#$&* course Mth 174 Questions from text assignment:*********************************************
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20:35:16 Formula: Int{p(x) e^(ax)} dx = (1/a) p(x) e^(ax) dx = (1/a) p(x) e^(ax) - (1/a^2) Int{p’(x) e^(ax)} dx = (1/a) p(x) e^(ax) - [(1/a^2) p’(x) e^(ax) - (1/a^3) 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) 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) Int{p’’’’(x) e^(ax)} dx}]} p(x) = x^4 p’(x) = 4x^3 p’’(x) = 12x^2 p’’’(x) = 24x p’’’’(x) = 24 a = 3 Int{p’’’’(x) e^(ax)} dx} = Int{24 e^(3x)} = 8 e^(3x) Int{x^4 e^(3x)} = (1/3) (x^4) e^(3x) - {(1/3^2) (4x^3) e^(3x) - [(1/3^3) (12x^2) e^(3x) - {(1/3^4) (24x) e^(3x) -(1/3^5) Int{(24) e^(3x)} dx}]} = (1/3) x^4 e^(3x) - {(4/9) 4x^3 e^(3x) - [(12/27) x^2 e^(3x) - {(24/81) x e^(3x) - (1/243) Int{(24) e^(3x)} dx}]} = (4/9) x^4 e^(3x) - {(4/9) 4x^3 e^(3x) - [(4/9) x^2 e^(3x) - {(8/27) x e^(3x) - (1/243) {8 e^(3x)}]} + C = (1/3) x^4 e^(3x) - (4/9) 4x^3 e^(3x) + (4/9) x^2 e^(3x) - (8/27) x e^(3x) + (8/243) e^(3x) + C 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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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I seem to have done something wrong to end up with several extra e^(3x)’s. ------------------------------------------------ Self-critique Rating: 2
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Formula 14: Int{p(x) e^(ax)} dx = (1/a) p(x) e^(ax) dx , etc. 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: Question 7.3 Problem 7 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} Formula used: Int{1/(x^2 + a^2)} = (1/a) arctan(x/a) Formula modified: Int{1/(x^2 + a^2)} = Int{1/(a^2(1+ x^2/(a^2))} Int{(1/a^2) (1+ x^2/(a^2))} = (1/a) arctan(x/a) x = z+2; a = 1 Int{(1/1^2) (1 + (z+2)^2/(1^2))} = 1/1 arctan[(z+2)/1) = arctan(z+2) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: Problem 7.4 Problem 1 7.4.1 (previously 7.4.6). Integrate 2y / ( y^3 - y^2 + y - 1)
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int{2y / ( y^3 - y^2 + y - 1)} = 2 * Int{y / ( y^3 - y^2 + y - 1)} y^3 - y^2 + y - 1 = y^2(y - 1) + (y - 1) = (y^2 + 1)(y - 1) Using partial fractions: (a y + b) /(y^2 + 1) + c /(y-1) = y / [(y^2+1)(y-1)] = [(a y + b)(y-1) + c(y^2+1)] / [(y^2+1)(y-1)] = y / [(y^2+1)(y-1)] To find a, b, c: (a y + b)(y-1) + c(y^2+1) = a (y^2) + (-a + b) y - b + c (y^2) + c = (a + c) y^2 + (-a + b) y + c - b = y a + c = 0 -a + b = 1 c - b = 0 b = c a = -c So… c + c = 1 2c = 1 c = 1/2. b = c = 1/2 a = -c = -1/2. (a y + b) /(y^2 + 1) + c /(y-1) = 1/2 * (-y + 1 ) / (y^2 + 1) + 1/2 * 1 / (y-1) = (-1/2 ) y / (y^2 + 1) + 1/2 * 1 / (y^2 + 1) + 1/2 * 1 / (y-1) 2 * Int{(-1/2 ) y / (y^2 + 1) + 1/2 * 1 / (y^2 + 1) + 1/2 * 1 / (y-1)} = 2{-1/4 ln(y^2 + 1) + 1/2 arcTan(y) + 1/2 ln | y - 1 |} + C = - 1/2 ln(y^2 + 1) + arcTan(y) + ln | y - 1 | + C confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Let's integrate just y / (y^3 - y^2 + y - 1), then double the result. 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 so that 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 ) / (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 -1/4 ln(y^2 + 1) + 1/2 arcTan(y) + 1/2 ln | y - 1 | Doubling the result to get the integral of the given function we have - 1/2 ln(y^2 + 1) + arcTan(y) + ln | y - 1 | + c, where c now stands for an arbitrary integration constant. DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: Section 7.4 Problem 7 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: Int{(z-1)/sqrt(2z-z^2)} u = 2z - z^2 du = (2 - 2z) dz (z - 1)dz = -du/2 (z-1)/sqrt(2z-z^2) = 1/sqrt(u) * (-du/2) = (-1/2)u^(-1/2) du Int{(-1/2)u^(-1/2) du} = -u^(1/2) + C = -sqrt(2z-z^2) + C confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If you let u = 2z - z^2 you get du = (2 - 2z) dz or -2(z-1) dz; thus (z-1) dz = -du / 2. So (z-1) / `sqrt(2z - z^2) dz = 1 / sqrt(u) * (-du/2) = - .5 u^-.5 du, which integrates to -u^.5. Translated in terms of the original variable z we get -sqrt(2z-z^2). If you let u = 2z - z^2 you get du = (2 - 2z) dz or -2(z-1) dz; thus (z-1) dz = -du / 2. So (z-1) / `sqrt(2z - z^2) dz = 1 / sqrt(u) * (-du/2) = - .5 u^-.5 du, which integrates to -u^.5. Translated in terms of the original variable z we get -sqrt(2z-z^2). DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: Section 7.4 Problem 9 7.4.9 (previously 7.4.36) partial fractions for 1 / (x (L-x))
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Partial Fractions for 1 / [x (L-x)] 1 / [x (L-x)] = a / x + b / (L-x) = [ a (L-x) + bx ] / [ x(L-x)] = [aL + (b-a)x] / [x(L-x)] aL = 1; b - a = 0 a = 1/L b = 1/L = 1/x + b/(L-x) = 1/L [1/x + 1 / (L-x)] Int{1/L [1/x + 1 / (L-x)]} = 1 / L [ln(x) - ln(L-x)] confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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) ). **
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int{(y+2) / (2y^2 + 3y + 1)}dy (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)] Formula 27.: (cx + d) / [(x - a)(x - b)] = [1/(a-b)](ln|x - a| - ln|x - b|) + C c = 1, d = 2, a = -1/2, b = -1 Int{(y+2) / (2y^2 + 3y + 1)}dy = [1/((-1/2)-(-1))](ln|x - (-1/2)| - ln|x - (-1)|) + C = [1/(1/2)](ln|x + 1/2| - ln|x + 1|) + C = 2 ln|x - (-1/2)| - ln|x - (-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 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. ********************************************* Question: `q006. Integrate (t + 3) / (t^2 - 5 t + 6). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int{(t + 3) / (t^2 - 5 t + 6)} = (t + 3) / [(t - 6)(t + 1)] Formula 27: Int{(cx + d) / [(x - a)(x - b)]} dx = [1/(a - b)] [(ac + d) ln|x - a| - (bc + d) ln|x - b|] + C c = 1; d = 3; a = 6; b = -1 Int{(t + 3) / [(t - 6)(t + 1)]} = [1/(6 - (-1))] [((6)(1) + (3)) ln|x - (6)| - ((-1)(1) + (3)) ln|x - (-1)|] + C = (1/7) (9 ln|x - 6| - 2 ln|x + 1|) + C = (9/7) ln|x - 6| - (2/7) ln|x + 1| + C confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!