Pre-Calc Homework Solutions 399

Pre-Calc Homework Solutions 399 - Chapter 9 Review 63. (a)...

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63. (a) Substituting x 2 for x in the Maclaurin series for sin x given at the end of Section 9.2, sin x 2 5 x 2 2 } x 3 6 ! } 1 } x 5 1 ! 0 } 2 1 ( 2 1) n } (2 x n 4 1 n 1 2 1)! } . Integrating term-by-term and observing that the constant term is 0, E x 0 sin t 2 dt 5 } x 3 3 } 2 } 7( x 3 7 !) } 1 } 11 x ( 1 5 1 !) } 2 1 ( 2 1) n } (4 n 1 x 3 4 ) n ( 1 2 n 3 1 1)! }1 (b) E 1 0 sin x dx 5 } 1 3 } 2 } 7( 1 3!) } 1 } 11 1 (5!) } 2 1 ( 2 1) n } (4 n 1 3) 1 (2 n 1 1)! . Since the third term is } 11 1 (5!) } 5 } 13 1 20 } , 0.001, it suffices to use the first two nonzero terms (through degree 7). (c) NINT(sin x 2 , x , 0, 1) < 0.31026830 (d) } 1 3 } 2 } 7( 1 3!) } 1} 11( 1 5!) } 2 } 15 1 (7!) } 5 } 2 8 5 3 8 1 , , 0 6 1 0 9 0 } < 0.31026816 This is within 1.5 3 10 2 7 of the answer in (c). 64. (a) Let f ( x ) 5 x 2 e x dx . E 1 0 x 2 e x dx 5 E 1 0 f ( x ) dx < } h 2 } 3 f (0) 1 2 f (0.5) 1 f (1) 4 5 } 1 4 } 3 0 1 2 } e 4 0.5 } 1 e 4 5 } e 8 0.5 } 1 } 4 e } < 0.88566 (b) x 2 e x 5 x 2 1 1 1 x 1 } x 2 2 ! } 1 1 } x n n ! } 1 2 5 x 2 1 x 3 1 } x 2 4 ! } 1 1 } x n n 1 ! 2 } 1 P 4 ( x ) 5 x 2 1 x 3 1 } x 2 4 } E 1 0 P 4 ( x ) 5 3 } x 3 3 } 1 } x 4 4 } 1 } 1 x 0 5 } 4 5 } 4 6 1 0 } < 0.68333 (c) Since f is concave up, the trapezoids used to estimate
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