int cal 10

# int cal 10 - pokharel(yp624 – HW09 – Radin –(57410 1...

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Unformatted text preview: pokharel (yp624) – HW09 – Radin – (57410) 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 + 4 x- 1 dx . 1. I = 15 2- 4 ln 3 2. I = 7 2 + 5 ln2 correct 3. I = 15 2 + 4 ln 3 4. I = 7 2 + 4 ln 3 2 5. I = 7 2- 4 ln2 6. I = 15 2- 5 ln 3 2 Explanation: After division x 2 + 4 x- 1 = ( x 2- 1) + 5 x- 1 = x 2- 1 x- 1 + 5 x- 1 = x + 1 + 5 x- 1 . In this case I = integraldisplay 3 2 parenleftBig x + 1 + 5 x- 1 parenrightBig dx = bracketleftBig 1 2 x 2 + x + 5 ln | x- 1 | bracketrightBig 3 2 = parenleftbigg 15 2- 4 parenrightbigg + 5 parenleftBig ln 2- ln 1 parenrightBig . Consequently, I = 7 2 + 5 ln2 . 002 10.0 points Evaluate the integral I = integraldisplay π/ 4 3 + 3 tan 2 x cos 2 x dx . 1. I = 3 2. I = 4 correct 3. I = 7 4. I = 6 5. I = 5 Explanation: Since 1 cos 2 x = sec 2 x , we see that 3 + 3 tan 2 x cos 2 x = (3 + 3 tan 2 x )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 = π 4 = ⇒ u = 1 . In this case, I = integraldisplay 1 (3 + 3 u 2 ) du = bracketleftBig 3 u + u 3 bracketrightBig 1 . Consequently, I = 4 . pokharel (yp624) – HW09 – Radin – (57410) 2 003 10.0 points Evaluate the integral I = integraldisplay π/ 6 1- tan 2 x sec 2 x dx . 1. I =- √ 3 4 2. I =- 1 2 3. I = √ 3 4 correct 4. I = 1 2 5. I =- 1 2 √ 2 6. I = 1 2 √ 2 Explanation: Since 1 sec 2 x = cos 2 x , tan x = sin x cos x , we see that 1- tan 2 x sec 2 x = (1- tan 2 x )cos 2 x = cos 2 x- sin 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 π/ 6 cos 2 x dx = bracketleftBig 1 2 sin 2 x bracketrightBig π/ 6 . keywords: 004 10.0 points Evaluate the definite integral I = integraldisplay 1 6 (1 + x 2 ) 3 / 2 dx . 1. I = 6 2. I = 3 √ 2 correct 3. I = 3 2 4. I = 6 √ 2 5. I = 3 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 = π 4 . Thus I = integraldisplay π/ 4 6 sec 2 u sec 3 u du = 6 integraldisplay π/ 4 cos u du = 6 bracketleftBig sin u bracketrightBig π/ 4 . Consequently, I = 3 √ 2 . pokharel (yp624) – HW09 – Radin – (57410) 3 005 10.0 points Determine the indefinite integral I = integraldisplay radicalbigg 3 + x 3- x dx . 1. I = 3 sin- 1 parenleftBig x 3 parenrightBig- radicalbig 9- x 2 + C correct 2. I = 3 parenleftBig tan- 1 parenleftBig x 3 parenrightBig- radicalbig 9- x 2 parenrightBig + C 3. I = 3 sin- 1 parenleftBig x 3 parenrightBig + radicalbig 9- x 2 + C 4. I = 3 parenleftBig sin- 1 parenleftBig x 3 parenrightBig- radicalbig 9- x 2 parenrightBig + C 5. I = 3 tan- 1 parenleftBig x 3 parenrightBig + radicalbig...
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int cal 10 - pokharel(yp624 – HW09 – Radin –(57410 1...

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