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M408L hwk3 solutions

# M408L hwk3 solutions - Pham Quoc Homework 3 Due Feb 6 2007...

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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 π/ 3 0 2 sin 2 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 sin 2 x = 2 sin x cos x , the integrand can be rewritten as 2 sin 2 x cos x = 4 sin x . But then I = Z π/ 3 0 4 sin x dx = h - 4 cos x i π/ 3 0 . Consequently, I = 2 . keywords: definite integral, trig functions double angle formula, 002 (part 1 of 1) 10 points Evaluate the integral I = Z 4 0 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 0 d dx (2 + 3 x 2 ) 1 / 2 dx = h (2 + 3 x 2 ) 1 / 2 i 4 0 . 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

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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 π/ 4 0 (6 sec 2 θ - 5 sin θ ) dθ. 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 6 tan θ + 5 cos θ i π/ 4 0 = 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 0 ( | 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 0 (1 - 2 x ) dx - Z 4 1 1 dx = I 1 - I 2 . Then I 1 = h x - x 2 i 1 0 = 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. 1. # barrels = 550 2. # barrels = 750 3. # barrels = 700 4. # barrels = 600 correct 5. # barrels = 650 Explanation: The total oil produced by the well in the second three months, i.e. , over the time inter- val (3 , 6], is given by the definite integral T = Z 6 3 f ( t ) dt = Z 6 3 108 ( t + 3) 2 dt. But Z 6 3 108 ( t + 3) 2 dt = h - 108 t + 3 i 6 3 = 6 .

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M408L hwk3 solutions - Pham Quoc Homework 3 Due Feb 6 2007...

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