course Mth 174 *********************************************
......!!!!!!!!...................................
20:35:16 SOLUTION: [1/3 * x^4e^3x] – [ 1/3^2 * 4x^3e^3x ] + [ 1/3^3 * 12 x^2e^3x] – [1/3^4 * 24xe^3x] + [1/3^5 * 24 e^3x] 1/3x^4 – 4/9x^3 + 12/27x^2 – 24/81x + 24/243e^3x + c [ 1/3x^4 – 4/9x^3 + 4/9x^2 – 8/27x + 8/81 e^3x +c ]
.............................................
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
.........................................
20:35:18
......!!!!!!!!...................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: Which formula from the table did you use?
......!!!!!!!!...................................
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Formula 14 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: You should have used formula 14, with a = 3 and p(x) = x^4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: problem 7.3.7 (previously 7.3.33 1 / [ 1 + (z+2)^2 ) ]) **** What is your integral? **** Which formula from the table did you use and how did you get the integrand into the form of this formula?
.......!!!!!!!!...................................
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Let u= z+2 so—1/(1+u^2) Formula 24 1/a^2 + x^2 1/a * arctan (x/a) If a = 1 and x = z+2 then 1/1 * arctan ( z+2/1) Which = arctan (z+2) confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: 7.4.1 (previously 7.4.6). Integrate 2y / ( y^3 - y^2 + y - 1)
.......!!!!!!!!...................................
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: [ y^3 – y^2 + y – 1 ] = y(y^2+1) – 1(y^2+1) = [(y-1)(y^2+1)] y/ (y-1)(y^2+1) (a y + b / y^2 + 1) + c/(y+1) = [-y/ y^2 +1 + 1/y^2+1 + 1/ y-1] confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 = 2, giving us 2 c = 2 so that c = 1. Thus b = c = 1 and a = -c = -1. Our integrand (a y + b) /(y^2 + 1) + c /(y-1) becomes (-y + 1 ) / (y^2 + 1) + 1/2 * 1 / (y-1), or -y / (y^2 + 1) + 1 / (y^2 + 1) + 1 / (y-1). An antiderivative is easily enough found with or without tables to be -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): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: 7.4.12 (previously 7.4.29 (4th edition)). Integrate (z-1)/`sqrt(2z-z^2) **** What did you get for your integral? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Let u = 2z – z^2 and du= (2 -2z)dz -2(z-1)dz (z-1)dz = -du/2 (z-1)/ sqrt [2z – z^2] dz = 1/sqrt(u) * -du/2 = - ½ u ^ - ½ du = - u ^ ½ == sqrt[2z – z^2 ] confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: 7.4.9 (previously 7.4.36) partial fractions for 1 / (x (L-x))
......!!!!!!!!...................................
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: L= 1 & a= 1/L & b = 1/L & (b-a)=0 1/ L [1/x + 1/(L-x) ] confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: 7.4.6 (previously 7.4.40 (3d edition #28)). integrate (y+2) / (2y^2 + 3y + 1)
......!!!!!!!!...................................
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1/(-1/2 – (-1)) [ (-1/2*1 + 2) ln | x- (-1/2) | - (-1*1 + 2) ln | x – (-1) | ] + c 1/(1/2) [3/2 ln | x + ½ | - 1 ln | x + 1 | + c confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ------------------------------------------------ Self-critique rating #$&* "