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HW05-solutions.pdf

HW05-solutions.pdf - young(toy68 HW05 villafuerte...

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Unformatted text preview: young (toy68) – HW05 – villafuerte altu – (53615) This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 002 1 10.0 points The shaded region in 10.0 points y Evaluate the integral I = π/2 Z x cos3 x dx . 0 π 2 1. I = 2 correct 3 2. I = 5 6 3. I = 1 6 π 3π 2 is bounded by the graph of f (x) = 2 sin3 (x) 4. I = 1 5. I = on [0, 3π/2] and the the x-axis. Find the area of this region. 1 3 1. area = 4π Explanation: Since cos2 x = 1 − sin2 x , we see that Z I = π/2 0 (1 − sin2 x) cos x dx . This suggests using the substitution u = sin x. For then du = cos x dx, while x = 0 x = =⇒ π 2 =⇒ u = 0, 2. area = 6π 3. area = 3 4. area = 6 5. area = 4 correct 6. area = 3π Explanation: The area of the shaded region is given by u = 1. I = In this case, I = Z 0 1 2 (1 − u ) du . Thus I = 1 u − u3 3 i1 0 = 2 . 3 3π/2 0 |2 sin3 (x)| dx which as the graph shows can in turn be written as Z π Z 3π/2 3 I = 2 sin (x) dx − 2 sin3 (x) dx . 0 h Z Since π sin2 (x) = 1 − cos2 (x) , young (toy68) – HW05 – villafuerte altu – (53615) we thus see that n Z π Z 3π/2 o I = − 2 sin(x)(1−cos2 (x)) dx . π 0 To evaluate these integrals, set u = cos(x). For then du = − sin(x) dx , 4. I = −1 5. I = − 3 2 Explanation: After division cos x − 2 sin x cos3 x in which case Z π 2 sin(x)(1 − cos2 (x)) dx = sec2 x − 2 tan x sec2 x 0 = −2 −1 Z 1 (1 − u2 ) du = 2 1 = 2 u − u3 3 while Z 3π/2 π h 2 i1 Z −1 = −1 = (1 − 2 tan x) sec2 x . 1 (1 − u2 ) du Thus 8 , 3 I = Z π/4 0 (1 − 2 tan x) sec2 x dx . Let u = tan x ; then 2 sin(x)(1 − cos2 (x)) dx = −2 Z 0 −1 (1 − u2 ) du du = sec2 x dx so I = Z 1 0 4 1 3 i0 = − . = −2 u − u 3 3 −1 h Consequently,  1 (1 − 2 u) du = u − u2 0 . Consequently, the shaded region has I = 0 . area = 4 . keywords: 003 10.0 points Evaluate the definite integral I = Z 0 1. I = − π/4 cos x − 2 sin x dx . cos3 x 1 2 2. I = −2 3. I = 0 correct 004 10.0 points Evaluate the indefinite integral Z I = 2 cos4 2t dt . 1. I = 2. I = 3. I = 4. I = 5. I =  1 1 3t+sin 4t+ sin 8t +C correct 4 8  1 1 3t + cos 4t + cos 8t + C 4 8  1 1 3t − sin 4t + sin 8t + C 4 8  1 1 3t + sin 4t − sin 8t + C 4 8   1 1 3t + cos 4t − cos 8t + C 4 8 young (toy68) – HW05 – villafuerte altu – (53615)  1 1 3t − cos 4t + cos 8t + C 4 8 6. I = Explanation: Since  1 1 + cos 2θ , cos θ = 2 3 √ 3. I = 4 2 4. I = 4 √ 5. I = 4 6 2 the integrand can be rewritten as 2 1 2 cos 2t = 1 + cos 4t 2  1 1 + 2 cos 4t + cos2 4t . = 2 6. I = 2 Explanation: By a double angle formula for cos(2x), 4 But in turn, this last expression can be rewritten as o 1 1n 1 + 2 cos 4t + 1 + cos 8t . 2 2 Thus 2 cos4 2t = and so 1 I = 2  13 1 + 2 cos 4t + cos 8t , 2 2 2 Z   3 1 + 2 cos 4t + cos 8t dt . 2 2 Consequently, I =  1 1 3t + sin 4t + sin 8t + C 4 8 with C an arbitrary constant. 005 10.0 points Evaluate the integral I = 2 Z 0 2π/3 p 1 + cos(θ) dθ . Hint: use a double angle formula to express 1 + cos(θ) in terms of cos2 (θ/2). √ 1. I = 2 6 correct √ 2. I = 2 2 cos(2x) = 2 cos2 (x) − 1 , so, taking x = θ/2, we see that s     p √ θ θ = 2 cos 1 + cos(θ) = 2 cos2 2 2 when cos(θ/2) ≥ 0, hence for 0 ≤ θ ≤ π. Thus  i   2π/3 √ Z 2π/3 √ h θ θ I = 2 2 dθ = 4 2 sin cos 2 2 0 0 Consequently, √ I = 2 6 . keywords: definite integral, double angle formula 006 10.0 points Find the value of Z π/4 I= 2 tan4 x dx . 0 1. I = 1 (3π − 2) 4 2. I = 1 (3π − 2) 6 3. I = 1 (3π − 8) 4 4. I = 1 (3π − 4) 6 5. I = 1 (3π − 8) correct 6 young (toy68) – HW05 – villafuerte altu – (53615) 1 (3π − 4) 4 6. I = 1 5 5 3. I = x2 + x sin(2x) + cos(2x) + C 2 2 4 correct Explanation: Since tan2 x = sec2 x − 1 , we see that tan4 x = tan2 x(sec2 x − 1) 2 2 4. I = 1 5 2 1 x + x sin(2x) + cos(2x) + C 2 2 4 5. I = 1 5 2 1 x − x cos(2x) − sin(2x) + C 2 2 4 Explanation: Since 2 = tan x sec x − tan x = tan2 x sec2 x − (sec2 x − 1) . Thus 2 2 tan x = (tan x − 1) sec x + 1 . In this case, I=2 cos2 (x) = 1 (1 + cos(2x)) 2 sin2 (x) = 1 (1 − cos(2x)) , 2 and 4 Z 0 π/4 4 we see that (tan2 x − 1) sec2 x dx + Z π/4 2 dx . Z 1 x (6 (1 + cos(2x)) I= 2 −4 (1 − cos(2x))) dx Z Z = x dx + 5 x cos(2x) dx 0 To evaluate the first of these integrals, set u = tan x. Then Z 1 i1 h1 4 2 (u2 − 1) du = 2 u3 − u = − . 3 3 0 0 On the other hand Z π/4 h iπ/4 1 = π. 2 dx = 2x 2 0 0 Consequently, I = 007 1 (3π − 8) . 6 10.0 points Determine the indefinite integral Z I = x (6 cos2 (x) − 4 sin2 (x)) dx . 1. I = 5 1 2 5 x − x cos(2x) − sin(2x) + C 2 2 4 2. I = 5 2 1 5 x − x sin(2x) + cos(2x) + C 2 2 4 1 = x2 + 5 2 Z x cos(2x) dx . But after integration by parts, Z 1 x cos(2x) dx = x sin(2x) 2 Z 1 sin(2x) dx − 2 1 1 = x sin(2x) + cos(2x) + C . 2 4 Consequently, I = 1 2 5 5 x + x sin(2x) + cos(2x) + C . 2 2 4 008 10.0 points The shaded region in young (toy68) – HW05 – villafuerte altu – (53615) y 5 On the other hand, sin2 (x) = 1 (1 − cos(2x)) . 2 Thus π 2 x π π/2  1 1 − cos(2x) + 2 sin(x) dx V = π 2 2 0 h1 iπ/2 1 = π x − sin(2x) − 2 cos(x) . 2 4 0 Z  Consequently, (not drawn to scale) is the one below the graph of f (x) = sin(x) on the interval [0, π/2]. Find the volume of the 3D solid obtained by rotating this region about the dashed line y = −1. 1. Volume = 1 1 π − cu.units 2 2 2. Volume = 1 2 π + 2π cu.units 2 1 3. Volume = π 2 + 2π cu.units correct 4 1 1 4. Volume = π − cu.units 4 2 Volume = keywords: 009 I = 1. I = 2. I = 6. Volume = 1 2 1 π − π cu.units 2 2 3. I = = π Z π/2  f (x)2 + 2f (x) dx 0 =π Z π/2 0  sin2 (x) + 2 sin(x) dx . Z 0 1 π + 2 cu.units 4 0 10.0 points Evaluate the integral 5. Volume = Explanation: The 3D solid obtained by rotating the shaded region about the line y = −1 is given by the integral Z π/2  I = π (f (x) + 1)2 − (1)2 dx 1 2 π + 2π cu.units . 4 4. I = 5. I = 6. I = π/3 sec(x) tan(x) dx . 3 + sec(x)   5 correct ln 4   5 − ln 6   5 − ln 4   4 ln 3   5 ln 6   4 − ln 3 Explanation: Since d (sec(x)) = sec(x) tan(x) , dx young (toy68) – HW05 – villafuerte altu – (53615) use of the substitution u = 3 + sec(x) and deal with I1 , I2 separately. For I1 , set u = cos(x). Then du = − sin(x) dx is suggested. For then du = sec(x) tan(x)dx , while x = 0 =⇒ π 3 x = =⇒ I = 5 4 Consequently, while u = 4, x = 0 π x = 2 u = 5. Thus Z h i5 1 du = ln(u) . u 4 10.0 points Evaluate the integral Z π/2 I = sin(x) (f ′ (cos(x)) − 2 sin(x)) dx =⇒ =⇒ u = 1, u = 0. Thus ′ f (u) du = 1 h = 0 Z I1 = −   5 I = ln . 4 010 6 Z 1 f ′ (u) du 0 i1 f (u) 0 = (f (1) − f (0)) = 3. On the other hand, sin2 (x) = so  1 1 − cos(2x) , 2 0 when f (0) = 2 and f (1) = 5. I2 = Z π/2 0 1. I = 3 − π = 2. I = 5 + π   1 − cos(2x) dx h x− iπ/2 1 1 = π. sin(2x) 0 2 2 Consequently, 3. I = 4 + π 1 I = 3− π . 2 1 4. I = 5 − π 2 1 5. I = 4 + π 2 011 Find the value of the integral 1 6. I = 3 − π correct 2 Explanation: To evaluate the integral we write I as the sum Z π/2 I = I1 + I2 = sin(x) f ′ (cos(x)) dx 0 −2 Z 0 10.0 points π/2 sin2 (x) dx I = Z 2 1. I = 3 2. I = 1 π 3 3. I = 3π 5 1 dx . 9 + (x − 2)2 young (toy68) – HW05 – villafuerte altu – (53615) 4. I = 1 6 1 π correct 5. I = 12 6. I = 1 π 6 7 4. I = Z sin5 (θ) sec6 (θ) dθ 5. I = Z sin5 (θ) dθ correct Explanation: Set x = sin(θ). Then Explanation: Set 3 tan u = x − 2. Then 9 + (x − 2)2 = 9 + (3 tan u)2 2 In this case 3 sec2 u du = dx . Consequently, Also x = 2 =⇒ u = 0, x = 5 =⇒ u = and I = π . 4 In this case Z Z π/4 1 π/4 3 sec2 u I = du = du. 9 sec2 u 3 0 0 1 h iπ/4 1 u π . = 3 12 0 10.0 points To which of the following does the integral Z x5 √ dx I = 1 − x2 reduce after an appropriate trig substitution? Z 1. I = sec5 (θ) sin6 (θ) dθ 2. I = tan(θ) sec5 (θ) dθ 3. I = Z sin5 (θ) sec5 (θ) dθ Z 013 sin5 (θ) dθ . 10.0 points Evaluate the integral I = Consequently, Z sin5 (θ) cos(θ) dθ . cos(θ) Z I = while 012 1 − sin2 (θ) = cos(θ) . 2 = 9(1 + tan u) = 9 sec u , I = q dx = cos(θ) dθ, Z 2 0 √ 1. I = 1 2 2. I = 1 π correct 2 3. I = 3 4 4. I = 3 π 4 3 dx . 16 − x2 5. I = π 6. I = 1 Explanation: Set x = 4 sin u; then dx = 4 cos u du young (toy68) – HW05 – villafuerte altu – (53615) 8 and 16 − x2 = 16(1 − sin2 u) = 8 cos2 u , while x = 0 x = 2 In this case Z I = 3 π/6 0 =⇒ u = 0, =⇒ π u = . 6 cos u du = 3 cos u Z 015 10.0 points Evaluate the integral Z I = 2 (4 + 0 p 4 − x2 ) dx . 1. I = 4 + π π/6 du . 0 2. I = 8 + 2π Consequently I = 1 π . 2 3. I = 4 − 2π 4. I = 8 + π correct 014 10.0 points Determine the integral Z 3t √ dt . I = 1 − t4  1. I = 3 tan−1 t2 + C  3 tan−1 t2 + C 2 3p 3. I = 1 − t4 + C 2 2. I = 4. I = 5. I = 8 − 2π 6. I = 4 − π Explanation: Since Z 2 Z I= 4 dx + 0 3 +C 1 − t4  6. I = 3 sin−1 t2 + C Explanation: Set t2 = sin(u). Then p 0 4 − x2 dx = I1 + I2 , we evaluate the two integrals separately. Now 3 −1 2  sin t + C correct 2 5. I = √ 2 I1 = 2 Z 4 dx = 8 . 0 On the other hand, to evaluate the second integral we set x = 2 sin(u). For then dx = 2 cos(u) du , while 4 − x2 = 4 cos2 (u) , x = 0 =⇒ u = 0, x = 2 =⇒ u = 2t dt = cos(u) du , in which case 1 I = 2 Z 3 cos(u) 3 du = u + C cos(u) 2 with C an arbitrary constant. Consequently, 3 I = sin−1 (t2 ) + C . 2 π . 2 In this case I2 = 4 Z π/2 cos2 (u) du 0 = 2 Z 0 π/2 (1 + cos(2u)) du . young (toy68) – HW05 – villafuerte altu – (53615) Thus h iπ/2 1 I2 = 2 u + sin(2u) = π. 0 2 Consequently, and so Z I= π/4 0 9 h 2 (sec θ − 1) dθ = tan θ − θ I = 1− . π 4 . 10.0 points Evaluate the integral Z 1 x2 I = dx . 2 3/2 0 (2 − x ) √ π 1. I = 2 3 + 3 π 2. I = 1 + 4 √ π 3. I = 2 3 − 3 √ π 4. I = 2 2 + 3 √ π 5. I = 2 − 4 π 6. I = 1 − correct 4 Explanation: √ Let x = 2 sin θ. Then √ dx = 2 cos θ dθ , 2 − x2 = 2 cos2 θ , while 0 Consequently, I = I1 + I2 = 8 + π . 016 iπ/4 x = 0 =⇒ θ = 0, x = 1 =⇒ θ = π . 4 In this case, Z π/4 √ 2 2 sin2 θ cos θ √ I = dθ 2 2 cos3 θ 0 Z π/4 Z π/4 sin2 θ dθ = tan2 θ dθ . = 2θ cos 0 0 Now tan2 θ = sec2 θ − 1 , d tan θ = sec2 θ , dθ 017 10.0 points Evaluate the integral I = 1 Z √ 0 1. I = √ 1 x2 +1 dx . 2−1 √ 2. I = ln( 2 − 1 ) 3. I = √ 2 (1 + 4. I = ln(1 + 5. I = √ 6. I = √ √ √ 2) 2 ) correct √ 2( 2−1) 2 ln(1 + √ 2) Explanation: Set x = tan(u) , then dx = sec2 (u) du , x2 + 1 = sec2 (u) , while x = 0 =⇒ u = 0, x = 1 =⇒ u = π . 4 In this case Z π/4 Z π/4 sec2 (u) I = du = sec(u) du . sec(u) 0 0 On the other hand, Z sec(u) du = ln (| sec(u) + tan(u)|) + C . young (toy68) – HW05 – villafuerte altu – (53615) Thus 10 On the other hand, by Pythagoras, I =  ln (| sec(u) + tan(u)|) π/4 . 0 Consequently, 018 √ 10.0 points x . x2 + 9 with C an arbitrary constant. Explanation: Set x = 3 tan θ. Then p dx = 3 sec2 θ dθ, x2 + 9 = 3 sec θ . In this case Z 2 sec2 θ I = dθ 3 tan θ sec θ Z sec θ 2 dθ = tan θ 3 019 10.0 points Evaluate the integral I = Z 1 (x2 0 2 dx . + 3)3/2 1. I = 2 2. I = 1 correct 3 3. I = 1 4. I = 1 2 5. I = 2 3 Explanation: √ Set x = 3 tan(u). Then dx = √ 3 sec2 (u) du , while Z csc θ dθ . But Z sin θ = √ =⇒ √ 2 x2 + 9 − 3 I = ln +C 3 x 2) . Determine the integral Z 2 √ dx . I = x x2 + 9 √ 2 x2 + 9 + 3 1. I = ln +C 3 x √ 3 x2 + 9 − 3 2. I = ln +C 2 x √ 2 x2 + 9 + x 3. I = ln +C 3 3 √ 3 x2 + 9 + x 4. I = ln +C 2 3 √ 3 x2 + 9 − x 5. I = ln +C 2 3 √ 2 x2 + 9 − 3 6. I = ln + C correct 3 x 2 3 x 3 Consequently, I = ln(1 + = tan θ = csc θ dθ = ln | csc θ − cot θ| + C . x = 0 =⇒ u = 0, x = 1 =⇒ u = π . 6 On the other hand, (x2 + 3)3/2 = 3(tan2 (u) + 1) √ = 3 3 sec3 (u) . 3/2 young (toy68) – HW05 – villafuerte altu – (53615) Thus 11 while I = 2 Z √ 3 sec2 (u) √ du 3 3 sec3 (u) π/6 0 2 = 3 Z π/6 cos(u) du = 0 iπ/6 2h . sin(u) 3 0 Consequently I = 1 . 3 x = 0 =⇒ x = 1 =⇒ In this case Z π/4 Z π/4 3 sec2 u I =1+ du = 1 + 3 du . sec2 u 0 0 Consequently, 3 I = 1+ π . 4 021 keywords: 020 10.0 points Evaluate the integral Z 1 2 x +4 I = dx . 2 0 1+x  π 1. I = 4 1 − 4 3 2. I = 1 + π correct 4 3 3. I = 1 − π 4 u = 0, π u = . 4 10.0 points Which one of the following functions is an antiderivative of f when 1 f (x) = 2 ? x − 4x+ 5 x − 5 1. F (x) = ln x + 2 2. F (x) = ln(x2 − 4 x + 5) 3. F (x) = − 5 (x − 2) − 4 x + 5)2 (x2 4. F (x) = sin−1 (x − 2) 5. F (x) = tan−1 (x − 2) correct Explanation: The indefinite integral Z Z I = f (x) dx = 4. I = 4(1 − π) 1 dx − 4x+ 5 consists of all antiderivatives of f . But by completing the square we see that 1 1 = x2 − 4 x + 5 (x2 − 4 x + 4) + 1  π 5. I = 4 1 + 4 6. I = 3 + 4π Explanation: Since x2 + 1 + 3 3 x2 + 4 = = 1 + , 1 + x2 1 + x2 1 + x2 the integral can be written as Z 1 Z 1 3  3 dx = 1+ dx . I = 1+ 2 1 + x2 0 1+x 0 In the last integral, set x = tan u. Then 2 dx = sec u du, = ThusZ I= x2 1 . (x − 2)2 + 1 1 dx = tan−1 (x − 2) + C . (x − 2)2 + 1 Consequently, only F (x) = tan−1 (x − 2) is an antiderivative of f , as can be checked by differentiating F . ...
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