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Pre-Calc Homework Solutions 218

# Pre-Calc Homework Solutions 218 - 218 Section 5.4 b b 39...

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39. Yes, E b a av ( f ) dx 5 E b a f ( x ) dx . This is because av ( f ) is a constant, so E b a av ( f ) dx 5 3 av ( f ) ? x 4 5 av ( f ) ? b 2 av ( f ) ? a 5 ( b 2 a ) av ( f ) 5 ( b 2 a ) 3 } b 2 1 a } E b a f ( x ) dx 4 5 E b a f ( x ) dx 40. (a) 300 mi (b) } 3 1 0 50 m m ph i } 1 } 5 1 0 50 m m ph i } 5 8 h (c) } 30 8 0 h mi } 5 37.5 mph (d) The average speed is the total distance divided by the total time. Algebraically, } d t 1 1 1 1 d t 2 2 } . The driver computed } 1 2 } 1 } d t 1 1 } 1 } d t 2 2 } 2 . The two expressions are not equal. 41. Time for first release 5 } 1 1 0 0 m 00 3 / m m 3 in } 5 100 min Time for second release 5 } 2 1 0 0 m 00 3 / m m 3 in } 5 50 min Average rate 5 } to t t o al ta r l e t l i e m as e ed } 5 } 2 1 0 5 0 0 0 m m in 3 } 5 13 } 1 3 } m 3 /min 42. E 1 0 sin x dx # E 1 0 x dx 5 3 } 1 2 } x 2 4 5 } 1 2 } 43. E 1 0 sec x dx \$ E 1 0 1 1 1 } x 2 2 } 2 dx 5 3 x 1 } x 6 3 } 4 5 } 7 6 } 44. (a) Area 5 } 1 2 } bh (b) } 2 h b } x 2 1 C (c) E b 0 y ( x ) dx 5 3 } 2 h b } x 2 4 5 } h 2 b b 2 } 5 } 1 2 } bh 45. av ( x k ) 5 } 1 k } E k 0 x k dx 5 } 1 k } 3 } k 1 1 1 } x k 1 1 4 5 } k ( k k k 1 1 1 1) } Graph y 1 5 } x ( x x x 1 1 1 1) } and y 2 5 x on a graphing calculator and find the point of intersection for x . 1.
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