# HW_3_Answers - Miller Kierste Homework 3 Due Feb 6 2007...

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Miller, Kierste – Homework 3 – Due: Feb 6 2007, 3:00 am – Inst: Gary Berg 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 1) 10 points Evaluate the integral I = Z π/ 8 0 2 sin 4 x cos 2 x dx. 1. I = 2 - 2 correct 2. I = 1 2 3. I = 1 4. I = 2 5. I = 1 2 (2 - 2) 6. I = 2(2 - 3) Explanation: Since sin 4 x = 2 sin 2 x cos 2 x, the integrand can be rewritten as 2 sin 4 x cos 2 x = 4 sin 2 x. But then I = Z π/ 8 0 4 sin 2 xdx = h - 2 cos 2 x i π/ 8 0 . Consequently, I = 2 - 2 . keywords: defnite integral, trig ±unctions double angle ±ormula, 002 (part 1 o± 1) 10 points Evaluate the integral I = Z 3 0 d dx (3 + 2 x 2 ) 1 / 2 dx. 1. I = 3 2. I = 21 + 3 3. I = 3 - 21 4. I = 21 - 3 correct 5. I = 21 Explanation: As an indefnite integral, Z d dx (3 + 2 x 2 ) 1 / 2 dx = (3 + 2 x 2 ) 1 / 2 + C where C is an arbitrary constant. Thus Z 3 0 d dx (3 + 2 x 2 ) 1 / 2 dx = h (3 + 2 x 2 ) 1 / 2 i 3 0 . Consequently, I = 21 - 3 . keywords: defnite integral, derivative, square root 003 (part 1 o± 1) 10 points Evaluate the defnite integral I = Z 4 1 x 3 + 1 x · dx. 1. I = 18 2. I = 19 3. I = 15 4. I = 17

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Miller, Kierste – Homework 3 – Due: Feb 6 2007, 3:00 am – Inst: Gary Berg 2 5. I = 16 correct Explanation: We frst expand x 3 + 1 x · = 3 x + 1 x , and then integrate term by term. This gives I = h 2 x 3 / 2 + 2 x 1 / 2 i 4 1 . Consequently, I = 16 . keywords: defnite integral, power ±unction, square root ±unction 004 (part 1 o± 1) 10 points Evaluate the integral I = Z π/ 4 0 (3 sec 2 θ - 9 sin θ ) dθ. 1. I = 12 - 9 2 2. I = - 6 - 3 2 3. I = - 6 + 3 2 4. I = 12 + 9 2 5. I = - 6 + 9 2 correct Explanation: By the Fundamental theorem o± calculus, I = h 3 tan θ + 9 cos θ i π/ 4 0 = 3 + 9 2 · - 9 = - 6 + 9 2 . keywords: integral, FTC, trig ±unction 005 (part 1 o± 1) 10 points Evaluate the defnite integral I = Z 6 0 ( | x - 4 |- x ) dx. 1. I = 7 2. I = - 8 correct 3. I = - 9 4. I = - 7 5. I = 8 6. I = 9 Explanation: Since | x - 4 | = 4 - x, x < 4, x - 4 , x 4, it ±ollows that | x - 4 |- x = 4 - 2 x, x < 4, - 4 , x 4. Thus we split the integral I into two parts I = Z 4 0 (4 - 2 x ) dx - Z 6 4 4 dx = I 1 - I 2 . Then I 1 = h 4 x - x 2 i 4 0 = 0 . Similarly, I 2 = h 4 x i 6 4 = 8 . Consequently, I = - 8 .
Miller, Kierste – Homework 3 – Due: Feb 6 2007, 3:00 am – Inst: Gary Berg 3 keywords: defnite integral, absolute value, piecewise-defned integrand 006 (part 1 o± 1) 10 points Oil is ²owing ±rom a well in a continuous stream at a rate o± f ( t ) = 54 ( t + 3) 2 barrels per month (in multiples o± 100). Find the total oil produced by the well in its second three months o± operation.

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## This note was uploaded on 05/08/2008 for the course M 408 L taught by Professor Cepparo during the Spring '08 term at University of Texas.

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HW_3_Answers - Miller Kierste Homework 3 Due Feb 6 2007...

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