course MTH 174 I am resubmitting this per the instructions in orientation about resubmitting work that has not been posted within a day. ×ó¯îÂòe~ÍÜðî½ç¾OwÌÊ°JÑš²[©±assignment #004 š¿Óî|ÂÆô Ðýé hŽrâÏWŽÊê° Physics II 09-22-2008
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19:40:17 query problem 7.2.12 (3d edition #11) antiderivative of sin^2 x
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RESPONSE --> This one was in section 7.3 and was #12. Section 7.2 was Integration by substitution and this seems better fitted to integration by parts. Here is the solution that I have: Using integration by parts (uv - integral uv' ): integral sin^2 x integral sin x * sin x let u = sin x and v' = sin x du/dx = cos x so du = cos x dx v = -cos x (+ C omitted here) uv - integral uv' sin x * - cos x - integral - cos^2 x dx -sin x cos x + integral (1 - sin^2 x) dx -sin x cos x + integral 1 dx - integral sin^2 x We see sin^2 x again. This integral is equation above is supposed to be equal to integral sin^2 x -sin x cos x + integral 1 dx - integral sin^2 x = integral sin^2 x -sin x cos x + x + C = 2 integral sin^2 x -1/2 sin x cos x + x/2 + C = integral sin^2 x
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19:40:33 what is the requested antiderivative?
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RESPONSE --> -1/2 sin x cos x + x/2 + C
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19:42:29 What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?
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RESPONSE --> The sin^2 x was broken into sin x and sin x. The technique used was integration by parts. The trick here was using the fact that integral of sin^2 x appeared again in the problem allowing the movement of it to the other side of the equation.
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19:58:35 query problem 7.2.16 antiderivative of (t+2) `sqrt(2+3t)
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RESPONSE --> This one also appeared in section 7.3 in the integration by parts. I used u = t+2 and v' = (2 + 3t)^(1/2) du/dt = 1 substituting in w = 2+3t and dw/dt = 3, so 1/3dw = dt v' = 1/3 w^(1/2) dw v = 1/3 * 2/3 w^(3/2) v = 2/9 (2 + 3t)^(3/2) uv - integral uv' (t+2)[2/9 (2 + 3t)^(3/2)] - integral 1 * 2/9 (2 + 3t)^(3/2) dt Again, letting w = 2+3t and 1/3dw = dt 2/9(t +2)(2 + 3t)^(3/2) - 2/9 integral 1/3 w^(3/2) dw 2/9(t +2)(2 + 3t)^(3/2) - 2/27 * 2/5 (2 + 3t)^(5/2) + C 2/9(t +2)(2 + 3t)^(3/2) - 4/135(2 + 3t)^(5/2) + C
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19:58:46 what is the requested antiderivative?
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RESPONSE --> 2/9(t +2)(2 + 3t)^(3/2) - 4/135(2 + 3t)^(5/2) + C
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20:00:05 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 u = t+2 and v' = (2 + 3t)^(1/2) and used integration by parts. Substitution was also used in finding the antiderivative for v and in the integral portion of the integration by parts.
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20:10:19 query problem 7.2.27 antiderivative of x^5 cos(x^3)
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RESPONSE --> This one also was in 7.3. Again, I used integration by parts (uv - integral uv'). integral x^5 cos(x^3) dx let u = x^3 du/dx = 3x^2 v' = x^2 cos(x^3) Substituting w = x^3, dw/dx = 3x^2 so 1/3dw = x^2dx v' = 1/3 cos w dw v = 1/3 sin (x^3) uv - integral uv' x^3 * 1/3 sin (x^3) - integral 3x^2 * 1/3 sin (x^3) dx x^3 * 1/3 sin (x^3) - integral x^2 sin (x^3) dx Again, substituting w = x^3: dw/dx = 3x^2 so 1/3dw = x^2dx x^3 * 1/3 sin (x^3) - integral 1/3 sin (w) dw 1/3 x^3 sin (x^3) + 1/3 cos (x^3) + C
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20:10:28 what is the requested antiderivative?
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RESPONSE --> 1/3 x^3 sin (x^3) + 1/3 cos (x^3) + C
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20:12:39 What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?
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RESPONSE --> let u = x^3 and v' = x^2 cos(x^3) The key was recognizing that the derivative of x^3 is 3x^2 after that, it was straight forward. The substitution of w = x^3 was used in the finding of the antiderivative of v' and in the integral in the integration by parts.
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20:49:51 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 am having problems visualizing this one. I'll describe what I have sketched out on my paper I have a trapezoidal graph of x vs t with a point ot (0,6) and a point at (1,5) and a slope between the two. At point (1,5) the slope is 2.
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20:50:12 What is the value of the requested integral?
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RESPONSE --> I am not sure.
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20:51:04 How did you use integration by parts to obtain this result? Be specific.
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RESPONSE --> I was unable to find the result.
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20:51:11 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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20:51:16 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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