M408L_EXAM02

# M408L_EXAM02 - Version 104 – EXAM 2 – Berg –(58460 1...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 104 – EXAM 2 – Berg – (58460) 1 This print-out should have 16 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 5 4 7 ( x- 3)(9- x ) dx . 1. I = 7 6 ln parenleftbigg 5 2 parenrightbigg correct 2. I = 7 6 ln parenleftbigg 7 4 parenrightbigg 3. I = 7 ln parenleftbigg 7 4 parenrightbigg 4. I = 1 6 ln parenleftbigg 5 2 parenrightbigg 5. I = 1 6 ln parenleftbigg 7 4 parenrightbigg 6. I = 7 ln parenleftbigg 5 2 parenrightbigg Explanation: To find A, B so that 7 ( x- 3)(9- x ) = A x- 3 + B 9- x , we first bring the right hand side to a common denominator. In this case, 7 ( x- 3)(9- x ) = A (9- x ) + B ( x- 3) ( x- 3)(9- x ) , and so A (9- x ) + B ( x- 3) = 7 . To find the values of A, B , we can make par- ticular choices of x since the equation holds for all x . When x = 3, therefore, A = 7 6 , and when x = 9, B = 7 6 . Thus I = integraldisplay 5 4 7 6 parenleftBig 1 x- 3 + 1 9- x parenrightBig dx. Hence after integration, I = bracketleftbigg 7 6 { ln( x- 3)- ln(9- x ) } bracketrightbigg 5 4 = 7 6 bracketleftbigg ln parenleftbigg x- 3 9- x parenrightbiggbracketrightbigg 5 4 . Consequently, I = 7 6 ln parenleftbigg 5 2 parenrightbigg . 002 10.0 points Evaluate the definite integral I = integraldisplay π/ 2 sin x 1 + cos 2 x dx . 1. I = 1 4 2. I = 1 3. I = 1 2 π 4. I = π 5. I = 1 2 6. I = 1 4 π correct Explanation: Let u = cos x . Then du =- sin x dx , while x = 0 = ⇒ u = 1 , x = π 2 = ⇒ u = 0 . In this case I =- integraldisplay 1 1 1 + u 2 du = integraldisplay 1 1 1 + u 2 du = bracketleftBig tan − 1 u bracketrightBig 1 . Version 104 – EXAM 2 – Berg – (58460) 2 Consequently, I = 1 4 π . 003 10.0 points Evaluate the definite integral I = integraldisplay 1 (8 x 2- 3) e x dx . 1. I = 5( e- 1) 2. I = 13 e- 5 3. I = 13 e + 5 4. I = 5 e + 13 5. I = 5 e- 13 correct Explanation: After Integration by Parts once, integraldisplay 1 (8 x 2- 3) e x dx = bracketleftBig (8 x 2- 3) e x bracketrightBig 1- 16 integraldisplay 1 xe x dx. To evaluate this last integral we Integrate by Parts once again. For then 16 integraldisplay 1 xe x dx = bracketleftBig 16 xe x bracketrightBig 1- 16 integraldisplay 1 e x . Consequently, I = bracketleftBig (8 x 2- 3- 16 x + 16) e x bracketrightBig 1 , and so I = 5 e- 13....
View Full Document

{[ snackBarMessage ]}

### Page1 / 8

M408L_EXAM02 - Version 104 – EXAM 2 – Berg –(58460 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online