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HW08-solutions

# HW08-solutions - kim(tk5895 HW08 Henry(54974 This print-out...

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kim (tk5895) – HW08 – Henry – (54974) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd 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 ) = cos 3 x on [0 , 3 π/ 2] and the the x -axis. ±ind the area oF this region. 1. area = 3 2. area = 2 π 3. area = 3 2 4. area = 3 π 5. area = 3 2 π 6. area = 2 correct Explanation: The area oF the shaded region is given by I = i 3 π/ 2 0 | cos 3 x | dx which as the graph shows can in turn be writ- ten as I = i π/ 2 0 cos 3 x dx - i 3 π/ 2 π/ 2 cos 3 x dx . Since cos 2 x = 1 - sin 2 x , we thus see that I = b i π/ 2 0 - i 3 π/ 2 π/ 2 B cos x (1 - sin 2 x ) dx . To evaluate these integrals, set u = sin x . ±or then du = cos x dx , in which case i π/ 2 0 cos x (1 - sin 2 x ) dx = i 1 0 (1 - u 2 ) du = ± u - 1 3 u 3 ² 1 0 = 2 3 , while i 3 π/ 2 π/ 2 cos x (1 - sin 2 x ) dx = i - 1 1 (1 - u 2 ) du = ± u - 1 3 u 3 ² - 1 1 = - 4 3 . Consequently, the shaded region has area = 2 . keywords: 002 10.0 points Evaluate the defnite integral I = i π/ 4 0 4 cos x + 5 sin x cos 3 x dx . 1. I = 9

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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 = i π/ 4 0 (4 + 5 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 = i 1 0 (4 + 5 u ) du = b 4 u + 5 2 u 2 B 1 0 . Consequently, I = 13 2 . 003 10.0 points Find the value of I = i π 3 0 tan 4 x dx . 1. I = π 6 - 8 3 9 2. I = π 4 - 2 3 3. I = π 4 + 2 3 4. I = π 3 3 5. I = π 3 correct 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 = i π 3 0 ±( tan 2 x - 1 ) sec 2 x + 1 ² dx = b 1 3 tan 3 x - tan x + x B π 3 0 . On the other hand, tan π 3 = 3 . Consequently, I = π 3 . 004 10.0 points Evaluate the de±nite integral I = i π/ 4 0 sin 2 θ ( 3 - 4 cos 2 θ ) dθ .
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 θ = 1 2 (1 + cos 2 θ ) , we see that 3 - 4 cos 2 θ = 3 - 2 (1 + cos 2 θ ) = 1 - 2 cos 2 θ Thus I = 1 2 i π/ 4 0 sin 2 θ (2 - 4 cos 2 θ ) dθ .

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