course 9/30/09 3:44pm If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htmYour solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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20:35:16 My solution: This integral fits formula 14 in the book which is p(x)e^(ax) x^4 is p(x) and e^(3x) is e^(ax), a=3 p’(x)=4x^3 p’’(x)=12x^2 p’’’(x)=24x p’’’’(x)=24 so, 1/a*p(x)e^(ax)- 1/a* int(p’(x)e^(ax) dx= 1/a *p(x)e^(ax)- 1/a^2*p’(x)e^(ax)+ 1/a^3*p’’(x)e^(ax)- 1/a^4*p’’’(x)e^(ax) + 1/a^5*p’’’’(x)e^(ax)= 1/3*x^4*e^(3x) – 1/9*4x^3e^(3x)+ 1/27*12x^2e^(3x) -1/81*24xe^(3x) +1/243*24e^(3x)= e^(3x)((1/3)x^4-(4/9)x^3+(4/9)x^2-(8/27)x+(8/81))+c confidence rating:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given solution: 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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20:35:18
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I used formula number 14 as mentioned in my solution. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You should have used formula 14, with a = 3 and p(x) = x^4. Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* 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: int(1/[1+(z+2)^2]) u=x+2 du=dz (1/[1+u^2]), this is derivative of arctan(u) so if we let u=z+2 gives us arctan(z+2) +c so now we can use formula 24 which is int(1/(x^2+a^2))dx= (1/a)arctan(x/a) +c int(1/(1+(z+2)^2)dz) x=z+2 dx=dz so in the formula a=1. 1/1*arctan(x/1)=arctan(x)=arctan(z+2) 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:ok ********************************************* Question: 7.4.1 (previously 7.4.6). Integrate 2y / ( y^3 - y^2 + y - 1)
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: factor denominator to (y^3 + y)-(y^2+1)=y(y^2+1)-1(y^2+1)=(y-1)(y^2+1) so now we have (ay+b)/(y^2+1) +c/(y-1)= y/((y^2+1)(y-1)) ((ay+b)(y-1) +c(y^2+1))/((y^2+1)(y-1))= y/((y^2+1)(y-1)) because the denominators are the same we now see (ay+b)(y-1)+c(y^2+1)=y ay^2+(-a+b)y-b+cy^2+c=y (a+c)y^2+(a-b)y+c-b=y so a+c=0 a-b=1 c-b=0 so now b=c and a=-c, so that makes c+c=1 which means 2c=1, so c=1/2 because b=c, b also equals ½ and because a= -c, a also equal -1/2 substituting that in gives us ((-1/2y + ½)/ (y^2+1)) + (½(1)/(y-1))=((-1/2y)/(y^2+1))+((1/2)/y^2+1)+(1/(y-1))= -1/2ln(y^2+1)+1/2arctan(y)+1/2 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)) = 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) = y, or a y^2 + (-a + b) y - b + c y^2 + c = y. Grouping the left-hand side: (a + c) y^2 + (a - b) y + c - b = 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. DER 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):ok Self-critique Rating:ok ********************************************* Question: 7.4.12 (previously 7.4.29 (4th edition)). Integrate (z-1)/`sqrt(2z-z^2) **** What did you get for your integral? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u=2z-z^2 du=2-2z dz z-1=-du/2 1/sqrt(u) * -du/z = 1/2u^(-1/2)du -u^(-1/2)= -sqrt(2z-z^2) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Given Solution: If you let u = 2z - z^2 you get du = (2 - 2z) dz or -2(z-1) dz; thus (z-1) dz = -du / 2. So (z-1) / `sqrt(2z - z^2) dz = 1 / sqrt(u) * (-du/2) = - .5 u^-.5 du, which integrates to -u^.5. Translated in terms of the original variable z we get -sqrt(2z-z^2). If you let u = 2z - z^2 you get du = (2 - 2z) dz or -2(z-1) dz; thus (z-1) dz = -du / 2. So (z-1) / `sqrt(2z - z^2) dz = 1 / sqrt(u) * (-du/2) = - .5 u^-.5 du, which integrates to -u^.5. Translated in terms of the original variable z we get -sqrt(2z-z^2). DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: 7.4.9 (previously 7.4.36) partial fractions for 1 / (x (L-x))
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Your solution: 1/(x(L-x)) using partial fractions equals (a/x) + (b/(L-x)) and if trying to get common denominator you get (a(L-x)+ b(x))/(x(L-x)))= (aL-ax+bx)/(x(L-x))=(aL+(b-a)x)/(x(L-x)) so, (aL+(b-a)x)/(x(L-x))=1/(x(L-x)) aL=1 b-a=0, so b=a a=1/L and b=1/L (1/x) + (b/(L-x)) = (1/L(1/x+1/(L-x)) 1/L(ln(x)-ln(L-x)) = 1/ L ln(x/(L-x)) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: a / x + b / (L-x) = [ a (L-x) + bx ] / [ x(L-x)] = [ a L + (b-a)x ] / [ x(L-x)]. This is equal to 1 / [ x(L-x) ]. So a L = 1 and (b-a) = 0. Thus a = 1 / L, and since b-a=0, b = 1/L. The original function is therefore 1 / x + b / (L-x) = 1 / L [ 1 / x + 1 / (L-x) ]. Integrating we get 1 / L ( ln(x) - ln(L-x) ) = 1 / L ln(x / (L-x) ). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: 7.4.6 (previously 7.4.40 (3d edition #28)). integrate (y+2) / (2y^2 + 3y + 1)
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Your solution: (y+2)/(2y^2+3y+1) = (y+2)/(2y+1)(y+1) (y+2)/(2(y+1/2)(y+1))= (1/2(y+2))/(y+1/2)(y+1) this now fits #27 of the table of indefinite integrals a=-1/2 b= -1 c=1 and d=2 so, the integral given in the back of the book is (1/a-b)((ac+d)ln(x-a)-(bc+d)ln(x-b)) (1/(-1/2+1)(-1/2*1+2)ln(x+1/2)-(-1*1+2)ln(x+1)+1=5ln(x+1/2)+3ln(x+1)+1 confidence rating:: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):ok ------------------------------------------------ Self-critique Rating:ok "