Cal 2 Test 2

# Cal 2 Test 2 - Version 080 – L EXAM 2 – Henry –(54974...

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Unformatted text preview: Version 080 – L EXAM 2 – Henry – (54974) 1 This print-out should have 18 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 2 √ 2 8 x √ x 2- 1 dx . 1. I = 4 3 2. I = 2 3 3. I = 2 3 π correct 4. I = 4 3 π 5. I = 1 6. I = π Explanation: Set x = sec u . Then dx = sec u tan u du , x 2- 1 = tan 2 u , while x = √ 2 = ⇒ u = π 4 , x = 2 = ⇒ u = π 3 . In this case, I = 8 integraldisplay π/ 3 π/ 4 sec u tan u sec u tan u du = integraldisplay π/ 3 π/ 4 8 du . Consequently, I = 8 parenleftBig π 3- π 4 parenrightBig = 2 3 π . 002 10.0 points Determine the integral I = integraldisplay x (ln x ) 2 dx . 1. I = x 2 parenleftBig (ln x ) 2 + ln x + 1 2 parenrightBig + C 2. I = 1 2 x 2 parenleftBig (ln x ) 2- ln x + 1 2 parenrightBig + C correct 3. I = x 2 parenleftBig (ln x ) 2 + ln x- 1 2 parenrightBig + C 4. I = 1 2 x 2 parenleftBig (ln x ) 2 + ln x + 1 2 parenrightBig + C 5. I = 1 2 x 2 parenleftBig (ln x ) 2- ln x- 1 2 parenrightBig + C 6. I = x 2 parenleftBig (ln x ) 2- ln x + 1 2 parenrightBig + C Explanation: After integration by parts, integraldisplay x (ln x ) 2 dx = 1 2 x 2 (ln x ) 2- integraldisplay x 2 1 x ln x dx = 1 2 x 2 (ln x ) 2- integraldisplay x ln x dx. But after integration by parts once again, integraldisplay x ln x dx = 1 2 x 2 ln x- 1 2 integraldisplay x 2 1 x dx = 1 2 x 2 ln x- 1 2 integraldisplay x dx = 1 2 x 2 ln x- 1 4 x 2 + C. Thus integraldisplay x (ln x ) 2 dx = 1 2 x 2 (ln x ) 2- 1 2 x 2 ln x + 1 4 x 2 + C. Consequently, I = 1 2 x 2 parenleftBig (ln x ) 2- ln x + 1 2 parenrightBig + C . Version 080 – L EXAM 2 – Henry – (54974) 2 keywords: integration by parts, log function 003 10.0 points Evaluate the integral I = integraldisplay π/ 4 sec 4 x dx . 1. I = 5 3 2. I = 1 3 3. I = 4 3 correct 4. I = 2 3 5. I = 1 Explanation: Since (tan x ) ′ = sec 2 x , sec 2 x = 1 + tan 2 x , use of the substitution u = tan x is suggested. For then du = sec 2 x dx , while x = 0 = ⇒ u = 0 , x = π 4 = ⇒ u = 1 . Thus I = integraldisplay π/ 4 (tan 2 x + 1) sec 2 x dx = integraldisplay 1 ( u 2 + 1) du = bracketleftBig 1 3 u 3 + u bracketrightBig 1 . Consequently, I = 4 3 . 004 10.0 points Evaluate the integral I = integraldisplay 2 1 4 x 3 + 4 x dx . 1. I = 1 2 ln parenleftBig 5 2 parenrightBig correct 2. I = 1 2 ln parenleftBig 8 5 parenrightBig 3. I = ln parenleftBig 5 2 parenrightBig 4. I = ln parenleftBig 8 5 parenrightBig 5. I = 1 4 ln parenleftBig 8 5 parenrightBig 6. I = 1 4 ln parenleftBig 5 2 parenrightBig Explanation: By partial fractions, 4 x 3 + 4 x = A x + Bx + C x 2 + 4 . To determine A, B, and C multiply through by x 3 + 4 x : for then 4 = A ( x 2 + 4) + x ( Bx + C ) = ( A + B ) x 2 + Cx + 4 A , which after comparing coefficients gives A =- B , C = 0 , A = 1 ....
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Cal 2 Test 2 - Version 080 – L EXAM 2 – Henry –(54974...

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