# hw07 - padilla(tp5647 – HW07 – cheng –(57455 1 This...

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Unformatted text preview: padilla (tp5647) – HW07 – cheng – (57455) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine the integral I = integraldisplay 1 1 + 25( x- 6) 2 dx . 1. I = sin − 1 5( x- 6) + C 2. I = 1 5 tan − 1 5( x- 6) + C correct 3. I = 5 sin − 1 parenleftBig x- 6 5 parenrightBig + C 4. I = tan − 1 5( x- 6) + C 5. I = 5 tan − 1 parenleftBig x- 6 5 parenrightBig + C 6. I = 1 5 sin − 1 5( x- 6) + C Explanation: Since d dx tan − 1 x = 1 1 + x 2 , the substitution u = 5( x- 6) is suggested. For then du = 5 dx , in which case I = 1 5 integraldisplay 1 1 + u 2 du = 1 5 tan − 1 u + C , with C an arbitrary constant. Consequently, I = 1 5 tan − 1 5( x- 6) + C . keywords: 002 10.0 points Determine the integral I = integraldisplay 1 8 √ 4- x 2 dx . 1. I = 4 3 π correct 2. I = 2 3. I = 8 3 4. I = 4 3 5. I = 8 3 π 6. I = 2 π Explanation: Since integraldisplay 1 √ 1- x 2 dx = sin − 1 x + C , we need to reduce I to an integal of this form by changing the x variable. Indeed, set x = 2 u . Then dx = 2 du while x = 0 = ⇒ u = 0 and x = 1 = ⇒ u = 1 2 . In this case I = 16 integraldisplay 1 / 2 1 2 √ 1- u 2 du = 8 integraldisplay 1 / 2 1 √ 1- u 2 du . Consequently, I = bracketleftBig 8 sin − 1 u bracketrightBig 1 / 2 = 4 3 π . keywords: 003 10.0 points padilla (tp5647) – HW07 – cheng – (57455) 2 Evaluate the definite integral I = integraldisplay π 2 sin θ 1 + cos 2 θ dθ . 1. I = 3 2 π 2. I = 3 4 π 3. I = 5 4 π 4. I = π correct 5. I = 7 4 π Explanation: Since d dθ cos θ =- sin θ , the substitution u = cos θ is suggested. For then du =- sin θ dθ , while θ = 0 = ⇒ u = 1 , θ = π = ⇒ u =- 1 , so that I =- 2 integraldisplay − 1 1 1 1 + u 2 du = 2 integraldisplay 1 − 1 1 1 + u 2 du , which can now be integrated using the fact that d du tan − 1 u = 1 1 + u 2 . Consequently, I = 2 bracketleftBig tan − 1 u bracketrightBig 1 − 1 = π since tan − 1 (- 1) =- π 4 , tan − 1 1 = π 4 . 004 10.0 points Determine the integral I = integraldisplay 1 2 tan − 1 x 1 + x 2 dx . 1. I = π 16 2. I = π 2 4 3. I = π 4 4. I = π 8 5. I = π 2 8 6. I = π 2 16 correct Explanation: Set u = tan − 1 x . Then du = 1 1 + x 2 dx , while x = 0 = ⇒ u = 0 , x = 1 = ⇒ u = π 4 . In this case, I = integraldisplay π/ 4 2 u du = bracketleftBig u 2 bracketrightBig π/ 4 . Consequently, I = 1 16 π 2 . keywords: 005 10.0 points Determine the integral I = integraldisplay 1- x √ 1- x 2 dx . padilla (tp5647) – HW07 – cheng – (57455) 3 1. I = 1 2 tan − 1 x + radicalbig 1- x 2 + C 2. I = 1 2 sin − 1 x- radicalbig 1- x 2 + C 3. I = tan − 1 x- radicalbig 1- x 2 + C 4. I = 1 2 tan − 1 x- 1 2 radicalbig 1- x 2 + C 5. I = sin − 1 x + 1 2 radicalbig 1- x 2 + C 6. I = sin − 1 x + radicalbig 1- x 2 + C correct Explanation: We deal with the two integrals I 1 = integraldisplay 1 √ 1- x 2 dx, I 2 = x √...
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hw07 - padilla(tp5647 – HW07 – cheng –(57455 1 This...

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