#$&* course Mth 272 7/4 2:16 pm 015.
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Given Solution: `a We let u = x du = dx dv = e^(-x)dx v = -e^(-x) Using u v - int(v du): (x)(-e^(-x)) - int(-e^(-x)) dx Integrate: x(-e^(-x)) - (e^(-x)) + C Factor out e^(-x): e^(-x) (-x-1) + C. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery problem 6.1.7 (was 6.2.3) integrate x^2 e^(-x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u = x^2 dv = e^(-x) dx v = -e^(-x) -x^2 e^(-x) - int(-e^(-x) * 2x dx) which equals 2 int(xe^(-x) dx) u = x dv = e^(-x) dx v = -e^(-x) -xe^(-x) - e^(-x) + c so -x^2 e^(-x) + 2(-xe^(-x) - e^(-x) + c) which simplifies to -e^(-x)(x^2 + 2x + 2) + c confidence rating #$&*: 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 1/u^3 du int(1/u^3 du) = -1/(2u^2) + c -1/(2 ln(x)) + c confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Let u = ln(x) so that du = 1 / x dx. This gives you 1 / u^3 * du and the rest is straightforward: 1/u^3 is a power function so int(1 / u^3 du) = -1 / (2 u^2) + c. Substituting u = ln(x) we have -1 / (2 ln(x)^2) + c. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery problem 6.1.46 (was 6.2.32) (was 6.2.34) integral of ln(1+2x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u = ln(1 + 2x) du = 2 / (1 + 2x) dx dv = dx v = x x(ln(1 + 2x)) - int (x(2/(1 + 2x))) which equals 2 int (x/(1 + 2x)) w = (1 + 2x) dw = 2 dx dx = dw / 2 x = (w - 1) / 2 (((w -1) / 2) / w) dw / 2 = (.25 -1/(4w)) dw w/4 - .25 ln(w) = 2x / 4 - .25 ln(1 + 2x) so x(ln(1 + 2x)) - 2(2x / 4 - .25 ln(1 + 2x) = x(ln(1 + 2x)) + ln(1 + 2x) / 2 - x f(1) - f(0) = .6479 - 0 = .6479 confidence rating #$&*: 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): I noticed a pattern of exponential antiderivatives where it keeps its exponent while transferring a coefficient or sign at the same time (e.g e^(-x) = -e^(-x)). That will prove useful later I think. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I noticed a pattern of exponential antiderivatives where it keeps its exponent while transferring a coefficient or sign at the same time (e.g e^(-x) = -e^(-x)). That will prove useful later I think.