#$&* course Mth 272 October 21 around 7:20 015.
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Given Solution: `a We let u = x du = dx dv = e^(-x)dx v = -e^(-x) Using u v - int(v du): (x)(-e^(-x)) - int(-e^(-x)) dx Integrate: x(-e^(-x)) - (e^(-x)) + C Factor out e^(-x): e^(-x) (-x-1) + C. Self-critique (if necessary): ok ------------------------------------------------ self-critique rating #$&*: 3 ********************************************* Question: `qQuery problem 6.1.7 integrate x^2 e^(-x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u=x^2 dv=e^(-x)dx v=-e^(-x) -x^(2)e^(-x)-(-e^(-x)(2xdx))=-x^(2)e^(-x)+2(xe^(-x)dx) u=x dv=e^(-x)dx v=-e^(-x) -x e^(-x)-(-e^(-x)dx)=-xe^(-x)-e^(-x)+c -x^(2)e^(-x)+2(-xe^-x-e^-x+c)=-e^(-x)(x^(2)+2x+2)+c confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a We perform two integrations by parts. First we use u=x^2 dv=e^-x)dx v= -e^(-x) to obtain -x^(2)e^(-x) - int [ -e^(-x) * 2x dx] =-x^(2)e^(-x) +2int[xe^(-x) dx] We then integrate x e^-x dx: u=x dv=e^(-x)dx v= -e^(-x) from which we obtain -x e^(-x) - int(-e^(-x) dx) = -x e^(-x) - e^(-x) + C Substituting this back into -x^(2)e^(-x) +2int[xe^(-x) dx] we obtain -x^(2)e^(-x) + 2 ( -x e^-x - e^-x + C) = -e^(-x) * [x^(2) + 2x +2] + C. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ self-critique rating #$&*: 3 ********************************************* Question: `qQuery problem 6.1.26 integral of 1 / (x (ln(x))^3) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u=ln(x) du=1/xdx 1/u^3du=-1/(2u^2)+c -1/(2ln(x)^2)+c confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Let u = ln(x) so that du = 1 / x dx. This gives you 1 / u^3 * du and the rest is straightforward: 1/u^3 is a power function so int(1 / u^3 du) = -1 / (2 u^2) + c. Substituting u = ln(x) we have -1 / (2 ln(x)^2) + c. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ self-critique rating #$&*: 3 ********************************************* Question: `qQuery problem 6.1.46 integral of ln(1+2x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u=ln(1+2x) du=2/(1+2x)dx dv=dx v=x w=1+2x dw=2dx dx=dw/2 x=(w-1)/2 x/(1+2x)dx (((w-1)/2)/w)dw/2=((1/4)-(1/(4w)))dw Antiderivative is (w/4)-(1/4)ln(w) (2x)/4-(1/4)ln(1+2x) xln(1+2x)-2(x/(1+2x)) xln(1+2x)-2((2x)/4-(1/4)ln(1+2x)) xln(1+2x)+ln(1+2x)/2-x 0ln(1+2(0))+ln(1+2(0))/2-0=0 1ln(1+2(1))+ln(1+2(1))/2-1=0.648 0+0.648=0.648 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Let u = ln ( 1 + 2x) du = 2 / (1 + 2x) dx dv = dx v = x. You get u v - int(v du) = x ln(1+2x) - int( x * 2 / (1+2x) ) = x ln(1+2x) - 2 int( x / (1+2x) ). The integral is done by substituting w = 1 + 2x, so dw = 2 dx and dx = dw/2, and x = (w-1)/2. Thus x / (1+2x) dx becomes { [ (w-1)/2 ] / w } dw/2 = { 1/4 - 1/(4w) } dw. Antiderivative is w/4 - 1/4 ln(w), which becomes (2x) / 4 - 1/4 ln(1+2x). So x ln(1+2x) - 2 int( x / (1+2x) ) becomes x ln(1+2x) - 2 [ (2x) / 4 - 1/4 ln(1+2x) ] or x ln(1+2x) + ln(1+2x)/2 - x. Integrating from x = 0 to x = 1 we obtain the result .648 approx. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ self-critique rating #$&*: 3 ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was surprised that I knew how to do every problem in this assignment. ------------------------------------------------ self-critique rating #$&*: 3 "