course Mth 174 Sorry for not turning in any work for the past week, I was on vacation. I will take my first test later this week and then use the following week to get caught back up.Also, I was wondering if you have been getting my emails about last semester's grade. My parents pay for my classes and they have started pecking at me about it.
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15:56:34 Query problem 7.3.17 (3d edition #15) x^4 e^(3x)
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RESPONSE --> I set x^4 = p(x) 3 = a I also used the table, III number 14. (1/3) x^4 e^3x - (1/9) 4x^3 e^3x + (1/27) 12x^2 e^3x - (1/81) 24x e^3x + (1/243) 24 e^3x = (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
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15:56:37 what it is your antiderivative?
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RESPONSE -->
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15:56:46 Which formula from the table did you use?
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RESPONSE --> Part III, number 14. int(p(x) e^ax dx = (1/a) p(x) e^ax - (1/a) int(p'(x) e^ax dx
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15:56:57 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 4 derivatives x^4 4x^3 12x^2 24x 24
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15:57:13 Query problem 7.3.33 (3d edition #30) 1 / [ 1 + (z+2)^2 ) ]
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RESPONSE --> I used part V number 24 to solve this. int(1 / (x^2 + a^2)) = (1/a) arctan (x/a) x^2 = 1 or 1^2 a^2 = (z + 2)^2 (1/(z+2)) arctan (1/(z + 2)) + C
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15:57:14 What is your integral?
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RESPONSE -->
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15:57:24 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 --> Part V, number 24
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15:57:40 7.4.6 (#8 in 3d edition). Integrate 2y / ( y^3 - y^2 + y - 1)
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RESPONSE --> I used Partial Fractions. I broke it down to 2y / [(y-1)(y^2 + 1)] 2y / [(y-1)(y^2 + 1)] = (Ay + B) / (y^2 + 1) + C / (y - 1) I multiplied both sides by (y-1)(y^2 + 1) to get 2y = (Ay + B)(y - 1) + C(y^2 + 1) 2y = Ay^2 - Ay + By - B + Cy^2 + C = (A + C)y^2 + (-A + B)y + (-B + C) Then I got a little confused. (A + C)y^2 = 2y (-A + B)y = 2y (-B + C) = 2y I won't go through the process of solving for anything on this answer, although I did it on paper. I became stuck after this.
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15:57:54 What is your result?
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RESPONSE -->
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15:58:05 How did you factor your denominator to get the integrand into a form amenable to partial fractions?
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RESPONSE --> My denominator was (y - 1)(y^2 + 1)
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15:58:51 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 --> 2y / [(y-1)(y^2 + 1)] = (Ay + B) / (y^2 + 1) + C / (y - 1)
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15:59:05 7.4.29 (3d edition #24). Integrate (z-1)/`sqrt(2z-z^2)
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RESPONSE --> To complete the square on in the denominator I rewrote it as z^2 - 2z + 0 I then completed the square to get (z^2 - 2z + 1) - 1 I then factored it to get (z - 1)^2 - 1 = 0 or 0 = 1 - (z - 1)^2 int( (z - 1) / sqrt[1 - (z-1)^2] dz I subsituted x - 1 = sin(z) w = sin(z) w' = cos(z) (z - 1) / sqrt[1 - (sin(z))^2] The Pythagorean identity, cos^2 + sin^2 = 1 so 1 - sin^2 = cos^2 [(z-1) / sqrt(cos^2(z))] * cos(z) int( (z - 1)(cos(z) / cos(z) z - 1 + C
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15:59:10 What did you get for your integral?
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RESPONSE -->
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15:59:11 What substitution did you use?
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RESPONSE -->
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16:20:06 7.4.40 (3d edition #28). integrate (y+2) / (2y^2 + 3y + 1)
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RESPONSE --> int( (y+2) / (2y^2 + 3y + 1) dy int( (y+2) / (2y + 1)(y + 1) dy (y+2) / (2y + 1)(y + 1) = A / (2y + 1) + B / (y + 1) Multiply by (2y + 1)(y + 1) (y + 2) = A(y+1) + B(2y + 1) (y + 2) = Ay+A + 2By + B (y + 2) = (A + 2B)y + A + B (A + 2B)y = y A = 2B A + B = 2 2B + B = 2 3B = 2 B = 2/3 A = 2(2/3) = 4/3 (y+2) / (2y^2 + 3y + 1) = (4/3) / (2y + 1) + (2/3) / (y + 1) This is as far as I know to go.
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16:20:08 What is your integral and how did you obtain it?
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RESPONSE -->
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16:20:42 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE -->
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