hw09 - padilla (tp5647) HW09 cheng (57455) 1 This print-out...

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Unformatted text preview: padilla (tp5647) HW09 cheng (57455) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Evaluate the definite integral I = integraldisplay 3 2 x 2 + 3 x- 1 dx . 1. I = 15 2 + 3 ln 3 2. I = 15 2- 3 ln 3 3. I = 7 2 + 3 ln 3 2 4. I = 15 2- 4 ln 3 2 5. I = 7 2 + 4 ln2 correct 6. I = 7 2- 3 ln2 Explanation: After division x 2 + 3 x- 1 = ( x 2- 1) + 4 x- 1 = x 2- 1 x- 1 + 4 x- 1 = x + 1 + 4 x- 1 . In this case I = integraldisplay 3 2 parenleftBig x + 1 + 4 x- 1 parenrightBig dx = bracketleftBig 1 2 x 2 + x + 4 ln | x- 1 | bracketrightBig 3 2 = parenleftbigg 15 2- 4 parenrightbigg + 4 parenleftBig ln 2- ln 1 parenrightBig . Consequently, I = 7 2 + 4 ln2 . 002 10.0 points Evaluate the integral I = integraldisplay / 4 3 tan 2 x- 2 cos 2 x dx . 1. I = 0 2. I = 1 3. I =- 1 correct 4. I =- 2 5. I = 2 Explanation: Since 1 cos 2 x = sec 2 x , we see that 3 tan 2 x- 2 cos 2 x = (3 tan 2 x- 2)sec 2 x . On the other hand, d dx tan x = sec 2 x . So if we set u = tan x , then du = sec 2 x dx while x = 0 = u = 0 , x = 4 = u = 1 . In this case, I = integraldisplay 1 (3 u 2- 2) du = bracketleftBig u 3- 2 u bracketrightBig 1 . Consequently, I =- 1 . padilla (tp5647) HW09 cheng (57455) 2 003 10.0 points Evaluate the integral I = integraldisplay / 6 tan 2 x- 1 sec 2 x dx . 1. I = 1 2 2 2. I = 3 4 3. I =- 1 2 2 4. I =- 3 4 correct 5. I = 1 2 6. I =- 1 2 Explanation: Since 1 sec 2 x = cos 2 x , tan x = sin x cos x , we see that tan 2 x- 1 sec 2 x = (tan 2 x- 1)cos 2 x = sin 2 x- cos 2 x . There are two ways to proceed now: 1. use double angle formulas : cos 2 x = 1 2 (1 + cos 2 x ) , sin 2 x = 1 2 (1- cos 2 x ) , 2. or the known trig identity : cos 2 x = cos 2 x- sin 2 x . But in either case it follows that I =- integraldisplay / 6 cos 2 x dx = bracketleftBig- 1 2 sin 2 x bracketrightBig / 6 . keywords: 004 10.0 points Evaluate the definite integral I = integraldisplay 1 6 (1 + x 2 ) 3 / 2 dx . 1. I = 3 2 2. I = 3 3. I = 6 2 4. I = 6 5. I = 3 2 correct 6. I = 3 2 2 Explanation: Set x = tan u . Then dx = sec 2 u du, (1 + x 2 ) 3 / 2 = sec 3 u , while x = 0 = u = 0 , x = 1 = u = 4 . Thus I = integraldisplay / 4 6 sec 2 u sec 3 u du = 6 integraldisplay / 4 cos u du = 6 bracketleftBig sin u bracketrightBig / 4 . Consequently, I = 3 2 . 005 10.0 points padilla (tp5647) HW09 cheng (57455) 3 Determine the indefinite integral I = integraldisplay radicalbigg 5 + x 5- x dx . 1. I = 5 sin- 1 parenleftBig x 5 parenrightBig + radicalbig 25- x 2 + C 2. I = 5 parenleftBig tan- 1 parenleftBig x 5 parenrightBig- radicalbig 25- x 2 parenrightBig + C 3. I = 5 sin- 1 parenleftBig x 5 parenrightBig- radicalbig 25- x 2 + C correct 4. I = 5 parenleftBig sin- 1 parenleftBig x 5 parenrightBig- radicalbig 25- x 2 parenrightBig + C 5. I = 5 tan- 1 parenleftBig x 5 parenrightBig + radicalbig...
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hw09 - padilla (tp5647) HW09 cheng (57455) 1 This print-out...

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