Math HW7 - Perez (nap563) HW07 Zheng (56555) 1 This...

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Unformatted text preview: Perez (nap563) HW07 Zheng (56555) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Determine the integral I = integraldisplay (5 cos - 2 cos 3 ) d . 1. I = 5 cos - 2 3 cos 3 + C 2. I = 5 sin + 2 3 sin 3 + C 3. I = 3 sin + 2 3 sin 3 + C correct 4. I = 3 cos - 2 3 cos 3 + C 5. I = 3 cos + 2 3 cos 3 + C 6. I = 5 sin - 2 3 sin 3 + C Explanation: Since cos 2 = 1- sin 2 , the integrand can be rewritten as 5 cos - 2 cos 3 = cos (5- 2 cos 2 ) = cos (5- 2(1- sin 2 )) = cos (3 + 2 sin 2 ) . Thus I = integraldisplay cos (3 + 2 sin 2 ) d . As the integrand is now of the form cos f (sin ) , f ( x ) = 3 + 2 x 2 , the substitution x = sin is suggested. For then dx = cos d , so that I = integraldisplay (3 + 2 x 2 ) dx = 3 x + 2 3 x 3 + C . Consequently I = 3 sin + 2 3 sin 3 + C . keywords: indefinite integral, trig function, Pythagorean identity, power cosine, 002 10.0 points Evaluate the integral I = integraldisplay / 2 sin 2 x cos 3 x dx . 1. I = 2 15 correct 2. I = 8 15 3. I = 4 15 4. I = 1 15 5. I = 2 5 Explanation: Since sin 2 x cos 3 x = (sin 2 x cos 2 x ) cos x = sin 2 x (1- sin 2 x )cos x = (sin 2 x- sin 4 x )cos x , the integrand is of the form cos xf (sin x ), sug- gesting use of the substitution u = sin x . For then du = cos x dx , while x = 0 = u = 0 x = 2 = u = 1 . Perez (nap563) HW07 Zheng (56555) 2 In this case I = integraldisplay 1 ( u 2- u 4 ) du . Consequently, I = bracketleftBig 1 3 u 3- 1 5 u 5 bracketrightBig 1 = 2 15 . keywords: Stewart5e, indefinite integral, powers of sin, powers of cos, trig substitu- tion, 003 10.0 points The shaded region in 2 3 2 x y is bounded by the graph of f ( x ) = 2 sin 3 x on [0 , 3 / 2] and the the x-axis. Find the area of this region. 1. area = 3 2. area = 4 correct 3. area = 6 4. area = 3 5. area = 6 6. area = 4 Explanation: The area of the shaded region is given by I = integraldisplay 3 / 2 | 2 sin 3 x | dx which as the graph shows can in turn be writ- ten as I = integraldisplay 2 sin 3 x dx- integraldisplay 3 / 2 2 sin 3 x dx . Since sin 2 x = 1- cos 2 x , we thus see that I = braceleftBig integraldisplay - integraldisplay 3 / 2 bracerightBig 2 sin x (1- cos 2 x ) dx . To evaluate these integrals, set u = cos x . For then du =- sin x dx , in which case integraldisplay 2 sin x (1- cos 2 x ) dx =- 2 integraldisplay 1 1 (1- u 2 ) du = 2 integraldisplay 1 1 (1- u 2 ) du = 2 bracketleftBig u- 1 3 u 3 bracketrightBig 1 1 = 8 3 , while integraldisplay 3 / 2 2 sin x (1- cos 2 x ) dx =- 2 integraldisplay 1 (1- u 2 ) du =- 2 bracketleftBig...
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This note was uploaded on 03/17/2010 for the course MATH 408 L taught by Professor Zheng during the Spring '10 term at University of Texas at Austin.

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Math HW7 - Perez (nap563) HW07 Zheng (56555) 1 This...

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