course Mth 174 Sorry about getting this in late. I am planning on taking the test Monday morning first thing. Quick question: Will I need the tables of indefinite integrals for Test 1?
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21:36:22 Query problem 7.3.17 (3d edition #15) x^4 e^(3x)
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RESPONSE --> Using the tables, this is what I found: Indefinite integral = (1/3)x^4 e^(3x) - (4/9)x^3 e^(3x) + (4/9)x^2 e^(3x) - (8/27)x e^(3x) + (8/81) e^(3x) + C
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21:36:27 what it is your antiderivative?
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RESPONSE --> See prior answer.
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21:36:51 Which formula from the table did you use?
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RESPONSE --> I used formula III.14.
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21:37:54 You should have used formula 14. What was your value of a, what was p(x), how many derivatives did you use and what were your derivatives of p(x)?
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RESPONSE --> a = 3 p(x) = x^4 I used four derivatives of p(x): p'(x) = 4x^3 p''(x) = 12x^2 p'''(x) = 24x p''''(x) = 24
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21:39:18 Query problem 7.3.33 (3d edition #30) 1 / [ 1 + (z+2)^2 ) ]
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RESPONSE --> I found the derivative to be: arctan(z+2) +C
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21:39:30 What is your integral?
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RESPONSE --> Well, the integral is listed as above.
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21:43:04 Which formula from the table did you use and how did you get the integrand into the form of this formula?
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RESPONSE --> I didn't. I used a substitution: u = z + 2 Which left the equation: 1 / (1 + u^2) I recognized this to be the derivative of the arctan function, and so I found: arctan(u) + C Putting the original value of u in: arctan(z + 2) + C
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22:29:46 7.4.6 (#8 in 3d edition). Integrate 2y / ( y^3 - y^2 + y - 1)
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RESPONSE --> I think I found an answer to this one. The functions I found are as follows: (-y + 1) / (y^2 + 1) + 1 / (y - 1)
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22:29:52 What is your result?
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RESPONSE --> See above.
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22:37:35 How did you factor your denominator to get the integrand into a form amenable to partial fractions?
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RESPONSE --> Looking for common factors doesn't work here, so I had to use for quadratic-style factoring. I looked at the factors of the highest power function: y * y^2 And the factors of the last number, -1: -1 * 1 So I can set those into two possible formats: (y^2 - 1)(y + 1) (y^2 + 1)(y - 1) The simplest way I know to find out which one fits is to just multiply it through. Any suggestions on this account would be greatly appreciated, as I am sure this isn't the best way.
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22:38:14 After the integrand was in the form 2y / [ (y-1) * (y^2 + 1) ], what form of partial fraction did you use?
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RESPONSE --> I had to use the form: (Ax + B) / (y^2 + 1) + C / (y - 1)
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22:58:15 7.4.29 (3d edition #24). Integrate (z-1)/`sqrt(2z-z^2)
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RESPONSE --> I'm guessing I really should wait for the next section to input my answer.
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23:30:44 What did you get for your integral?
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RESPONSE --> Initially I only worked on finding a possible substitution by going to the inmost function and completing the square. My result: w = 1 - (z - 1)^2 But I am not completely sure where to go from there. Using substition, here are my steps: dw = 2(z -1) dz dz = (1/2) dw / (z - 1) Substituting into the original equation: = (z - 1) / 'sqrt (w) * (1/2) dw / (z - 1) = (1/2) dw / 'sqrt (w) Finding the antiderivative of this: antiderivative = 'sqrt (w) Expanded: = 'sqrt ( 1 - (z - 1)^2 ) + C Is that right?
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23:30:51 What substitution did you use?
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RESPONSE --> See prior answer.
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23:32:19 7.4.40 (3d edition #28). integrate (y+2) / (2y^2 + 3y + 1)
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RESPONSE --> To be answered in the next box...
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23:41:43 What is your integral and how did you obtain it?
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RESPONSE --> I found the integral to be: (3/2) ln |2y + 1| - ln |y + 1| I obtained it by first factoring the denominator, 2y^2 +3y +1: (y + 2) / ( (2y + 1)(y+1) ) and then splitting it up into partial fractions: A / (2y + 1) + B / (y + 1) Solving for A and B: A = 3 B = -1 This gave me the result: 3 / (2y + 1) - 1 / (y + 1) I then found the antiderivative to be: (3/2) ln |2y + 1| - ln |y + 1|
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23:42:50 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Whew. That was wild. But it was still incredibly interesting to see how substition can be used to chain derivatives and antiderivatives together until a complex equation is simple, and then simply reverse the process to find the answer.
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