course Mth 272 7/28 10:30 am 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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery problem 6.2.3 integrate x^2 e^(-x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve this integral, we first need to identify u, du, dv, and v. We find that u = x^2 and du = dx = 2x. Dv = e^-x and v = -e^-x. Now to solve the integral, we can simply put these values into the given formula. So the integral = (x^2)(e^-x) Int(2x(-e^-x)). We can use integration by parts again to solve the integral in our answer. For this integral, we find that u = x and du = dx. Dv = -e^-x, and v= e^-x. Now we can make our whole expression = (x^2)(e^-x) ((x)(e^-x) In(e^-x)dx, which simplifies to (x^2)(e^-x) ((x)(2e^-x). 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: ********************************************* Question: `qQuery problem 6.2.18 integral of 1 / (x (ln(x))^3) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve the integral, we can first identify u and du. We find that u = lnx and du = 1/x. Now we can take the integral of our new expression, 1/u^3 or u^-3, using our integration rules. We find that our integral is -1/2u^2 + C. Now we can finally rewrite our expression as -1/2(lnx)^2+C when we substitute u back in. 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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery problem 6.2.32 (was 6.2.34) integral of ln(1+2x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve the integral, we first need to identify u, du, dv, and v. We find that u = ln(1+2x) and du = 2/(1+2x). Dv = dx, so v=x. Now using the formula, we find that our integral = (ln(1+2x))(x) Int(x (1/(1+2x)). So now we can take the integral of the second half of this expression using the formula. For this integral, we find that u = 1+2x and du = 2. Dv = du/2 and v = (u-1)/2. This makes our new integral = u/4 lnu, which when u is substituted back in equals (1+2x)/4-ln(1+2x). When we put our whole expression together, we find that it = x ln(1+2x)-2((2x)/4-1/4ln(1+2x)). Now to find the area from 0 to 1, we can use the Fundamental Theorem of Calculus. Area = (1) ln(1+2(1))-2((2(1))/4-1/4ln(1+2(1))) (0 ln(1+2(0))-2((2(0))/4-1/4ln(1+2(0))) = .648. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I had a lot of trouble getting through the steps in this problem at first, and need to do more practice problems using this formula.