HW02-solutions - wei (jw35975) HW02 kalahurka (55230) 1...

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Unformatted text preview: wei (jw35975) HW02 kalahurka (55230) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points If g ( x ) is continuous on [a,b] and m is a constant, determine the integral I = integraldisplay b a m g ( x ) dx without further restrictions on g and m . 1. I = g ( b )- g ( a ) b- a 2. I = m ( b- a ) 3. I = m ( g ( b )- g ( a )) 4. I = m ( G ( b )- G ( a )) , G ( x ) = g ( x ) correct 5. I = m ( G ( a )- G ( b )) , G ( x ) = g ( x ) 6. I = bracketleftBig m g ( x ) 2 2 bracketrightBig b a Explanation: One version of the Fundamental Theorem of Calculus says that g has an anti-derivative G , while another version then says that integraldisplay b a g ( x ) dx = G ( b )- G ( a ) . But if G ( x ) is an anti-derivative of g ( x ), then mG ( x ) is an anti-derivative of mg ( x ) for any constant m . Consequently, again by the Fun- damental Theorem of Calculus, I = m ( G ( b )- G ( a )) . 002 10.0 points A function h has graph 2- 2 2- 2 on (- 4 , 4). If f is defined on (- 4 , 4) by f ( x ) = - 1 ,- 4 < x <- 3 , integraldisplay x 3 h ( t ) dt,- 3 x < 4 , which of the following is the graph of f ? 1. 2- 2 2- 2- 4 2. 2- 2 2- 2 correct wei (jw35975) HW02 kalahurka (55230) 2 3. 2- 2 2- 2 4. 2- 2 2- 2 5. 2- 2 2 4- 2 Explanation: Since f (- 3) = integraldisplay 3 3 h ( t ) dt = 0 , two of the five graphs can be eliminated im- mediately. On the other hand, by the Fun- damental Theorem of Calculus, f ( x ) = h ( x ) on (- 3 , 4); in particular, the critical points of f occur at the x-intercepts of the graph of h . As these x-intercepts occur at- 1 , , 2, this eliminates a third graph. Thus the remain- ing two possible graphs for f both have the same critical points and to decide which one is the graph of f we can use the first derivative test because the graph of f will have a local maximum at an x-intercept of the graph of h where it changes from positive to negative values, and a local minimum at an x-intercept where h changes from negative to positive val- ues. Consequently, the graph of f must be 2- 2 2- 2 keywords: 003 10.0 points10....
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HW02-solutions - wei (jw35975) HW02 kalahurka (55230) 1...

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