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 vt C vC e Ce t + =− += −+ −+ = 17 2 17 2 17 47 17 1 2 ln 0 2 2 30 17 30 30 17 ⇒= C ve t t (b) lim( ) t t e →∞ −− = 30 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. st t t dt t t C () + =+ + + 2 2 3 8 1 3 8 1 sC C t t . 0 0 3 8 01 91 3 8 1 1 3 3 + +=⇒= + + Thus, 2. s . 1 1 3 8 11 1 16 3 3 + This is the same as the answer we found in Example 2a. 3. s . 5 5 3 8 51 14 4 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 ππ 2 0 2 ,, . or Test one point on each subinterval: for xx = 3 2 21 π ,s in ; for x x = = 4 4 ; ; for and for 3 4 . The function changes sign at 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 + = 3 2 32 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 , 231 1 = . 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 ,, , . or Test one point on each subinterval: for x = 8 , x x cos , , cos , 2 2 16 2 2 2 2 == = = xxx cos ; , cos . . 24 2 0 5 8 and for The function changes sign at 4 3 4 5 4 . and The graph is + + 04 ± 4 3 ± 4 5 ± 4 f ( x ) x 6. xe x = 0 when x = 0. On the rest of the interval, x is always positive. 7. x x x 2 1 00 + when . Test one point on each subinterval: for x x x x x x + = + = 1 1 1 2 1 1 1 2 22 , . 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 02 == ± 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 19 2 4 413 2 2 2 2 ,; . , . ; for x x x x x x = ≈− 0 2 4 1 2 2 4 2 2 2 2 . , . ; and for x x x = = 5 2 2 4 17 9 2 2 ,. The function changes sign at –2, 22 , and 2. The graph is ++ +– –3 –2 23 2 2 f ( x ) x 9. sec sin cos cos 11 1 1 2 +− () = + x x is undefined when xk k ≈+ + 0 9633 2 1783 .. ππ 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 sin . . The sign alternates over successive subintervals. The function changes sign at 0.9633 + k π or 2.1783 + k , where k is an integer. The graph is –– –2.1783 –0.9633 0.9633 2.1783 f ( x ) x 10. On the interval, . 1 0 1 3 1 2 x x or Test one point on each subinterval: for x = 0.1, . ; 1 054 x x x = 015 1 037 ., s i n .; and for x = 0.2, . .
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## FDWKCalcSM_ch7 - Section 7.1 313 69. (a) Separate the...

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