review_sols - Name Sec No Math 10B Review Assignment...

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Name: PID: Sec. No: Sec. Time: Math 10B Review Assignment September 2, 2010 Please turn in on Thursday in class. This assignment may replace your worst homework. Show all of your work, no credit will be given for unsupported answers. Please write neatly and make sure the answer is clear. 1. Please compute the following indefinite integrals. (a) x ( x + 4)( x - 3) dx Solution: This problem requires partial fractions. The decomposition rule tells us x ( x + 4)( x - 3) = A x + 4 + B ( x - 3) If you work out the algebra, you should get that A = 4 / 7 and B = 3 / 7 Therefore x ( x + 4)( x - 3) dx = (4 / 7) 1 x + 4 dx + (3 / 7) 1 x - 3 dx = (4 / 7)ln | x + 4 | + (3 / 7)ln | x - 3 | + C (b) x + 1 x - 4 dx Solution: You can use substitution or integration by parts here. I’ll use substitution. Begin by letting w = x - 4, which means dw = dx . The substitution also provides the necessary information that x + 1 = ( w + 4) + 1 = w + 5. Now plugging everything in gives you that x + 1 x - 4 dx = w + 5 w dw = w 1 / 2 dw + 5 w - 1 / 2 dw = (2 / 3) w 3 / 2 + 10 w 1 / 2 + C = (2 / 3)( x - 4) 3 / 2 + 10( x - 4) 1 / 2 + C
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(c) x ln(3 x ) dx Solution: This problem requires integration by parts. Make the following table u = ln(3 x ) v = x u = 3 / (3 x ) v = (1 / 2) x 2 Using the integration by parts formula gives us x ln(3 x ) dx = uv dx = uv - u v dx + C = (1 / 2) x 2 ln(3 x ) - (1 / 2) x dx + C = (1 / 2) x 2 ln(3 x ) - (1 / 4) x 2 + C 2. Find the volume of the figure formed by revolving the region between the functions f ( x ) = x and h ( x ) = x on [0 , 1] around the line y = 3. Solution: Below is a picture showing the region S you must rotate to get the figure. x y 0 1 y = 3 h ( x ) = x f ( x ) = x R = 3 - x 1 r = 3 - x 2 x 1 x 2 S The integral you need to compute is 1 0 ( π (3 - x ) 2 - π (3 - x ) 2 ) dx = 1 0 ( π (9 - 6 x + x 2 ) - π (9 - 6 x + x )) dx = π 1 0 ( - 7 x + 6 x + x 2 ) dx = π (( - 7 / 2) + 4 + (1 / 3)) = 5 π/ 6
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3. Solve the following initial value problem ds dt = s 2 sin( t ) s ( π ) = 1 Solution: We have to use separation of variables for this one. Start by getting the
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