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Unformatted text preview: Rehman (aar638) – HW04 – sachse – (56620) 1 This printout should have 21 questions. Multiplechoice 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 antiderivative 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 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 ) . Rehman (aar638) – HW04 – sachse – (56620) 2 Thus I = 2 integraldisplay π 6 (3 sin x 2 cos x ) dx = 2 bracketleftBig 3 cos x 2 sin x bracketrightBig π 6 = 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 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 dθ tan θ = sec 2 θ , we see that I = integraldisplay π/ 6 parenleftBig 3 sec 2 θ + 2 sin θ parenrightBig dθ = bracketleftBig 3 tan θ 2 cos θ bracketrightBig π/ 6 = parenleftBig 3 √ 3 √ 3 parenrightBig + 2 . Consequently, I = 2 . 005 10.0 points Find the value of the integral I = integraldisplay 4 vextendsingle vextendsingle 3 x x 2 vextendsingle vextendsingle dx ....
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This note was uploaded on 02/01/2011 for the course MATH 408L taught by Professor Gogolev during the Fall '09 term at University of Texas.
 Fall '09
 GOGOLEV
 Calculus

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