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Unformatted text preview: kim (tk5895) HW08 Henry (54974) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points The shaded region in 2 3 2 x y is bounded by the graph of f ( x ) = cos 3 x on [0 , 3 / 2] and the the xaxis. Find the area of this region. 1. area = 3 2. area = 2 3. area = 3 2 4. area = 3 5. area = 3 2 6. area = 2 correct Explanation: The area of the shaded region is given by I = integraldisplay 3 / 2  cos 3 x  dx which as the graph shows can in turn be writ ten as I = integraldisplay / 2 cos 3 x dx integraldisplay 3 / 2 / 2 cos 3 x dx . Since cos 2 x = 1 sin 2 x , we thus see that I = braceleftBig integraldisplay / 2 integraldisplay 3 / 2 / 2 bracerightBig cos x (1 sin 2 x ) dx . To evaluate these integrals, set u = sin x . For then du = cos x dx , in which case integraldisplay / 2 cos x (1 sin 2 x ) dx = integraldisplay 1 (1 u 2 ) du = bracketleftBig u 1 3 u 3 bracketrightBig 1 = 2 3 , while integraldisplay 3 / 2 / 2 cos x (1 sin 2 x ) dx = integraldisplay 1 1 (1 u 2 ) du = bracketleftBig u 1 3 u 3 bracketrightBig 1 1 = 4 3 . Consequently, the shaded region has area = 2 . keywords: 002 10.0 points Evaluate the definite integral I = integraldisplay / 4 4 cos x + 5 sin x cos 3 x dx . 1. I = 9 kim (tk5895) HW08 Henry (54974) 2 2. I = 21 4 3. I = 3 2 4. I = 13 2 correct 5. I = 11 4 Explanation: After division we see that 4 cos x + 5 sin x cos 3 x = 4 sec 2 x + 5 tan x sec 2 x = (4 + 5 tan x ) sec 2 x . Thus I = integraldisplay / 4 (4 + 5 tan x ) sec 2 x dx . Let u = tan x ; then du = sec 2 x dx , while x = 0 = u = 0 , x = 4 = u = 1 . In this case I = integraldisplay 1 (4 + 5 u ) du = bracketleftbigg 4 u + 5 2 u 2 bracketrightbigg 1 . Consequently, I = 13 2 . 003 10.0 points Find the value of I = integraldisplay 3 tan 4 x dx . 1. I = 6 8 3 9 2. I = 4 2 3 3. I = 4 + 2 3 4. I = 3 3 5. I = 3 correct 6. I = 6 + 8 3 9 Explanation: Since tan 2 x = sec 2 x 1 , we see that tan 4 x = tan 2 x ( sec 2 x 1 ) = tan 2 x sec 2 x tan 2 x . Thus by trig identities yet again, tan 4 x = ( tan 2 x 1 ) sec 2 x + 1 . In this case, I = integraldisplay 3 bracketleftbig( tan 2 x 1 ) sec 2 x + 1 bracketrightbig dx = bracketleftbigg 1 3 tan 3 x tan x + x bracketrightbigg 3 . On the other hand, tan 3 = 3 . Consequently, I = 3 . 004 10.0 points Evaluate the definite integral I = integraldisplay / 4 sin 2 ( 3 4 cos 2 ) d . kim (tk5895) HW08 Henry (54974) 3 1. I = 3 2. I = 2 3. I = 0 correct 4. I = 1 5. I = 3 2 Explanation: Since cos 2 = 1 2 (1 + cos 2 ) , we see that 3 4 cos 2 = 3 2 (1 + cos 2 ) = 1 2 cos 2 Thus I = 1 2 integraldisplay / 4 sin 2 (2 4 cos 2 ) d ....
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 Fall '08
 Cepparo
 Calculus

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