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HW09-solutions

# HW09-solutions - kim(tk5895 – HW09 – Henry –(54974 1...

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Unformatted text preview: kim (tk5895) – HW09 – Henry – (54974) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine the indefinite integral I = integraldisplay 1 √ x 2 + 4 x − 21 dx 1. I = sin- 1 parenleftBig x + 2 5 parenrightBig + C 2. I = ln vextendsingle vextendsingle vextendsingle x − 2 + radicalbig x 2 + 4 x − 21 vextendsingle vextendsingle vextendsingle + C 3. I = ln vextendsingle vextendsingle vextendsingle x + 2 + radicalbig x 2 + 4 x − 21 vextendsingle vextendsingle vextendsingle + C correct 4. I = ln vextendsingle vextendsingle vextendsingle x + 2 + radicalbig x 2 − 4 x − 21 vextendsingle vextendsingle vextendsingle + C 5. I = ln vextendsingle vextendsingle vextendsingle x − 2 + radicalbig x 2 − 4 x − 21 vextendsingle vextendsingle vextendsingle + C 6. I = sin- 1 parenleftBig x + 5 2 parenrightBig + C Explanation: By completing the square we see that x 2 + 4 x − 21 = ( x 2 + 4 x + 4 ) − 25 = ( x + 2) 2 − 25 . This suggests the substitution x + 2 = 5 sec θ , for then dx = 5 sec θ tan θ dθ , while ( x + 2) 2 − 25 = 25tan 2 θ . In this case I = integraldisplay 5 sec θ tan θ 5 tan θ dθ = integraldisplay sec θ dθ = ln | sec θ + tan θ | + C . Now x + 2 = 5 sec θ = ⇒ tan θ = radicalbig ( x + 2) 2 − 25 5 , so I = ln vextendsingle vextendsingle vextendsingle x + 2 + radicalbig ( x + 2) 2 − 25 5 vextendsingle vextendsingle vextendsingle + C . Consequently I = ln vextendsingle vextendsingle vextendsingle x + 2 + radicalbig x 2 + 4 x − 21 vextendsingle vextendsingle vextendsingle + C . 002 10.0 points Evaluate the definite integral I = integraldisplay 1- 1 e 3 arctan y 1 + y 2 dy . 1. I = − 1 3 e 3 π/ 4 − 1 3 e- 3 π/ 4 2. I = 1 4 e 3 π/ 4 − 1 4 e- 3 π/ 4 3. I = 1 4 e 3 π/ 4 + 1 4 e- 3 π/ 4 4. I = − 1 4 e 3 π/ 4 + 1 4 e- 3 π/ 4 5. I = 1 3 e 3 π/ 4 − 1 3 e- 3 π/ 4 correct 6. I = 1 3 e 3 π/ 4 + 1 3 e- 3 π/ 4 Explanation: Set u = arctan y . Then du = 1 1 + y 2 dy , in which case I = integraldisplay π/ 4- π/ 4 e 3 u du = bracketleftBig e 3 u 3 bracketrightBig π/ 4- π/ 4 . kim (tk5895) – HW09 – Henry – (54974) 2 Consequently, I = 1 3 ( e 3 π/ 4 − e- 3 π/ 4 ) . 003 10.0 points Evaluate the definite integral I = integraldisplay 2 1 x 2 + 2 x + 1 dx . 1. I = − 1 2 + 2 ln3 2. I = 1 2 − 3 ln 3 2 3. I = − 1 2 − 3 ln2 4. I = 1 2 + 2 ln2 5. I = − 1 2 − 2 ln3 6. I = 1 2 + 3 ln 3 2 correct Explanation: After division x 2 + 2 x + 1 = ( x 2 − 1) + 3 x + 1 = x 2 − 1 x + 1 + 3 x + 1 = x − 1 + 3 x + 1 . In this case I = integraldisplay 2 1 parenleftBig x − 1 + 3 x + 1 parenrightBig dx = bracketleftBig 1 2 x 2 − x + 3 ln | x + 1 | bracketrightBig 2 1 = parenleftBig − parenleftBig − 1 2 parenrightBigparenrightBig + 3 parenleftBig ln3 − ln2 parenrightBig ....
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HW09-solutions - kim(tk5895 – HW09 – Henry –(54974 1...

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