HW09 Solution

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badruddin (ssb776) – 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 + 5 x - 1 dx . 1. I = 7 2 + 5 ln 3 2 2. I = 15 2 - 6 ln 3 2 3. I = 15 2 + 5 ln 3 4. I = 7 2 - 5 ln 2 5. I = 7 2 + 6 ln 2 correct 6. I = 15 2 - 5 ln 3 Explanation: After division x 2 + 5 x - 1 = ( x 2 - 1) + 6 x - 1 = x 2 - 1 x - 1 + 6 x - 1 = x + 1 + 6 x - 1 . In this case I = integraldisplay 3 2 parenleftBig x + 1 + 6 x - 1 parenrightBig dx = bracketleftBig 1 2 x 2 + x + 6 ln | x - 1 | bracketrightBig 3 2 = parenleftbigg 15 2 - 4 parenrightbigg + 6 parenleftBig ln 2 - ln 1 parenrightBig . Consequently, I = 7 2 + 6 ln 2 . 002 10.0 points Evaluate the integral I = integraldisplay π/ 4 0 2 - 3 tan 2 x cos 2 x dx . 1. I = 2 2. I = 1 correct 3. I = 4 4. I = 5 5. I = 3 Explanation: Since 1 cos 2 x = sec 2 x , we see that 2 - 3 tan 2 x cos 2 x = (2 - 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 0 (2 - 3 u 2 ) du = bracketleftBig 2 u - u 3 bracketrightBig 1 0 . Consequently, I = 1 .

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badruddin (ssb776) – HW09 – Radin – (57410) 2 003 10.0 points Evaluate the integral I = integraldisplay π/ 6 0 1 - tan 2 x sec 2 x dx . 1. I = - 1 2 2 2. I = 3 4 correct 3. I = - 3 4 4. I = 1 2 2 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 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 0 cos 2 x dx = bracketleftBig 1 2 sin 2 x bracketrightBig π/ 6 0 . keywords: 004 10.0 points Evaluate the definite integral I = integraldisplay 1 0 6 (1 + x 2 ) 3 / 2 dx . 1. I = 6 2. I = 3 2 2 3. I = 3 2 4. I = 3 5. I = 3 2 correct 6. I = 6 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 0 6 sec 2 u sec 3 u du = 6 integraldisplay π/ 4 0 cos u du = 6 bracketleftBig sin u bracketrightBig π/ 4 0 . Consequently, I = 3 2 . 005 10.0 points
badruddin (ssb776) – HW09 – Radin – (57410) 3 Determine the indefinite integral I = integraldisplay radicalbigg 2 + x 2 - x dx . 1. I = 2 sin - 1 parenleftBig x 2 parenrightBig - radicalbig 4 - x 2 + C correct 2. I = 2 tan - 1 parenleftBig x 2 parenrightBig - radicalbig 4 - x 2 + C 3. I = 2 tan - 1 parenleftBig x 2 parenrightBig + radicalbig 4 - x 2 + C 4. I = 2 sin - 1 parenleftBig x 2 parenrightBig + radicalbig 4 - x 2 + C 5. I = 2 parenleftBig tan - 1 parenleftBig x 2 parenrightBig - radicalbig 4 - x 2 parenrightBig + C 6. I = 2 parenleftBig sin - 1 parenleftBig x 2 parenrightBig - radicalbig 4 - x 2 parenrightBig + C Explanation: Rationalizing the numerator we see that radicalbigg 2 + x 2 - x = 2 + x 2 - x 2 + x = 2 + x 4 - x 2 , in which case I = integraldisplay 2 + x 4 - x 2 dx .

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