course Mth 272 014.
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Given Solution: `a To integrate (2t-1) / (t^2-t+2) use u = t^2 - t + 2. We find that du = (2 t - 1) dt, which is there just waiting for you in the integrand. This gives you integrand du / u. The integral is ln | u | + c. Substituting we get int( (2t-1) / (t^2 - 1 + 2) with respect to t) = ln | t^2 - t + 2 | + c. The absolute value is important because t^2 - t + 2 can be negative. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): correct ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery problem 6.1.32 (was 6.1.26) integral of 1 / (`sqrt(x) + 1) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u= sqrt(x) +1 (u-1)^2= x 2(u-1)/u du =(u/u)- (1/u) =1- 1/u =u- ln |u| 2(sqrt(x)+1- ln |sqrt(x)+1| + C confidence rating: I feel pretty confident, but there was many steps. I would say 2. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a If we let u = (sqrt x) + 1 we can solve for x to get x = (u-1)^2 so that dx = 2(u-1) du. So the integral of 1 / (`sqrt(x) + 1) dx becomes the integral of (2 ( u - 1 ) / u) du. Integrating ( u - 1) / u with respect to u we express this as ( u - 1) / u = (u / u) - (1 / u) = 1 - (1 / u). An antiderivative is u - ln | u |. Substituting u = (sqrt x) + 1 and adding the integration constant c we end up with (sqrt x) + 1 - ln | (sqrt x) + 1 | + c. Our integral is of 2 (u-1)/u, double the expression we just integrated, so our result will also be double. We get 2 (sqrt x) + 2 - 2 ln | (sqrt x) + 1 | + c. Since c is an arbitrary constant, 2 + c is also an arbitrary constant so the final solution can be expressed as 2 (sqrt x) - 2 ln | (sqrt x) + 1 | + c. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My answer wasnt simplified fully, but still correct. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery problem 6.1.56 (was 6.1.46) area bounded by x (1-x)^(1/3) and y = 0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: zeros for x are 0 and 1 (1-u)*u^1/3 =u^1/3 - u^4/3 integrate= 3/4 u^4/3 - 3/7 u^7/3 plug U value in...... 3/4(1-x)^4/3 - 3/7(1-x)^7/3 confidence rating: I am confident with my answer. 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Begin by sketching a graph to see that there is a finite region bounded by the graph and by y = 0. Then find the points where the graph crosses the line y = 0 but finding where the expression takes the value 0: x(1-x)^(1/3) = 0 when x = 0 or when 1-x = 0. Thus the graph intersects the x axis at x = 0 and x = 1. We therefore integrate x (1-x)^(1/3) from 0 to 1. We let u = 1-x so du = dx, and x = 1 - u. This transforms the integrand to (1 - u) * u^(1/3) = u^(1/3) - u^(4/3), which can be integrated term by term, with each term being a power function. Our antiderivative is 3/4 u^(4/3) - 3/7 u^(7/3), which translates to 3/4 (1-x)^(4/3) - 3/7 ( 1-x)^(7/3). The result is 9/28 = .321 approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not plug in the vales to reach a result, but what I had was correct. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery problem P = int(1155/32 x^3(1-x)^(3/2), x, a, b). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1155/32(u^3/2+3u^5/2+3u^7/2+u^9/2) 1155/32(2/5u^5/2+ 6/7u^7/2+2/3u^9/2+2/11u^11/2) this is the integral, but I am not sure what to do at this point confidence rating: about a 2, b/c what I have should be right. I just do not know where to go from where I am. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a For reference int(1155/32 x^3(1-x)^(3/2), x, a, b) means 'the integral of 1155 / 32 x^3 ( 1 - x)^(3/2), integrated with respect to x between limits x = a and x = b'. That would be written with an integral sign with limits a and b, then 1155/32 x^3(1-x)^(3/2) dx. It's easiest to integrate if you change the variable to u = 1 - x, which changes the integrand to 1155/32 (1 + u)^3 * u^(3/2). Expand the cube and multiply through by u^(3/2) to get a sum of four power functions, easily integrated The result is of course a bit messy: First we expand the cube: (1 + u)^3 = 1 + 3 u + 3 u^2 + u^3, so 1155/32 ((1 + u)^3)(u^(3/2)) = 1155 / 32 ( 1 + 3 u + 3 u^2 + u^3) * u^(3/2) = 1155 / 32 ( u^(3/2) + 3 u^(5/2) + 3 u^(7/2) + u^(9/2)). An antiderivative is 1155 / 32 ( 2/5 u^(5/2) + 6/7 u^(7/2) + 6/9 u^(9/2) + 2/11 u^(11/2) ). Since u = 1 - x the limits on the u integral would be 1 - a and 1 - b. The result would therefore be 1155 / 32 ( 2/5 (1-b)^(5/2) + 6/7 (1-b)^(7/2) + 6/9 (1-b)^(9/2) + 2/11 (1-b)^(11/2) ) - 1155 / 32 ( 2/5 (1-a)^(5/2) + 6/7 (1-a)^(7/2) + 6/9 (1-a)^(9/2) + 2/11 (1-a)^(11/2) ), which is slightly simplified to 1155 / 32 ( 2/5 (1-b)^(5/2) + 6/7 (1-b)^(7/2) + 6/9 (1-b)^(9/2) + 2/11 (1-b)^(11/2) ) - 2/5 (1-a)^(5/2) - 6/7 (1-a)^(7/2) - 6/9 (1-a)^(9/2) - 2/11 (1-a)^(11/2) ). Factoring (1 - b)^(5/2) from the first four terms and (1 - a)^(5/2) from the last four terms, and simplifying yields ((1 - a)^(5/2)(105 a^3 + 70 a^2 + 40 a + 16) - (1 - b)^(5/2) (105 b^3 + 70 b^2 + 40 b + 16))/16. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Wow The answer was full of numbers. I guess that I was bound to mess up somewhere. Overall, I was close. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the probability that a sample will contain between 0% and 25% iron? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: after looking at the answer from the previous problem, I made sure my formula was right. With that, one should just plug in 0 and .25 for the x values. My results x=.25 .19486-.31316+.18268-.03737= (.02701)*-1155/32= -.97489 x=0 .4-.85714+.66667-.18182= (.02771)*-1155/32= -1.0002 ANSWER= .02531, which is 2.5% confidence rating: I am confident with my results. 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The probability of an occurrence between 0 and .25 is found by integrating the expression from x = 0 to x = .25: Let u = 1-x so du = -dx and x = 1-u. Express in terms of u: -(1155/32) * int ( (1-u)^3 (u)^(3/2) du ) Expand the integrand: -(1155/32) * int( (1 - 3 u + 3 u^2 - u^3) * u^(3/2) ) = -1155/32 * int( u^(3/2) - 3 u^(5/2) + 3 u^(7/2) - u^(9/2) ) . An antiderivative of u^(3/2) - 3 u^(5/2) + 3 u^(7/2) - u^(9/2) is 2/5 u^(5/2) - 3 * 2/7 u^(7/2) + 3 * 2/9 * u^(9/2) - 2/11 u^(11/2) . So we obtain for the indefinite integral -1155/32 * ( 2/5 u^(5/2) - 3 * 2/7 u^(7/2) + 3 * 2/9 * u^(9/2) - 2/11 u^(11/2) ). Express in terms of x: -1155/32 * ( 2/5 ( 1 - x )^(5/2) - 3 * 2/7 ( 1 - x )^(7/2) + 3 * 2/9 * ( 1 - x )^(9/2) - 2/11 ( 1 - x )^(11/2) ) Evaluate this antiderivative at the limits of integration 0 and .25 . You get a probability of .0252, approx, which is about 2.52%, for the integral from 0 to .25; this is the probability of a result between 0 and 25%. To get the probability of a result between 50% and 100% integrate between .50 and 1. You get .736, which is 73.6% &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My answer was correct. However, I got zero the first time because I did not round far enough. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the probability that a sample will contain between 50% and 100% iron? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: one should just plug in .50 and 1 into the equation given from the first part. Since using 1 in the equation gives you 0, just plug in .5 and that is the result. My results 1155/32(.071-.076+.029-.004) =1155/32(.02) =.72, which is 72% confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ pretty confident. 3
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Given Solution: `a To get the probability of a result between 50% and 100% integrate between .50 and 1. You get .736, which is 73.6% &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): my answer was very close. I must have not rounded enough.....Again ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. " xxxx