HW04-solutions

# HW04-solutions - griffin(ang684 – HW04 – Gilbert...

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Unformatted text preview: griffin (ang684) – HW04 – Gilbert – (56650) 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 1 f ( x ) dx = 31 2 , find the value of F (8)- F (0). 1. F (8)- F (0) = 35 2 2. F (8)- F (0) = 18 3. F (8)- F (0) = 17 4. F (8)- F (0) = 33 2 5. F (8)- F (0) = 16 correct Explanation: We already know that the area under the graph on the interval 1 ≤ x ≤ 8 is equal to 31 2 , alternatively, by the Fundamental Theorem of Calculus we can say that F (8)- F (1) = 31 2 . On the other hand, integraldisplay 8 f ( x ) dx = integraldisplay 1 f ( x ) dx + integraldisplay 8 1 f ( x ) dx. Thus we need to find integraldisplay 1 f ( x ) dx = F (1)- F (0) . Now integraldisplay 1 f ( x ) dx = integraldisplay 1 x dx = 1 2 bracketleftBig x 2 bracketrightBig 1 = 1 2 . Consequently, F (8)- F (0) = 31 2 + 1 2 = 16 . keywords: velocity, distance, graph analysis, fundamental theorem 002 10.0 points Evaluate the definite integral I = integraldisplay π 6 3 sin 2 x- 2 cos 2 x cos x dx . 1. I = 6- √ 3 2. I = 3- √ 2 3. I = 7- 3 √ 3 4. I = 6 + √ 3 5. I = 5- 3 √ 3 correct 6. I = 6 + √ 2 Explanation: Since sin 2 x = 2 sin x cos x , the integrand can be rewritten as 6 sin x cos x- 2 cos 2 x cos x = 2(3 sin x- cos x ) . griffin (ang684) – HW04 – Gilbert – (56650) 2 Thus I = 2 integraldisplay π 6 (3 sin x- cos x ) dx = 2 bracketleftBig- 3 cos x- sin x bracketrightBig π 6 = 2 parenleftbigg- 3 2 √ 3- 1 2 parenrightbigg + 6 . Consequently, I = 5- 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 = 23 3. I = 26 4. I = 24 5. I = 25 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. I =- 1 4. I = 2 √ 3 + 2 5. I = 2 √ 3 + 1 6. I =- 2 7. I = 2 √ 3- 2 correct 8. I = 1 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 √ 3- 2 . 005 10.0 points Find the value of the integral I = integraldisplay 5 vextendsingle vextendsingle 4 x- x 2 vextendsingle vextendsingle dx ....
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## This note was uploaded on 09/09/2009 for the course M 408L taught by Professor Gilbert during the Spring '09 term at University of Texas-Tyler.

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HW04-solutions - griffin(ang684 – HW04 – Gilbert...

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