Query 14

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Given Solution: `a To integrate (2t-1) / (t^2-t+2) use u = t^2 - t + 2. We find that du = (2 t - 1) dt, which is there just waiting for you in the integrand. This gives you integrand du / u. The integral is ln | u | + c. Substituting we get int( (2t-1) / (t^2 - 1 + 2) with respect to t) = ln | t^2 - t + 2 | + c. The absolute value is important because t^2 - t + 2 can be negative. STUDENT QUESTION: Having a hard time still seeing how to get started
INSTRUCTOR RESPONSE Understood. This is a common problem. Once you see what to do it's fairly straightforward, but how do you see what needs to be done?
Generally I recommend checking to see if any other part of the integrand has a derivative which is equal to, or a multiple of, a factor of the integrand. Just what does this mean? In the present case the integrand is (2t-1)/(t^2-t+2). What are the factors of the integrand? The integrand has only one factor, which is (2 t - 1). What are the 'other parts' of the integrand, and what are their derivatives? The only thing you have besides the factor (2 t - 1) is the denominator t^2 - t + 2. The derivative of this 'part' is 2 t - 1. Do any of those derivatives match any of the factors? Yes. The derivative is 2 t - 1, and the only factor is 2 t - 1. If the answer to the above is 'yes', then let u be the part whose derivative is equal to the factor and proceed. Ok, so u = t^2 - t + 1. Thus du/dt = 2 t - 1, so du = (2 t - 1) dt. The expression (2 t - 1) / (t^2 - t + 2) dt has 'numerator' (2 t - 1) dt and denominator u. So the expression can be written du / u. An antiderivative of du / u is ln | u |, and the general antiderivative is ln | u | + c. Thus we get our general solution ln | t^2 - t + 2 | + c. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3

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Given Solution: `a If we let u = (sqrt x) + 1 we can solve for x to get x = (u-1)^2 so that dx = 2(u-1) du. So the integral of 1 / (`sqrt(x) + 1) dx becomes the integral of (2 ( u - 1 ) / u) du. Integrating ( u - 1) / u with respect to u we express this as ( u - 1) / u = (u / u) - (1 / u) = 1 - (1 / u). An antiderivative is u - ln | u |. Substituting u = (sqrt x) + 1 and adding the integration constant c we end up with (sqrt x) + 1 - ln | (sqrt x) + 1 | + c. Our integral is of 2 (u-1)/u, double the expression we just integrated, so our result will also be double. We get 2 (sqrt x) + 2 - 2 ln | (sqrt x) + 1 | + c. Since c is an arbitrary constant, 2 + c is also an arbitrary constant so the final solution can be expressed as 2 (sqrt x) - 2 ln | (sqrt x) + 1 | + c. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3

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Given Solution: `a Begin by sketching a graph to see that there is a finite region bounded by the graph and by y = 0. Then find the points where the graph crosses the line y = 0 but finding where the expression takes the value 0: x(1-x)^(1/3) = 0 when x = 0 or when 1-x = 0. Thus the graph intersects the x axis at x = 0 and x = 1. We therefore integrate x (1-x)^(1/3) from 0 to 1. We let u = 1-x so du = dx, and x = 1 - u. This transforms the integrand to (1 - u) * u^(1/3) = u^(1/3) - u^(4/3), which can be integrated term by term, with each term being a power function. Our antiderivative is 3/4 u^(4/3) - 3/7 u^(7/3), which translates to 3/4 (1-x)^(4/3) - 3/7 ( 1-x)^(7/3). The result is 9/28 = .321 approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3

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Given Solution: `a For reference int(1155/32 x^3(1-x)^(3/2), x, a, b) means 'the integral of 1155 / 32 x^3 ( 1 - x)^(3/2), integrated with respect to x between limits x = a and x = b'. That would be written with an integral sign with limits a and b, then 1155/32 x^3(1-x)^(3/2) dx. It's easiest to integrate if you change the variable to u = 1 - x, which changes the integrand to 1155/32 (1 + u)^3 * u^(3/2). Expand the cube and multiply through by u^(3/2) to get a sum of four power functions, easily integrated The result is of course a bit messy: First we expand the cube: (1 + u)^3 = 1 + 3 u + 3 u^2 + u^3, so 1155/32 ((1 + u)^3)(u^(3/2)) = 1155 / 32 ( 1 + 3 u + 3 u^2 + u^3) * u^(3/2) = 1155 / 32 ( u^(3/2) + 3 u^(5/2) + 3 u^(7/2) + u^(9/2)).
An antiderivative is
1155 / 32 ( 2/5 u^(5/2) + 6/7 u^(7/2) + 6/9 u^(9/2) + 2/11 u^(11/2) ).
Since u = 1 - x the limits on the u integral would be 1 - a and 1 - b.
The result would therefore be
1155 / 32 ( 2/5 (1-b)^(5/2) + 6/7 (1-b)^(7/2) + 6/9 (1-b)^(9/2) + 2/11 (1-b)^(11/2) ) - 1155 / 32 ( 2/5 (1-a)^(5/2) + 6/7 (1-a)^(7/2) + 6/9 (1-a)^(9/2) + 2/11 (1-a)^(11/2) ), which is slightly simplified to
1155 / 32 ( 2/5 (1-b)^(5/2) + 6/7 (1-b)^(7/2) + 6/9 (1-b)^(9/2) + 2/11 (1-b)^(11/2) ) - 2/5 (1-a)^(5/2) - 6/7 (1-a)^(7/2) - 6/9 (1-a)^(9/2) - 2/11 (1-a)^(11/2) ).
Factoring (1 - b)^(5/2) from the first four terms and (1 - a)^(5/2) from the last four terms, and simplifying yields
((1 - a)^(5/2)(105 a^3 + 70 a^2 + 40 a + 16) - (1 - b)^(5/2) (105 b^3 + 70 b^2 + 40 b + 16))/16. STUDENT COMMENT I understand the integration but the expanding of the cube is were I got lost. INSTRUCTOR RESPONSE Use the distributive and commutative laws to expand the cube: (1 + u)^3 = (1 + u) * (1 + u) ^2 (1 + u)^2 = (1 + u) ( 1 + u) = 1 ( 1 + u) + u * (1 + u) = 1 + u + u + u^2 = u^2 + 2 u + 1, so (1 + u)^3 = (1 + u) * (1 + u) ^2 = (1 + u) * (u^2 + 2 u + 1) = 1 * (u^2 + 2 u + 1) + u * (u^2 + 2 u + 1) = u^2 + 2 u + 1 + u^3 + 2 u^2 + u = u^3 + 3 u^2 + 3 u + 1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:

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Given Solution: `a The probability of an occurrence between 0 and .25 is found by integrating the expression from x = 0 to x = .25: Let u = 1-x so du = -dx and x = 1-u. Express in terms of u: -(1155/32) * int ( (1-u)^3 (u)^(3/2) du ) Expand the integrand: -(1155/32) * int( (1 - 3 u + 3 u^2 - u^3) * u^(3/2) ) = -1155/32 * int( u^(3/2) - 3 u^(5/2) + 3 u^(7/2) - u^(9/2) ) . An antiderivative of u^(3/2) - 3 u^(5/2) + 3 u^(7/2) - u^(9/2) is 2/5 u^(5/2) - 3 * 2/7 u^(7/2) + 3 * 2/9 * u^(9/2) - 2/11 u^(11/2) . So we obtain for the indefinite integral -1155/32 * ( 2/5 u^(5/2) - 3 * 2/7 u^(7/2) + 3 * 2/9 * u^(9/2) - 2/11 u^(11/2) ). Express in terms of x: -1155/32 * ( 2/5 ( 1 - x )^(5/2) - 3 * 2/7 ( 1 - x )^(7/2) + 3 * 2/9 * ( 1 - x )^(9/2) - 2/11 ( 1 - x )^(11/2) ) Evaluate this antiderivative at the limits of integration 0 and .25 . You get a probability of .0252, approx, which is about 2.52%, for the integral from 0 to .25; this is the probability of a result between 0 and 25%. To get the probability of a result between 50% and 100% integrate between .50 and 1. You get .736, which is 73.6% &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:

Query 14

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&#Good responses. Let me know if you have questions. &#