# And then convinced of the error he abruptly dismissed

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Chapter 34 / Exercise 34-13
Principles of Instrumental Analysis
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and then, convinced of the error, he abruptly dismissed the meeting – to the astonishment of most of the audience. It was an evidence of intellectual courage as well as honesty and doubtless won for him the supreme admiration of every person in the group – an admiration which was in no ways diminished, but rather increased, when at a later meeting he announced that after all he had been able to prove the step to be correct. —H. E. Slaught
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Chapter 34 / Exercise 34-13
Principles of Instrumental Analysis
Skoog Expert Verified
124 The AP CALCULUS PROBLEM BOOK 4.19 Multiple Choice Problems on Integrals 1102 (AP) . For any real number b , b 0 | 2 x | dx is A) b | b | B) b 2 C) b 2 D) b | b | E) None of these 1103 (AP) . Let f and g have continuous first and second derivatives everywhere. If f ( x ) g ( x ) for all real x , which of the following must be true? I) f ( x ) g ( x ) for all real x II) f ′′ ( x ) g ′′ ( x ) for all real x III) 1 0 f ( x ) dx 1 0 g ( x ) dx A) None B) I only C) III only D) I and II E) I, II, and III 1104 (AP) . Let f be a continuous function on the closed interval [0 , 2]. If 2 f ( x ) 4, then the greatest possible value of 2 0 f ( x ) dx is A) 0 B) 2 C) 4 D) 8 E) 16 1105 (AP) . If f is the continuous, strictly increasing function on the interval [ a, b ] as shown below, which of the following must be true? I) b a f ( x ) dx < f ( b )( b a ) II) b a f ( x ) dx > f ( a )( b a ) III) b a f ( x ) dx = f ( c )( b a ) for some c in [ a, b ]. a b f ( x ) A) I only B) II only C) III only D) I and II E) I, II, and III 1106 (AP) . Which of the following definite integrals is not equal to zero? A) π π sin 3 x dx B) π π x 2 sin x dx C) π 0 cos x dx D) π π cos 3 x dx E) π π cos 2 x dx Archimedes will be remembered when Aeschylus is forgotten, because languages die and mathematical ideas do not. —G. H. Hardy
CHAPTER 4. INTEGRALS 125 1107. π / 2 π / 6 cot x dx = A) ln 1 2 B) ln 2 C) 1 2 D) ln( 3 1) E) None of these 1108. 3 2 | x + 1 | dx = A) 5 2 B) 17 2 C) 9 2 D) 11 2 E) 13 2 1109. 2 1 (3 x 2) 3 dx = A) 16 3 B) 63 4 C) 13 3 D) 85 4 E) None of these 1110. π / 2 π / 4 sin 3 θ cos θ d θ = A) 3 16 B) 1 8 C) 1 8 D) 3 16 E) 3 4 1111. 1 0 e x (3 e x ) 2 dx = A) 3 ln( e 3) B) 1 C) 1 3 e D) e 1 2(3 e ) E) e 2 3 e 1112. 0 1 e x dx = A) 1 e B) 1 e e C) e 1 D) 1 1 e E) e + 1 1113. 1 0 x x 2 + 1 dx = A) π 4 B) ln 2 C) 1 2 (ln 2 1) D) 3 2 E) ln 2 Anyone who cannot cope with mathematics is not fully human. At best he is a tolerable subhuman who has learned to wear shoes, bathe, and not make messes in the house. —Robert A. Heinlein
126 The AP CALCULUS PROBLEM BOOK 1114. The acceleration of a particle moving along a straight line is given by a = 6 t . If, when t = 0 its velocity v = 1 and its distance s = 3, then at any time t the position function is given by A) s = t 3 + 3 t + 1 B) s = t 3 + 3 C) s = t 3 + t + 3 D) s = 1 3 t 3 + t + 3 E) s = 1 3 t 3 + 1 2 t 2 + 3
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