course Mth 174 Question: problem 7.3.3 (previously 7.3.15) x^4 e^(3x) **** what it is your antiderivative?......!!!!!!!!...................................
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20:35:18
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Formula #14. Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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))dz Using formula #24 with a=1 and x=z+2 = (1)arctan((z+2)/1) + C = arctan(z+2) + C Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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: 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))dy (= I) = Int.((2y)/((y-1)(y^2+1))dy = (Ay+B)/(y^2+1) + C/(y-1) = Ay^2-Ay+By-B+Cy^2+C = y^2(A+C)-y(A+B)-B+C A+C=0 and –A+B=1 and C-B=0 So A = -1/2 and B=1/2 and C=1/2 Therefore, I = Int.(((-1/2y+1/2)/y^2+1)+((1/2)/(y-1)))dy = -1/2*ln Abs(y^2+1) + 1/2*arctan(y) + 1/2*ln Abs(y-1) +C Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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: Int.((z-1)/sqrt(2z-z^2))dz with u = 2z-z^2 so du = 2-2z dz so (1-z)dz = -du/2 = Int.(1/sqrt(u))(-du/2) = Int.(-1/2*u^(-1/2))du = -u^(-1/2) = -sqrt(2z-z^2) + C Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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: 7.4.9 (previously 7.4.36) partial fractions for 1 / (x (L-x))
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int.(1/(x(L-x)))dx (= I) = A/x +B/(L-x) = A(L-x) + Bx = (x(-A+B) + AL)/(x(L-x)) So, AL = 1 and (-A+B) = 0. Therefore, A = 1/L and B = 1/L and L = 1 So, I = Int.(1/(L(1/x + 1/(L-x))))dx = 1/L*ln(x/(L-x)) + C Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int.((y+2)/(y^2 + 3y + 1))dy = ½*Int.((y+2)/((y+1/2)(y+1)))dy Using formula 27, a = -1/2, b = -1, c = 1, d = 2. So, 1/(-1/2+1)((-1/2+2)*lnAbs.(y+1/2) – (-1 + 2)*lnAbs(y+1)) + C Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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 "