#$&* course mth 272 Question: `q5.6.26 (was 5.6.24 trap rule n=4, `sqrt(x-1) / x on [1,5] YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: `a Dividing [1, 5] into four intervals each will have length ( 5 - 1 ) / 4 = 1. The four intervals are therefore [1, 2], [2, 3], [3, 4], [4,5]. The function values at the endpoints f(1) = 0, f(2) = .5, f(3) = .471, f(4) = .433 and f(5) = .4. The average altitudes of the trapezoids are therefore (0 + .5) / 2 = 0.25, (.5 + .471)/2 = 0.486, (.471 + .433) / 2 = 0.452 and (.433 + .4) / 2 = 0.417. Trapeoid areas are equal to ave height * width; since the width of each is 1 the areas will match the average heights 0.25, 0.486, 0.452, 0.417. The sum of these areas is the trapezoidal approximation 1.604. ** DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q5.6.32 (was 5.6.24 est pond area by trap and midpt (20 ft intervals, widths 50, 54, 82, 82, 73, 75, 80 ft). How can you tell from the shape of the point whether the trapezoidal or midpoint estimate will be greater? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (0 + 50 / 2) = 25 (50 + 54) / 2 = 52 (54 + 82) / 2 = 68 (82 + 82) / 2 = 82 (82 + 73) / 2 = 77.5 (73 + 75) / 2 = 74 (75 + 80) / 2 = 77.5 20 ft (25+52+68+82+77.5+74+77.5+40) ft = 9920 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Since widths at 20-ft intervals are 50, 54, 82, 82, 73, 75, 80 ft the pond area can be approximated by a series of trapezoids with these altitudes. The average altitudes are therefore respectively (0 + 50 / 2) = 25 (50 + 54) / 2 = 52 (54 + 82) / 2 = 68 (82 + 82) / 2 = 82 etc., with corresponding areas 25 * 20 = 500 52 * 20 = 1040 etc., all areas in ft^2. The total area, according to the trapezoidal approximation, will therefore be 20 ft (25+52+68+82+77.5+74+77.5+40) ft = 9920 square feet. The midpoint widths would be calculated based on widths at positions 10, 30, 50, 70, ..., 150 ft. Due to the convex shape of the pond these estimates and will lie between the estimates made at 0, 20, 40, ..., 160 feet. The convex shape of the pond also ensures that the midpoint of each rectangle will 'hump above' the trapezoidal approximation, so the midpoint estimate will exceed the trapezoidal estimate. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:
#$&* course mth 272 Question: `q5.7.4 (was 5.7.4 vol of solid of rev abt x axis, y = `sqrt(4-x^2)What is the volume of the solid?
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Given Solution: `a*& If the curve is revolved about the x axis the radius of the circle at position x will be the y value radius = `sqrt(4 - x^2). The cross-section at that position will therefore be pi * radius^2 = pi ( sqrt(4 - x^2) ) ^ 2 = pi ( 4 - x^2). The volume of a 'slice' of thickeness `dx will therefore be approximately pi ( 4 - x^2 ) * `dx, which leads us to the integral int(pi ( 4 - x^2) dx, x from 0 to 2). An antiderivative of 4 - x^2 = 4 x - x^3 / 3. Between x = 0 and x = 2 the value of this antiderivative changes by 4 * 2 - 2^3 / 3 = 16 / 3, which is the integral of 4 - x^2 from x = 0 to x = 2. Multiplying by pi to get the desired volume integral we obtain volume = pi ( 16/3) = 16 pi / 3. *&*& DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q5.7.16 (was 5.7.16) vol of solid of rev abt x axis region bounded by y = x^2, y = x - x^2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^2 = x - x^2 x(2x-1) = 0 x = 0, x = ½ x - x^2, x^2 `pi (x-x^2)^2 - `pi(x^2)^2 `pi ( (x-x^2)^2 - x^2) -`pi/40 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a y = x^2 and y = x - x^2 intersect where x^2 = x - x^2, i.e., where 2x^2 - x = 0 or x(2x-1) = 0. This occurs when x = 0 and when x = 1/2. So the region runs from x=0 to x = 1/2. Over this region y = x^2 is less than y = x - x^2. When the region is revolved about the x ais we will get an outer circle of radius x - x^2 and an inner circle of radius x^2. The area between the inner and outer circle is `pi (x-x^2)^2 - `pi(x^2)^2 or `pi ( (x-x^2)^2 - x^2). We integrate this expression from x = 0 to x = 1/2. The result is -`pi/40. ** DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q5.7.18 (was 5.7.18) vol of solid of rev abt y axis y = `sqrt(16-x^2), y = 0, 0
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Given Solution: `a At x = 0 we have y = 4, and at x = 4 we have y = 0. So the y limits on the integral run from 0 to 4. At a given y value we have y = sqrt(16 - x^2) so that y^2 = 16 - x^2 and x = sqrt(16 - y^2). At a given y the solid will extend from x = 0 to x = sqrt(16 - y^2), so the radius of the solid will be sqrt(16 - y^2). So we integrate pi x^2 = pi ( sqrt(16 - y^2) ) ^2 = pi ( 16 - y^2) from y = 0 to y = 4. An antiderivative of pi ( 16 - y^2) with respect to y is pi ( 16 y - y^3 / 3). We get int( pi ( 16 - y^2), y, 0, 4) = pi ( 16 * 4 - 4^3 / 3) - pi ( 16 * 0 - 0^3 / 3) = 128 pi / 3. This is the volume of the solid of revolution. ** DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q5.7.31 (was 5.7.30) fuel take solid of rev abt x axis y = 1/8 x^2 `sqrt(2-x) What is the volume of the solid? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: radius=1/8 x^2 sqrt(2-x) area=pi * (1/8 x^2 sqrt(2-x) )^2 = pi/64 * x^4(2-x) = pi/64 * ( 2 x^4 - x^5) vol=int(pi/64 * ( 2 x^4 - x^5), x, 0, 2) 2 x^5 / 5 - x^6 / 6 pi/64 * [ 2 * 2^5 / 5 - 2^6 / 6 - ( 2 * 0^5 / 5 - 0^6 / 6) ] = pi / 64 * ( 64 / 30) = pi/30 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The radius of the solid at position x is 1/8 x^2 sqrt(2-x) so the cross-sectional area at that point is c.s. area = pi * (1/8 x^2 sqrt(2-x) )^2 = pi/64 * x^4(2-x) = pi/64 * ( 2 x^4 - x^5). The volume is therefore vol = int(pi/64 * ( 2 x^4 - x^5), x, 0, 2). An antiderivative of 2 x^4 - x^5 is 2 x^5 / 5 - x^6 / 6 so the integral is pi/64 * [ 2 * 2^5 / 5 - 2^6 / 6 - ( 2 * 0^5 / 5 - 0^6 / 6) ] = pi / 64 * ( 64 / 30) = pi/30. ** DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course mth 272 Question: `qQuery problem 6.1.5 (was 6.1.4) integral of (2t-1)/(t^2-t+2)YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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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 ********************************************* Question: `qQuery problem 6.1.32 (was 6.1.26) integral of 1 / (`sqrt(x) + 1) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u = (sqrt x) + 1 x = (u-1)^2 dx = 2(u-1) du 1 / (`sqrt(x) + 1) dx (2 ( u - 1 ) / u) du ( u - 1) / u = (u / u) - (1 / u) = 1 - (1 / u) u - ln | u | (sqrt x) + 1 - ln | (sqrt x) + 1 | + c 2 (sqrt x) + 2 - 2 ln | (sqrt x) + 1 | + c 2 (sqrt x) - 2 ln | (sqrt x) + 1 | + c confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a If we let u = (sqrt x) + 1 we can solve for x to get x = (u-1)^2 so that dx = 2(u-1) du. So the integral of 1 / (`sqrt(x) + 1) dx becomes the integral of (2 ( u - 1 ) / u) du. Integrating ( u - 1) / u with respect to u we express this as ( u - 1) / u = (u / u) - (1 / u) = 1 - (1 / u). An antiderivative is u - ln | u |. Substituting u = (sqrt x) + 1 and adding the integration constant c we end up with (sqrt x) + 1 - ln | (sqrt x) + 1 | + c. Our integral is of 2 (u-1)/u, double the expression we just integrated, so our result will also be double. We get 2 (sqrt x) + 2 - 2 ln | (sqrt x) + 1 | + c. Since c is an arbitrary constant, 2 + c is also an arbitrary constant so the final solution can be expressed as 2 (sqrt x) - 2 ln | (sqrt x) + 1 | + c. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qquery problem 6.1.56 (was 6.1.46) area bounded by x (1-x)^(1/3) and y = 0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x(1-x)^(1/3) = 0 the graph intersects the x axis at x = 0 and x = 1 u = 1-x du = dx x = 1 - u (1 - u) * u^(1/3) = u^(1/3) - u^(4/3) 3/4 (1-x)^(4/3) - 3/7 ( 1-x)^(7/3)=9/28=.321 confidence rating #$&*: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 ********************************************* Question: `qQuery problem P = int(1155/32 x^3(1-x)^(3/2), x, a, b). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u = 1 - x 1155/32 (1 + u)^3 * u^(3/2) (1 + u)^3 = 1 + 3 u + 3 u^2 + u^3 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)) 1155 / 32 ( 2/5 u^(5/2) + 6/7 u^(7/2) + 6/9 u^(9/2) + 2/11 u^(11/2) ) 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) ) ((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 confidence 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:3 ********************************************* Question: `qWhat is the probability that a sample will contain between 0% and 25% iron? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u = 1-x du = -dx x = 1-u -(1155/32) * int ( (1-u)^3 (u)^(3/2) du ) -(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) ) 2/5 u^(5/2) - 3 * 2/7 u^(7/2) + 3 * 2/9 * u^(9/2) - 2/11 u^(11/2) -1155/32 * ( 2/5 u^(5/2) - 3 * 2/7 u^(7/2) + 3 * 2/9 * u^(9/2) - 2/11 u^(11/2) ) -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) ) 73.6% confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:3 `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!