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Unformatted text preview: munoz (gm7794) – HW08 – radin – (54915) 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 π 2 π 3 π 2 x y is bounded by the graph of f ( x ) = 4 cos 3 x on [0 , 3 π/ 2] and the the xaxis. Find the area of this region. 1. area = 12 2. area = 12 π 3. area = 8 correct 4. area = 6 5. area = 8 π 6. area = 6 π Explanation: The area of the shaded region is given by I = integraldisplay 3 π/ 2  4 cos 3 x  dx which as the graph shows can in turn be writ ten as I = integraldisplay π/ 2 4 cos 3 x dx integraldisplay 3 π/ 2 π/ 2 4 cos 3 x dx . Since cos 2 x = 1 sin 2 x , we thus see that I = braceleftBig integraldisplay π/ 2 integraldisplay 3 π/ 2 π/ 2 bracerightBig 4 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 π/ 2 4 cos x (1 sin 2 x ) dx = 4 integraldisplay 1 (1 u 2 ) du = 4 bracketleftBig u 1 3 u 3 bracketrightBig 1 = 8 3 , while integraldisplay 3 π/ 2 π/ 2 4 cos x (1 sin 2 x ) dx = 4 integraldisplay 1 1 (1 u 2 ) du = 4 bracketleftBig u 1 3 u 3 bracketrightBig 1 1 = 16 3 . Consequently, the shaded region has area = 8 . keywords: 002 10.0 points Evaluate the definite integral I = integraldisplay π/ 4 2 cos x + 4 sin x cos 3 x dx . 1. I = 6 munoz (gm7794) – HW08 – radin – (54915) 2 2. I = 4 correct 3. I = 3 4. I = 1 5. I = 0 Explanation: After division we see that 2 cos x + 4 sin x cos 3 x = 2 sec 2 x + 4 tan x sec 2 x = (2 + 4 tan x ) sec 2 x . Thus I = integraldisplay π/ 4 (2 + 4 tan x ) sec 2 x dx . Let u = tan x ; then du = sec 2 x dx , while x = 0 = ⇒ u = 0 , x = π 4 = ⇒ u = 1 . In this case I = integraldisplay 1 (2 + 4 u ) du = bracketleftbig 2 u + 2 u 2 bracketrightbig 1 . Consequently, I = 4 . 003 10.0 points Find the value of I = integraldisplay π 3 tan 4 x dx . 1. I = π 6 8 √ 3 9 2. I = π √ 3 3 3. I = π 4 2 3 4. I = π 3 correct 5. I = π 4 + 2 3 6. I = π 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 π 3 bracketleftbig( tan 2 x 1 ) sec 2 x + 1 bracketrightbig dx = bracketleftbigg 1 3 tan 3 x tan x + x bracketrightbigg π 3 . On the other hand, tan π 3 = √ 3 . Consequently, I = π 3 . 004 10.0 points Evaluate the definite integral I = integraldisplay π/ 4 sin 2 θ ( 3 4 sin 2 θ ) dθ . munoz (gm7794) – HW08 – radin – (54915) 3 1. I = 3 2. I = 3 2 3. I = 1 correct 4. I = 0 5. I = 2 Explanation: Since sin 2 θ = 1 2 (1 cos 2 θ ) , we see that 3 4 sin 2 θ = 3 2 (1 cos 2 θ ) = 1 + 2 cos 2 θ Thus I = 1 2 integraldisplay π/ 4 sin 2 θ (2 + 4 cos 2 θ ) dθ ....
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 Fall '08
 Cepparo
 Calculus, Logarithm, Classless InterDomain Routing, dx, Munoz

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