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FDWKCalcSM_ch7

# FDWKCalcSM_ch7 - Section 7.1 313 69(a Separate the...

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Section 7.1 313 69. (a) Separate the variables to obtain dv v dt v t C v Ce Ce t + = − + = − + + = + = 17 2 17 2 17 47 17 1 2 ln 0 2 2 30 17 30 30 17 = − + = − = − C v e v e t t (b) lim( ) t t e →∞ = − 30 17 17 2 feet per second (c) = − 20 30 17 2 e t t = ln . 10 2 1 151 seconds Chapter 7 Applications of Definite Integrals Section 7.1 Integral as Net Change (pp. 378–389) Exploration 1 Revisiting Example 2 1. s t t t dt t t C ( ) ( ) = + = + + + 2 2 3 8 1 3 8 1 s C C s t t t ( ) ( ) . 0 0 3 8 0 1 9 1 3 8 1 1 3 3 = + + + = = = + + + Thus, 2. s ( ) . 1 1 3 8 1 1 1 16 3 3 = + + + = This is the same as the answer we found in Example 2a. 3. s ( ) . 5 5 3 8 5 1 1 44 3 = + + + = This is the same answer we found in Example 2b. Quick Review 7.1 1. On the interval, sin 2 x = 0 when x = − π π 2 0 2 , , . or Test one point on each subinterval: for x x = − = 3 2 2 1 π , sin ; for x x x x = − = − = = π π 4 2 1 4 2 1 , sin ; , sin ; for and for x x = − = − 3 4 2 1 π , sin . The function changes sign at π π 2 0 2 , , . and The graph is + + –3 2 0 2 2 f ( x ) x 2. x 2 – 3 x + 2 = ( x – 1)( x – 2) = 0 when x = 1 or 2. Test one point on each subinterval: for x = 0, x 2 – 3 x + 2 = 2; for x x x = + = − 3 2 3 2 1 4 2 , ; and for x = 3, x 2 – 3 x + 2 = 2. The function changes sign at 1 and 2. The graph is + + –2 1 2 4 f ( x ) x 3. x 2 – 2 x + 3 = 0 has no real solutions, since b 2 4 ac = ( ) 2 2 – 4(1)(3) = –8 < 0. The function is always positive. The graph is + –4 2 f ( x ) x 4. 2 3 x 3 2 x + 1 = ( ) x 1 2 (2 x + 1) = 0 when x = − 1 2 or 1. Test one point on each subinterval: for x = –1, 2 3 x 3 2 x + 1 = –4; for x = 0, 2 3 x – 3 x 2 + 1 = 1; and x = − 3 2 , 2 3 1 1 3 2 x x + = . The function changes sign at 1 2 . The graph is + + –2 2 1 1 2 f ( x ) x 5. On the interval, x cos 2 x = 0 when x = 0 4 3 4 5 4 , , , . π π π or Test one point on each subinterval: for x = π 8 , x x x x x x cos , , cos , 2 2 16 2 2 2 2 = = = − = π π π π for for x x x x x cos ; , cos . . 2 4 2 0 58 = = ≈− π andfor The function changes sign at π π π 4 3 4 5 4 , , . and The graph is + + 0 4 4 3 4 5 4 f ( x ) x 6. xe x = 0 when x = 0. On the rest of the interval, xe x is always positive. 7. x x x 2 1 0 0 + = = when . Test one point on each subinterval: for x x x x x x = − + = − = + = 1 1 1 2 1 1 1 2 2 2 , ; , . for The function changes sign at 0. The graph is + –5 0 30 f ( x ) x

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314 Section 7.1 8. x x x 2 2 2 4 0 2 = = ± when and is undefined when x = ± 2. Test one point on each subinterval: for x x x x x x = − = = − 5 2 2 4 17 9 1 9 2 4 4 13 2 2 2 2 , ; . , . ; for for for x x x x x x = = = ≈− 0 2 4 1 2 1 9 2 4 4 13 2 2 2 2 , ; . , . ; andfor x x x = = 5 2 2 4 17 9 2 2 , . The function changes sign at –2, 2 2 , and 2. The graph is + + + –3 –2 2 3 2 2 f ( x ) x 9. sec sin cos cos 1 1 1 1 2 + ( ) = + ( ) x x is undefined when x k k + + 0 9633 2 1783 . . π π or for any integer k . Test for x = + ( ) ≈ − 0 1 1 0 2 4030 2 :sec sin . . Test for x = ± + ( ) 1 1 1 1 32 7984 2 :sec sin . . The sign alternates over successive subintervals. The function changes sign at 0.9633 + k π or 2.1783 + k π , where k is an integer. The graph is + + –2.1783 –0.9633 0.9633 2.1783 f ( x ) x 10. On the interval, sin .
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FDWKCalcSM_ch7 - Section 7.1 313 69(a Separate the...

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