#$&* course Mth 272 6/30 1 am Question: `qQuery problem 6.1.6 (was 6.2.2) integrate x e^(-x)
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Given Solution: `a We let u = x du = dx dv = e^(-x)dx v = -e^(-x) Using u v - int(v du): (x)(-e^(-x)) - int(-e^(-x)) dx Integrate: x(-e^(-x)) - (e^(-x)) + C Factor out e^(-x): e^(-x) (-x-1) + C. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: Ok ********************************************* Question: `qQuery problem 6.1.7 (was 6.2.3) integrate x^2 e^(-x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u= x^2 du= 2x dx dv= e^x v= e^x x^2 * e^x - integral(2x * e^x dx) x^2 * e^x - 2x*e^x + 2*integral(e^x dx) e^-x [-x^2 - 2x - 2] + C confidence rating #$&*: 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): 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= lnx du= 1/x integral 1/u^3 du -1/2(ln(x)^2) +C confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Let u = ln(x) so that du = 1 / x dx. This gives you 1 / u^3 * du and the rest is straightforward: 1/u^3 is a power function so int(1 / u^3 du) = -1 / (2 u^2) + c. Substituting u = ln(x) we have -1 / (2 ln(x)^2) + c. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: Ok ********************************************* Question: `qQuery problem 6.1.46 (was 6.2.32) (was 6.2.34) integral of ln(1+2x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u= ln(2x+1) du= 2dx/ (2x+1) uv= x*ln(2x+1) lnt(dx)- ln t (1/(2x+1)/dx)) x*ln(2x+1)- x + ln(2x+1)/2 confidence rating #$&*: 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): 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): When looking at notes, I started remembering all of this from calc. ------------------------------------------------ Self-critique Rating: Ok"