course MTH 174 12:28 a.m. 6/16/10. I realize this is a little late but I put it off to take a pre-test. I don't have time to submit the pretest though because it is so late and I'm taking the test tomorrow afternoon so I don't think you could get it back to me before then anyway. If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: e^3x((1/3 x^4) - (4/9 x^3) + (4/9 x^2) - (8/27 x ) + (8/81)) + c. Confidence: 3 The integral is of x^4 e^(3 x). x^4 is a polynomial, and e^(3 x) is of the form e^(a x). So the integrand is of the form p(x) e^(a x) with p(x) = x^4 and a = 3. The correct formula to use is #14 We obtain p ' (x) = 4 x^3 p '' (x) = 12 x^2 p ''' (x) = 24 x p '''' (x) = 24. Thus the solution is 1 / a * p(x) e^(a x) - 1 / a^2 * p ' (x) e^(a x) + 1 / a^3 * p''(x) e^(a x) - 1 / a^4 * p ''' (x) e^(a x) + 1 / a^5 * p''''(x) e^(a x) = 1 / 3 * x^4 e^(3 x) - 1 / 3^2 * 4 x^3 e^(3 x) + 1 / 3^3 * 12 x^2 e^(3 x) - 1 / 3^4 * 24 x e^(3 x) + 1 / 3^5 * 24 e^(3 x) = ((1/3) x^4 - (4/9) x^3 + (4/9) x^2 - (8/27)x + (8/81)) e^(3x) + C
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: Which formula from the table did you use?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I used #14. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Given Solution: You should have used formula 14, with a = 3 and p(x) = x^4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Didn't mention the a and p(x) part but you have to know that to use the formula anyway. ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: problem 7.3.7 (previously 7.3.33 1 / [ 1 + (z+2)^2 ) ]) **** What is your integral? **** Which formula from the table did you use and how did you get the integrand into the form of this formula?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My integral was: Int(1 / (1 + (z+2)^2 ))= ArcTan (z+2) + c. I used formula #24 and used a substitution with u= z+2 and u'= 1 dz. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Given Solution: If you let u = x+2 then du = dz and the integrand becomes 1 / (1 + u^2). This is the derivative of arctan(u), so letting u = z+2 gives us the correct result • arctan(z+2) + C Applying the formula: z is the variable of integration in the given problem, x is the variable of integration in the table. a is a constant, so a won't be z + 2. By Formula 24 the antiderivative of 1 / (a^2 + x^2) is 1/a * arcTan(x/a). Unlike some formulas in the table, this formula is easy to figure out using techniques of integration you should already be familiar with: 1 / (a^2 + x^2) = 1 / (a^2( 1 + x^2/a^2) ) = 1/a^2 ( 1 / (1 + (x/a)^2). Let u = x / a so du = dx / a; you get integrand 1 / a * (1 / (1 + u^2) ) , which has antiderivative 1/a * arcTan(u) = 1/a * arcTan(x/a). You don't really need to know all that, but it should clarify what is constant and what is variable. Starting with int(1/ (1+(z+2)^2) dz ), let x = z + 2 so dx = dz. You get int(1/ (1+x^2) dx ), which is formula 24 with a = 1. The result is 1/1 * arcTan(x/1), or just arcTan(x). Since x = z + 2, the final form of the integral is arcTan(z+2). Self-critique (if necessary): OK Self-critique rating #$&* ********************************************* Question: 7.4.1 (previously 7.4.6). Integrate 2y / ( y^3 - y^2 + y - 1)
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using partial fractions to evaluate this first gives: 2y / ( y^3 - y^2 + y - 1)= (Ay+B)/(y^2+1) + C/(y-1). Rearrangement produces the three linear equations we need to solve for A, B, and C. A+C=2, -A+B=0, -B+C=0. Solving these simultaneously gives: A= -1, B=1, C=1. Now we can form the three individual integrals with these three numerators. Int(1/(y^2+1)) - Int(y/(y^2+1)) + Int(1/(y-1)) = ArcTan y - 1/2 ln | y^2+1| + ln |y-1| + c. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The denominator factors by grouping: y^3 - y^2 + y – 1 = (y^3 + y) – (y^2 + 1) = y ( y^2 + 1) – 1 ( y^2 + 1) = (y – 1) ( y^2 + 1). Using partial fractions you would then have (a y + b) /(y^2 + 1) + c /(y-1) = y / ( (y^2+1)(y-1)) where we need to evaluate a, b and c. Putting the left-hand side over a common denominator (multiply the first fraction by (y-1)/(y-1) and the second by (y^2+1) / (y^2 + 1)) we obtain: [ (a y + b)(y-1) + c(y^2+1) ] / ((y^2+1)(y-1)) = 2 y / ((y^2+1)(y-1)). The denominators are identical so the numerators are equal, giving us (a y + b)(y-1) + c(y^2+1) = 2 y, or a y^2 + (-a + b) y - b + c y^2 + c = 2 y. Grouping the left-hand side: (a + c) y^2 + (a - b) y + c - b = 2 y. Since this must be so for all y, we have a + c = 0 (this since the coefficient of y^2 on the right is 0) -a + b = 1 (since the coefficient of y on the right is 1) c - b = 0 (since there is no constant term on the right). From the third equation we have b = c; from the first a = -c. So the second equation becomes c + c = 1, giving us 2 c = 1 and c = 1/2. Thus b = c = 1/2 and a = -c = -1/2. Our integrand (a y + b) /(y^2 + 1) + c /(y-1) becomes (-1/2 y + 1/2 ) / (y^2 + 1) + 1/2 * 1 / (y-1), or -1/2 y / (y^2 + 1) + 1/2 * 1 / (y^2 + 1) + 1/2 * 1 / (y-1). An antiderivative is easily enough found with or without tables to be -1/2 ln(y^2 + 1) + 1/2 arcTan(y) + 1/2 ln | y - 1 | + c. STUDENT COMMENTS: I have had some trouble figuring out this section. I couldn't figure out how to break down this denominator to make a partial fraction. Also I do not understand a step in the process in general. In the #10 class notes it explains another problem: This one is 1/[(x -3)(x +5)]. I understand this part of the notes: """"We know that when a fraction with denominator x-3 is added to a fraction with denominator x+5 we will obtain a fraction whose denominator is (x-3)(x+5). We conclude that it must be possible to express the given fraction as the sum A / (x-3) + B / (x+5), where A and B are numbers to be determined. Setting the original fraction equal to the sum we obtain an equation for A and B, as expressed in the third line. Multiplying both sides of the equation by (x-3) (x+5) we obtain the equation in the fourth line, which we rearrange to the form in the fifth line by collecting the x terms and the constant terms on the left-hand side and factoring."" This part I do not understand: """"Since the right-hand side does not have an x term, we see that A + B = 0"" How did you find that this equals 0? INSTRUCTOR RESPONSE: The equation for this function would be • A / (x-3) + B / (x+5) = 1/[(x -3)(x +5)] To simplify the left-hand side need to obtain a common denominator. We multiply the first term by (x + 5) / (x + 5), and the second term by (x - 3) / (x - 3): A / (x-3) * (x + 5) / (x + 5) + B / (x+5) * (x - 3) / (x - 3) = 1/[(x -3)(x +5)] so that A ( x + 5) / ( (x - 3) ( x + 5) ) + B ( x - 3) / ( (x - 3) (x + 5) ) = 1/[(x -3)(x +5)] . Adding the fractions on the left-hand side: ( A ( x + 5) + B ( x - 3) ) / ( ( x - 3) ( x + 5) ) = 1/[(x -3)(x +5)] . Simplifying the numerator we have ( (A + B) x + (5 A - 3 B) ) / ( ( x - 3) ( x + 5) ) = 1/[(x -3)(x +5)]. The denominators are equal, so the equation is solved if the numerators are equal: (A + B) x + (5 A - 3 B) = 1. It is this last equation which lacks an x term on the right-hand side. To maintain equality the left-hand side must also have no x term, which can be so only if A + B = 0. The other term 5 A - 3 B is equal to 1. Thus we have the simultaneous equations A + B = 0 5 A - 3 B = 1. These equations are easily solve, yielding the solution A = 1/8, B = -1/8. CONTINUED STUDENT COMMENT: I understand this: """"we see that therefore 5A - 3B = 1, so we have two equations in two unknowns A and B."""" I could not figure out how you found A and B as shown below: Solving these equations we obtain B = -1/8, A = 1/8, as indicated. We conclude that the expression to be integrated is A / (x-3) + B / (x+5) = 1/8 * 1/(x-3) - 1/8 * 1/(x+5). INSTRUCTOR RESPONSE The system A + B = 0 5 A - 3 B = 1. can be solved by elimination or substitution. Using substitution: Solve the first equation for A, obtaining A = -B. Substitute this value of A into the second equation. obtaining 5 * (-B) + (-3 B) = 1 so that -8 B = 1 and B = -1/8. Go back to the fact that A = -B to obtain A = - (-1/8) = 1/8. To solve by elimination you could add 3 times the first equation to the second, eliminating B and obtaining 8 A = 1, so that A = 1/8. Substituting this back into the first equation we obtain 1/8 + B = 0 so that B = -1/8. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): There is an error in the given solution. Very early on you put: (a y + b) /(y^2 + 1) + c /(y-1) = y / ( (y^2+1)(y-1)) where we need to evaluate a, b and c. You should have 2y/((y^2+1)(y-1)) This change leads to the three equations you used and not the three in my solution and so the A, B, C values are all ± 1/2 and this affects the final integral by having ± 1/2 in front of all three functions and not just (-1/2 ln |y^2+1|) like it should be.
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Given Solution: (y+2) / (2y^2 + 3 y + 1) = (y + 2) / ( (2y + 1) ( y + 1) ) = (y + 2) / ( 2(y + 1/2) ( y + 1) ) = 1/2 * (y + 2) / ( (y + 1/2) ( y + 1) ) The expression (y + 2) / ( (y + 1/2) ( y + 1) ) is of the form (cx + d) / ( (x - a)(x - b) ) with c = 1, d = 2, a = -1/2 and b = -1. Its antiderivative is given as 1 / (a - b) [ (ac + d) ln | x - a | - (bc + d) ln | x - b | ] + C. The final result is obtained by substitution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Something is wrong here. I have rechecked my answer several times and can't come up with your solution. I think something went awry in the given solution where the 1/2 is factored out somehow. I even resorted to Wolfram|Alpha to work out the partial fraction and it always shows my solution as being correct. I don't know though maybe our solutions are just rearranged or something. Self-critique rating #$&* 2