#$&*
course 272 11/19 8:30pm 017..............................................
Given Solution: `a First we factor x the demoninator: 3/ [ x(x-3) ] Use partial fractions: 3/ [ x(x-3) ] = A/x + B/(x-3) Multiply both sides by common denominator x(x-3) to get 3 = A(x-3) + B(x) or 3 = (A+B) x - 3 A, which is the same as 0 x + 3 = (A + B) x - 3 A. The coefficients of x on both sides must be the same so we have A + B = 0 (coefficients of x) and -3 A = 3 . From the second we get A = -1. Substituting this into the first we solve to get B = 1. So our integrand is 3 / (x^2 - 3x) = -1 / (x-3) + 1 / x. The integration is straightforward. We get ln |x-3| - ln |x| + c , which we rewrite using the laws of logarithms as ln | (x-3) / x | + c. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3
********************************************* Question: `qQuery problem 6.2.27 (7th edition 6.3.29) (was 6.3.27) integrate (x+2) / (x^2 - 4x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: int (x+2) / (x^2 - 4x) factor out the x : (x+2) / x(x - 4) = A/x + B/(x-4) x+2 = A(x-4) + B(x)….(x+2) = (A+B)x - 4A A+B = 1 -4A= 2 so A = (-1/2) B = (3/2) Int (-1/2) |x + (3/2) | (x-4) = (3/2) ln |x-4| - (1/2) ln |x| + c = (1/2) ln (|x-4|^3 / |x|) + c confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: `a We factor x out of the denominator to get (x+2)/ [ x(x-4) ] Use partial fractions: (x+2)/x(x-4) = A/x + B/(x-4) Multiply both sides by common denominator: x+2 = A(x-4) + B(x) or x+2 = (A+B) x - 4 A. Thus A + B = 1 and -4 A = 2 so A = -1/2 and B = 3/2. Our integrand becomes (-1/2) / x + (3/2) / (x-4). The general antiderivative is easily found to be 3/2 ln |x-4| - 1/2 ln |x| + c, which can be expressed as 1/2 ln ( |x-4|^3 / | x | ) + c &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: 3
********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. I am comfortable with these calculations. It is nice putting together pieces we have been learning throughout this class, I really like the structure of the queries to piece together concepts. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. I am comfortable with these calculations. It is nice putting together pieces we have been learning throughout this class, I really like the structure of the queries to piece together concepts. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! `gr31#$&* course 272 11/19 8:30pm 017.
.............................................
Given Solution: `a First we factor x the demoninator: 3/ [ x(x-3) ] Use partial fractions: 3/ [ x(x-3) ] = A/x + B/(x-3) Multiply both sides by common denominator x(x-3) to get 3 = A(x-3) + B(x) or 3 = (A+B) x - 3 A, which is the same as 0 x + 3 = (A + B) x - 3 A. The coefficients of x on both sides must be the same so we have A + B = 0 (coefficients of x) and -3 A = 3 . From the second we get A = -1. Substituting this into the first we solve to get B = 1. So our integrand is 3 / (x^2 - 3x) = -1 / (x-3) + 1 / x. The integration is straightforward. We get ln |x-3| - ln |x| + c , which we rewrite using the laws of logarithms as ln | (x-3) / x | + c. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qQuery problem 6.2.27 (7th edition 6.3.29) (was 6.3.27) integrate (x+2) / (x^2 - 4x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: int (x+2) / (x^2 - 4x) factor out the x : (x+2) / x(x - 4) = A/x + B/(x-4) x+2 = A(x-4) + B(x)….(x+2) = (A+B)x - 4A A+B = 1 -4A= 2 so A = (-1/2) B = (3/2) Int (-1/2) |x + (3/2) | (x-4) = (3/2) ln |x-4| - (1/2) ln |x| + c = (1/2) ln (|x-4|^3 / |x|) + c confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a We factor x out of the denominator to get (x+2)/ [ x(x-4) ] Use partial fractions: (x+2)/x(x-4) = A/x + B/(x-4) Multiply both sides by common denominator: x+2 = A(x-4) + B(x) or x+2 = (A+B) x - 4 A. Thus A + B = 1 and -4 A = 2 so A = -1/2 and B = 3/2. Our integrand becomes (-1/2) / x + (3/2) / (x-4). The general antiderivative is easily found to be 3/2 ln |x-4| - 1/2 ln |x| + c, which can be expressed as 1/2 ln ( |x-4|^3 / | x | ) + c &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. I am comfortable with these calculations. It is nice putting together pieces we have been learning throughout this class, I really like the structure of the queries to piece together concepts. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. I am comfortable with these calculations. It is nice putting together pieces we have been learning throughout this class, I really like the structure of the queries to piece together concepts. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!