course Mth 174 This assignment was pretty tough. I will need to spend more time with substitution and integration by parts. ݄𗰴Ш䙺assignment #004
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21:28:07 query problem 7.2.12 (3d edition #11) antiderivative of sin^2 x
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RESPONSE --> I used the inegration by parts method. u = sin x u' = cos x v' = sin x v = -cos x int(sin x * sin x) = sin x * -cos x - int(cos x * -cos x) int(sin^2 x) = sin x * -cos x - int(cos^2 x) That was as far as I got.
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21:28:32 what is the requested antiderivative?
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RESPONSE --> sin x * -cos x - int(cos^2 x)
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21:29:22 What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?
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RESPONSE --> I broke it into parts. u = sin x u' = cos x v' = sin x v = -cos x
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21:59:09 query problem 7.2.16 antiderivative of (t+2) `sqrt(2+3t)
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RESPONSE --> I am still really bad at substitution, which is what I tried to use to answer this problems. w = 2 + 3t w' = 3 t = (w-2) / 3 t' = 1/3 (w - 2) / 3 + 2 * sqrt(w) + C
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21:59:11 what is the requested antiderivative?
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RESPONSE -->
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21:59:13 What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?
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RESPONSE -->
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22:10:16 query problem 7.2.27 antiderivative of x^5 cos(x^3)
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RESPONSE --> I used substitution again on this one. w = x^3 w' = 3x^2 x = w^(1/3) x' = (1/3) * w^(4/3) int[w^(1/3)^(5) * cos^w] int[w^(5/3) * cos^w] (3/8)w^(8/3) * (sin^(w + 1)) / (w +1) + C
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22:17:36 query problem 7.2.52 (3d edition #50) f(0)=6, f(1) = 5, f'(1) = 2; find int( x f'', x, 0, 1).
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RESPONSE --> I would guess that the int(x * f''(x)) would be (1/2)x^2 * f'(x) + C
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22:17:38 What is the value of the requested integral?
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RESPONSE -->
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22:18:16 How did you use integration by parts to obtain this result? Be specific.
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RESPONSE --> I used substitution. I know this chapter is about integration by parts, but I am still trying to be good at substitution
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22:18:19 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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