#$&* course MTH 272 12/10 about 12:15 pm 015.
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Given Solution: `a We let u = x du = dx dv = e^(-x)dx v = -e^(-x) Using u v - int(v du): (x)(-e^(-x)) - int(-e^(-x)) dx Integrate: x(-e^(-x)) - (e^(-x)) + C Factor out e^(-x): e^(-x) (-x-1) + C. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery problem 6.1.7 (was 6.2.3) integrate x^2 e^(-x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u=x^2 du=2x dx dv=e^(-x) dx v = -e^(-x) Int. x^2e^(-x) = -x^2e^(-x) - int. ((-e^-x)*2x dx) u=2x du= 2 dx dv= -e^(-x) dx v=e^(-x) Then becomes x^2e^(-x) - [2xe^(-x) - int.(-2e^(-x) dx)] which equals -x^2e^(-x) - 2xe^(-x) - 2e^(-x) + c = -e^(-x)(x^2+2x+2) + c confidence rating #$&*: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:OK ********************************************* Question: `qQuery problem 6.1.26 (was 6.2.18) integral of 1 / (x (ln(x))^3) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u=ln x du = 1/x dx Int.(1/u^3) du = Int. u^(-3) du = (u^(-2))/2 +c = -1/(2*u^2) + c = -1/(2*lnx^2) + c confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Let u = ln(x) so that du = 1 / x dx. This gives you 1 / u^3 * du and the rest is straightforward: 1/u^3 is a power function so int(1 / u^3 du) = -1 / (2 u^2) + c. Substituting u = ln(x) we have -1 / (2 ln(x)^2) + c. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery problem 6.1.46 (was 6.2.32) (was 6.2.34) integral of ln(1+2x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u=ln(1+2x) du=2/(1+2x) dx dv=dx v=x Int.(ln(1+2x) dx) = x*ln(1+2x) - int.(x*(2/(1+2x)) dx U=1+2x du = 2 dx x=(u-1)/2 dx=du/2 X*ln (1+2x) - 2*int.((u-1)/2 / u * du/2) = x*ln (1+2x) - 2*int.((u-1)/4 du) x*ln (1+2x) - 2*int.(((1/4) - (1/4u) du) = x*ln (1+2x) - 2*(((u/4) - (1/4)(ln u) + c x*ln (1+2x) - 2*((((1+2x)/4) - (1/4)(ln (1+2x)) + c = x*ln (1+2x) - ((2x+1)/2) + (1/2)*ln(1+2x) + c Plug in the limits of 0 to 1 and subtract to get .14798 - -0.5 = 0.647918 confidence rating #$&*: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:OK ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!