Homework 4 - Rehman(aar638 HW04 sachse(56620 This print-out...

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Rehman (aar638) – HW04 – sachse – (56620) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The graph of f is shown in the figure 2 4 6 8 2 4 6 If F is an anti-derivative of f and integraldisplay 8 2 f ( x ) dx = 21 , find the value of F (8) - F (1). 1. F (8) - F (1) = 93 4 correct 2. F (8) - F (1) = 24 3. F (8) - F (1) = 105 4 4. F (8) - F (1) = 99 4 5. F (8) - F (1) = 51 2 Explanation: We already know that the area under the graph on the interval 2 x 8 is equal to 21, alternatively, by the Fundamental Theorem of Calculus we can say that F (8) - F (2) = 21 . On the other hand, integraldisplay 8 1 f ( x ) dx = integraldisplay 2 1 f ( x ) dx + integraldisplay 8 2 f ( x ) dx. Thus we need to find integraldisplay 2 1 f ( x ) dx = F (2) - F (1) . Now integraldisplay 2 1 f ( x ) dx = integraldisplay 2 1 3 2 x dx = 3 4 bracketleftBig x 2 bracketrightBig 2 1 = 9 4 . Consequently, F (8) - F (1) = 21 + 9 4 = 93 4 . keywords: velocity, distance, graph analysis, fundamental theorem 002 10.0 points Evaluate the definite integral I = integraldisplay π 6 0 3 sin 2 x - 4 cos 2 x cos x dx . 1. I = 3 + 4 2 2. I = 6 + 2 3 3. I = 3 - 4 2 4. I = 4 - 3 3 correct 5. I = 8 - 3 3 6. I = 6 - 2 3 Explanation: Since sin 2 x = 2 sin x cos x , the integrand can be rewritten as 6 sin x cos x - 4 cos 2 x cos x = 2(3 sin x - 2 cos x ) .
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Rehman (aar638) – HW04 – sachse – (56620) 2 Thus I = 2 integraldisplay π 6 0 (3 sin x - 2 cos x ) dx = 2 bracketleftBig - 3 cos x - 2 sin x bracketrightBig π 6 0 = 2 parenleftbigg - 3 2 3 - 1 parenrightbigg + 6 . Consequently, I = 4 - 3 3 . 003 10.0 points Evaluate the definite integral I = integraldisplay 4 1 x parenleftBig 3 + 4 x parenrightBig dx . 1. I = 22 correct 2. I = 21 3. I = 24 4. I = 20 5. I = 23 Explanation: We first expand x parenleftBig 3 + 4 x parenrightBig = 3 x + 4 x , and then integrate term by term. This gives I = bracketleftBig 2 x 3 / 2 + 8 x 1 / 2 bracketrightBig 4 1 . Consequently, I = 22 . 004 10.0 points Evaluate the integral I = integraldisplay π/ 6 0 parenleftBig 3 cos 2 θ + 2 sin θ parenrightBig dθ . 1. I = 2 3 - 1 2. I = 2 3 + 1 3. I = - 1 4. I = 2 3 - 2 5. I = 2 3 + 2 6. I = 1 7. I = - 2 8. I = 2 correct Explanation: Since 1 cos 2 θ = sec 2 θ , d tan θ = sec 2 θ , we see that I = integraldisplay π/ 6 0 parenleftBig 3 sec 2 θ + 2 sin θ parenrightBig = bracketleftBig 3 tan θ - 2 cos θ bracketrightBig π/ 6 0 = parenleftBig 3 3 - 3 parenrightBig + 2 . Consequently, I = 2 . 005 10.0 points Find the value of the integral I = integraldisplay 4 0 vextendsingle vextendsingle 3 x - x 2 vextendsingle vextendsingle dx . 1. I = 13 2 2. I = 20 3
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Rehman (aar638) – HW04 – sachse – (56620) 3 3. I = 6 4. I = 19 3 correct 5. I = 37 6 Explanation: The graph of f ( x ) = 3 x - x 2 is a parabola 3 4 (not drawn to scale) which lies above the x - axis on [0 , 3] and below on [3 , 4]. Thus | 3 x - x 2 | = braceleftBigg 3 x - x 2 , 0 x 3 , x 2 - 3 x, 3 x 4 . So I = integraldisplay 3 0 (3 x - x 2 ) dx + integraldisplay 4 3 ( x 2 - 3 x ) dx = bracketleftbigg 3 2 x 2 - 1 3 x 3 bracketrightbigg 3 0 + bracketleftbigg 1 3 x 3 - 3 2 x 2 bracketrightbigg 4 3 .
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