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
Unformatted text preview: husain (aih243) HW14 Gilbert (56215) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Evaluate the definite integral I = integraldisplay 1 (2 e x e x ) 2 dx . 1. I = 2 e 2 + 7 + e 2 2. I = 2 e 2 7 e 2 3. I = 2 e 2 + 4 + e 2 4. I = 2 e 2 11 2 + 1 2 e 2 5. I = 2 e 2 + 4 1 2 e 2 6. I = 2 e 2 11 2 1 2 e 2 correct Explanation: After expansion (2 e x e x ) 2 = 4 e 2 x 4 + e 2 x , in which case I = integraldisplay 1 ( 4 e 2 x 4 + e 2 x ) dx = bracketleftBig 2 e 2 x 4 x 1 2 e 2 x bracketrightBig 1 . Consequently, I = 2 e 2 11 2 1 2 e 2 . 002 10.0 points Determine the indefinite integral I = integraldisplay 4 x 2 x 2 x dx. 1. I = 2 x 2 x 2 ln  x  2. I = 2 x 2 x 2 ln  x  + C correct 3. I = 4 x 2 + x + 2 ln  x  + C 4. I = 4 x 2 x 2 ln  x  + C 5. I = 4 x 2 + x + 2 ln  x  6. I = 2 x 2 + x + 2 ln  x  Explanation: After division, 4 x 2 x 2 x = 4 x 1 2 x , so we can now integrate term by term using the fact that integraldisplay x r dx = x r +1 r + 1 + C, ( r negationslash = 1) and that integraldisplay 1 x dx = ln  x  + C . Thus integraldisplay 4 x 2 x 2 x dx = 4 integraldisplay x dx integraldisplay dx 2 integraldisplay 1 x dx . Consequently, I = 2 x 2 x 2 ln  x  + C with C an arbitrary constant. 003 10.0 points Evaluate the definite integral I = integraldisplay 3 / 4 / 4 7 cos x 3 sin x 7 sin x + 3 cos x dx . 1. I = ln parenleftbigg 5 2 parenrightbigg correct 2. I = ln(3) husain (aih243) HW14 Gilbert (56215) 2 3. I = ln parenleftbigg 5 2 parenrightbigg 4. none of these 5. I = ln (3) Explanation: Since the integrand is of the form f ( x ) f ( x ) , f ( x ) = 7 sin x + 3 cos x this suggests the substitution u = 7 sin x + 3 cos x . For then du = (7 cos x 3 sin x ) dx, while x = 4 = u = 10 2 , x = 3 4 = u = 4 2 . Thus I = integraldisplay 4 / 2 10 / 2 1 u du = bracketleftBig ln  u  bracketrightBig 4 / 2 10 / 2 . Consequently, I = ln parenleftBig 2 5 parenrightBig = ln parenleftbigg 5 2 parenrightbigg ....
View Full
Document
 Spring '06
 McAdam

Click to edit the document details