{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW02-solutions

# HW02-solutions - wei(jw35975 – HW02 – kalahurka...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

HW02-solutions - wei(jw35975 – HW02 – kalahurka...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online