M408L hwk3 solutions - Pham, Quoc Homework 3 Due: Feb 6...

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Unformatted text preview: Pham, Quoc Homework 3 Due: Feb 6 2007, 3:00 am Inst: Eric Katerman 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Evaluate the integral I = Z / 3 2sin2 x cos x dx. 1. I = 1 2. I = 4(2- 3) 3. I = 2- 2 4. I = 2(2- 2) 5. I = 4 6. I = 2 correct Explanation: Since sin2 x = 2sin x cos x, the integrand can be rewritten as 2sin2 x cos x = 4sin x. But then I = Z / 3 4sin xdx = h- 4cos x i / 3 . Consequently, I = 2 . keywords: definite integral, trig functions double angle formula, 002 (part 1 of 1) 10 points Evaluate the integral I = Z 4 d dx (2 + 3 x 2 ) 1 / 2 dx. 1. I = 6 2 2. I = 2 3. I = 7 2 4. I = 5 2 5. I = 4 2 correct Explanation: As an indefinite integral, Z d dx (2 + 3 x 2 ) 1 / 2 dx = (2 + 3 x 2 ) 1 / 2 + C where C is an arbitrary constant. Thus Z 4 d dx (2 + 3 x 2 ) 1 / 2 dx = h (2 + 3 x 2 ) 1 / 2 i 4 . Consequently, I = 4 2 . keywords: definite integral, derivative, square root 003 (part 1 of 1) 10 points Evaluate the definite integral I = Z 4 1 x 3- 4 x dx. 1. I = 10 2. I = 6 correct 3. I = 8 4. I = 7 Pham, Quoc Homework 3 Due: Feb 6 2007, 3:00 am Inst: Eric Katerman 2 5. I = 9 Explanation: We first expand x 3- 4 x = 3 x- 4 x , and then integrate term by term. This gives I = h 2 x 3 / 2- 8 x 1 / 2 i 4 1 . Consequently, I = 6 . keywords: definite integral, power function, square root function 004 (part 1 of 1) 10 points Evaluate the integral I = Z / 4 (6sec 2 - 5sin ) d. 1. I = 11- 5 2 2. I = 1 + 6 2 3. I = 11 + 5 2 4. I = 1 + 5 2 correct 5. I = 1- 6 2 Explanation: By the Fundamental theorem of calculus, I = h 6tan + 5cos i / 4 = 6 + 5 2 - 5 = 1 + 5 2 . keywords: integral, FTC, trig function 005 (part 1 of 1) 10 points Evaluate the definite integral I = Z 4 ( | x- 1 |- x ) dx. 1. I = 3 2. I =- 3 correct 3. I = 2 4. I =- 2 5. I =- 4 6. I = 4 Explanation: Since | x- 1 | = 1- x, x < 1, x- 1 , x 1, it follows that | x- 1 |- x = 1- 2 x, x < 1,- 1 , x 1. Thus we split the integral I into two parts I = Z 1 (1- 2 x ) dx- Z 4 1 1 dx = I 1- I 2 . Then I 1 = h x- x 2 i 1 = 0 . Similarly, I 2 = h x i 4 1 = 3 . Consequently, I =- 3 . keywords: definite integral, absolute value, piecewise-defined integrand Pham, Quoc Homework 3 Due: Feb 6 2007, 3:00 am Inst: Eric Katerman 3 006 (part 1 of 1) 10 points Oil is flowing from a well in a continuous stream at a rate of f ( t ) = 108 ( t + 3) 2 barrels per month (in multiples of 100). Find the total oil produced by the well in its second three months of operation....
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This homework help was uploaded on 03/19/2008 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas at Austin.

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M408L hwk3 solutions - Pham, Quoc Homework 3 Due: Feb 6...

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