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# solution9_pdf - So(dds785 – HW08 – radin –(54915 1...

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Unformatted text preview: So (dds785) – HW08 – radin – (54915) 1 This print-out should have 16 questions. Multiple-choice 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 ) = 2 sin 3 x on [0 , 3 π/ 2] and the the x-axis. Find the area of this region. 1. area = 6 2. area = 3 π 3. area = 3 4. area = 4 correct 5. area = 4 π 6. area = 6 π Explanation: The area of the shaded region is given by I = integraldisplay 3 π/ 2 | 2 sin 3 x | dx which as the graph shows can in turn be writ- ten as I = integraldisplay π 2 sin 3 x dx- integraldisplay 3 π/ 2 π 2 sin 3 x dx . Since sin 2 x = 1- cos 2 x , we thus see that I = braceleftBig integraldisplay π- integraldisplay 3 π/ 2 π 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 π 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 π/ 2 π 2 sin x (1- cos 2 x ) dx =- 2 integraldisplay- 1 (1- u 2 ) du =- 2 bracketleftBig u- 1 3 u 3 bracketrightBig- 1 =- 4 3 . Consequently, the shaded region has area = 4 . keywords: 002 10.0 points Evaluate the definite integral I = integraldisplay π/ 4 3 cos x + 6 sin x cos 3 x dx . 1. I = 6 correct So (dds785) – HW08 – radin – (54915) 2 2. I = 9 2 3. I = 3 2 4. I = 9 5. I = 0 Explanation: After division we see that 3 cos x + 6 sin x cos 3 x = 3 sec 2 x + 6 tan x sec 2 x = (3 + 6 tan x ) sec 2 x . Thus I = integraldisplay π/ 4 (3 + 6 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 (3 + 6 u ) du = bracketleftbig 3 u + 3 u 2 bracketrightbig 1 . Consequently, I = 6 . 003 10.0 points Find the value of I = integraldisplay π 6 tan 4 x dx . 1. I = π 6- 8 √ 3 27 correct 2. I = π 6 + 8 √ 3 27 3. I = π √ 3 3 4. I = π 4 + 2 3 5. I = π 4- 2 3 6. I = π 3 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 6 bracketleftbig( tan 2 x- 1 ) sec 2 x + 1 bracketrightbig dx = bracketleftbigg 1 3 tan 3 x- tan x + x bracketrightbigg π 6 . On the other hand, tan π 6 = √ 3 3 . Consequently, I = π 6- 8 √ 3 27 . 004 10.0 points Evaluate the definite integral I = integraldisplay π/ 4 sin 2 θ ( 4- 2 sin 2 θ ) dθ . So (dds785) – HW08 – radin – (54915) 3 1. I = 5 4 2. I = 9 4 3. I = 7 4 correct 4. I = 2 5. I = 11 4 Explanation: Since sin 2 θ = 1 2 (1- cos 2 θ ) , we see that 4- 2 sin 2 θ = 4- (1- cos 2 θ ) = 3 + cos 2 θ Thus I = 1 2 integraldisplay π/ 4 sin 2 θ (6 + 2 cos 2 θ ) dθ ....
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solution9_pdf - So(dds785 – HW08 – radin –(54915 1...

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