HW 07 solution - gutierrez (ig3472) HW07 Neitzke (56585) 1...

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Unformatted text preview: gutierrez (ig3472) HW07 Neitzke (56585) 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 (cos + 4 cos 3 ) d . 1. I = cos + 4 3 cos 3 + C 2. I = 5 cos - 4 3 cos 3 + C 3. I = sin + 4 3 sin 3 + C 4. I = 5 cos + 4 3 cos 3 + C 5. I = 5 sin - 4 3 sin 3 + C correct 6. I = sin - 4 3 sin 3 + C Explanation: Since cos 2 = 1- sin 2 , the integrand can be rewritten as cos + 4 cos 3 = cos (1 + 4 cos 2 ) = cos (1 + 4(1- sin 2 )) = cos (5- 4 sin 2 ) . Thus I = integraldisplay cos (5- 4 sin 2 ) d . As the integrand is now of the form cos f (sin ) , f ( x ) = 5- 4 x 2 , the substitution x = sin is suggested. For then dx = cos d , so that I = integraldisplay (5- 4 x 2 ) dx = 5 x- 4 3 x 3 + C . Consequently I = 5 sin - 4 3 sin 3 + C . keywords: indefinite integral, trig function, Pythagorean identity, power cosine, 002 10.0 points Evaluate the integral I = integraldisplay / 2 sin 3 x cos 2 x dx . 1. I = 2 5 2. I = 1 15 3. I = 4 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 = 2 = u = 0 . gutierrez (ig3472) HW07 Neitzke (56585) 2 In this case I =- integraldisplay 1 ( u 2- u 4 ) du . 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 2 3 2 x y is bounded by the graph of f ( x ) = sin 3 x on [0 , 3 / 2] and the the x-axis. Find the area of this region. 1. area = 2 2. area = 2 correct 3. area = 3 2 4. area = 3 2 5. area = 3 6. area = 3 Explanation: The area of the shaded region is given by I = integraldisplay 3 / 2 | sin 3 x | dx which as the graph shows can in turn be writ- ten as I = integraldisplay sin 3 x dx- integraldisplay 3 / 2 sin 3 x dx . Since sin 2 x = 1- cos 2 x , we thus see that I = braceleftBig integraldisplay - integraldisplay 3 / 2 bracerightBig 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 sin x (1- cos 2 x ) dx =- integraldisplay 1 1 (1- u 2 ) du = integraldisplay 1 1 (1- u 2 ) du = bracketleftBig u- 1 3 u 3 bracketrightBig 1 1 = 4 3 , while integraldisplay 3 / 2 sin x (1- cos 2 x ) dx =- integraldisplay...
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HW 07 solution - gutierrez (ig3472) HW07 Neitzke (56585) 1...

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