphs that an even function is a function in which f (− x ) = f ( x ) for all x in the domain—that is, the graph of the curve is unchanged when x is replaced with − x . The graphs of even functions are symmetric about the y -axis. An odd function is one in which f (− x ) =− f ( x ) for all x in the domain, and the graph of the function is symmetric about the origin. Integrals of even functions, when the limits of integration are from − a to a , involve two equal areas, because they are symmetric about the y -axis. Integrals of odd functions, when the limits of integration are similarly [− a , a ], evaluate to zero because the areas above and below the x -axis are equal.

Rule: Integrals of Even and Odd Functions For continuous even functions such that f (− x ) = f ( x ), ⌠ ⌡ − a a For continuous odd functions such that f (− x ) =− f ( x ), ∫ − a a

f ( x ) dx =2 ∫ 0 a

f ( x ) dx .

f ( x ) dx =0.

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