# cal7 - choi(kc24547 – HW07 – BERG –(56525 1 This...

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Unformatted text preview: choi (kc24547) – HW07 – BERG – (56525) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine the integral I = integraldisplay (4 cos θ- 3 cos 3 θ ) dθ . 1. I = sin θ + sin 3 θ + C correct 2. I = cos θ + cos 3 θ + C 3. I = 4 sin θ- sin 3 θ + C 4. I = 4 cos θ- cos 3 θ + C 5. I = 4 sin θ + sin 3 θ + C 6. I = cos θ- cos 3 θ + C Explanation: Since cos 2 θ = 1- sin 2 θ , the integrand can be rewritten as 4 cos θ- 3 cos 3 θ = cos θ (4- 3 cos 2 θ ) = cos θ (4- 3(1- sin 2 θ )) = cos θ (1 + 3 sin 2 θ ) . Thus I = integraldisplay cos θ (1 + 3 sin 2 θ ) dθ . As the integrand is now of the form cos θ f (sin θ ) , f ( x ) = 1 + 3 x 2 , the substitution x = sin θ is suggested. For then dx = cos θ dθ , so that I = integraldisplay (1 + 3 x 2 ) dx = x + x 3 + C . Consequently I = sin θ + sin 3 θ + C . keywords: indefinite integral, trig function, Pythagorean identity, power cosine, 002 10.0 points Evaluate the integral I = integraldisplay π/ 2 sin 3 x cos 2 x dx . 1. I = 2 5 2. I = 4 15 3. I = 1 15 4. I = 2 15 correct 5. I = 8 15 Explanation: Since sin 3 x cos 2 x = sin x (sin 2 x cos 2 x ) = sin x (1- cos 2 x )cos 2 x = sin x (cos 2 x- cos 4 x ) , the integrand is of the form sin xf (cos x ), sug- gesting use of the substitution u = cos x . For then du =- sin x dx , while x = 0 = ⇒ u = 1 x = π 2 = ⇒ u = 0 . In this case I =- integraldisplay 1 ( u 2- u 4 ) du . choi (kc24547) – HW07 – BERG – (56525) 2 Consequently, I = bracketleftBig- 1 3 u 3 + 1 5 u 5 bracketrightBig 1 = 2 15 . keywords: Stewart5e, indefinite integral, powers of sin, powers of cos, trig substitu- tion, 003 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. Find the area of this region. 1. area = 3 2 2. area = 3 2 π 3. area = 2 π 4. area = 2 correct 5. area = 3 π 6. area = 3 Explanation: The area of the shaded region is given by I = integraldisplay 3 π/ 2 | cos 3 x | dx which as the graph shows can in turn be writ- ten as I = integraldisplay π/ 2 cos 3 x dx- integraldisplay 3 π/ 2 π/ 2 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 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 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 π/ 2 π/ 2 cos x (1- sin 2 x ) dx = integraldisplay − 1 1 (1- u 2 ) du = bracketleftBig u- 1 3 u 3 bracketrightBig...
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cal7 - choi(kc24547 – HW07 – BERG –(56525 1 This...

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