Assignment 32

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Given Solution: `a An antiderivative would be x^2 y + y^3 / 3. Evaluating this antiderivative at endpoints y = x^2 and y = `sqrt(x) we get [ x^2 * `sqrt(x) + (`sqrt(x) ) ^ 3 / 3 ] - [ x^2 * x^2 + (x^2)^3 / 3 ] = x^(5/2) + x^(3/2) / 3 - ( x^4 + x^6 / 3) = x^(5/2) + x^(3/2) / 3 - x^4 - x^6 / 3. This can be simplified in various ways, but the most standard form is just decreasing powers of x: - x^6/3 - x^4 + x^(5/2) + x^(3/2)/3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):My answer is the same as your expect times -1. See my question in the solution. I worked it out as I assumed but understand the process of your solution. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qQuery problem 7.8.14 integrate `sqrt(1-x^2) from 0 to x wrt y then from 0 to 1 wrt x YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For the inner integral you get (1-x^2)^(1/2)*y so plugging in the limits and subtracting you get x*(1-x^2)^(1/2) - 0 = x*(1-x^2)^(1/2) Integrating this result, you get U=(1-x^2) du=-2x (-1/2)*int. (u^(1/2)*du) = (-1/2)*(2/3)(u^(3/2))=(-2/6)*(u^(3/2))=(-(1-x^2)^(3/2))/3) Plugging in the limits you get (-(1-1^2)^(3/2))/3) - (-(1-0^2)^(3/2))/3)=0 - (-1/3) = 1/3 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: `a The limits on the first integral are 0 and x. The result of the first integral is then to be integrated with respect to x. The solution: ** Since the function has no y dependence the whole expression is treated as a constant and an antiderivative with respect to y is just `sqrt(1 - x^2) * y. The definite integral from 0 to x is therefore `sqrt(1 - x^2) * x - `sqrt(1 - x^2) * 0 = x sqrt(1-x^2). We now need to integrate the resulting expression x sqrt(1 - x^2) with respect to x, from x = 0 to x = 1. An antiderivative of x sqrt( 1 - x^2 ) is easily found by letting u = 1 - x^2, so that du = -2 x dx and x dx = -du / 2. The integrand becomes -sqrt(u), with antiderivative -2/3 u^(3/2). Substituting 1 - x^2 for u we get antiderivative -2/3 (1 - x^2)^(3/2). Between x = 0 and x = 1 this expression changes from -2/3 to 0, a change of 2/3. The definite integral is therefore 2/3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):My answer varies from yours. After the second integration you have -2/3 u^(3/2) and I have -1/3 u^(3/2). I believe mine is correct but please correct me if I am wrong. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qQuery problem 7.8.32 sketch region and reverse order of integration for integral of 1 from 0 to 4-y^2, then from -2 to 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The inner integral becomes x and plugging in the limits you get 4-y^2. Integrating this you get 4y-(y^3 / 3). Plugging in the limits(-2 to 2) you get (8-(8/3))-(-8+(8/3)) = 32/3 0<=x<=4-y^2 and -2<=y<=2 Solve this for y and get y=+/-sqrt(4-x). This is a parabola opening to the left with the -x term. When x=0 you get +/-2, y intercepts. When y=0 you get 4, x-intercept. So the original region defined was from x=0 to 4. So reversing the integration you have for the inner integral (1,y,-sqrt(4-x),sqrt(4-x)) and the outer integral with respect to x from 0 to 4. Inner integral you get sqrt(4-x) - (-sqrt(4-x))=2*sqrt(4-x) Next you get 2*(-1*((4-x)^(3/2)/3)=(-4/3)*(4-x)^(3/2) Plugging in the limits you get (-4/3)*(4-4)^(3/2)-((-4/3)*(4-0)^(3/2)=0-(-32/3)=32/3 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: `a We first integrate 1 from 0 to 4 - y^2, with respect to x. An antiderivative of 1 with respect to x is x; substituting the limits 0 and 4 - y^2 we get [ 4 - y^2 ] - 0, or just 4 - y^2. Then integrating from y = -2 to 2, with respect to y, the antiderivative of 4 - y^2 is 4 y - y^3 / 3; substituting the limits -2 and 2 we get [ 4 * 2 - 2^3 / 3 ] - [ 4 * (-2) - (-2)^3 / 3 ] = 8 - 8/3 - [ -8 - (-8 / 3) ] = 16 - 16/3 = 32 /3. The region of integration is -2 <= y <= 2, 0 <= x <= 4 - y^2. This region lies between the parabola x = 4 - y^2 and the y axis. The vertex is (4, 0), the parabola opens to the left, and it intercepts the y axis at the points (0, 2) and (0, -2). For a given x value between 0 and 4, a vertical line through (x, 0) intersects this region at the points where x = 4 - y^2. The y coordinates of these points are easily found by solving x = 4 - y^2 for y, obtaining y = +-sqrt(4 - x). Thus the vertical line intersects the region between (x, -sqrt(4-x)) and (x, sqrt(4 - x)). If x < 0, or if x > 4 the point (x, 0) lies outside the region. Thus the region can be described by 0 <= x <= 4, -sqrt(4-x) <= y <= sqrt(4 - x) and we integrate 1 first with respect to y, from limit -sqrt(4 - x) to sqrt( 4 - x), then from x = 0 to 4. The 'inner integral' with respect to y yields antiderivative y, which evaluated between the limits -sqrt(4 - x) and sqrt( 4 - x) gives us 2 sqrt(4 - x). This result is then integrated between x = 0 and x = 4. Our antiderivative is easily found to be -2 * (2/3 (4 - x)^(3/2)) = -4/3 (4 - x)^(3/2). Evaluating between the limits x = 0 and x = 4 we obtain -4/3 (4 - 4)^(3/2) - (-4/3 * (4 - 0)^(3/2)) = 32/3. This result agrees with our first integral, as expected. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery problem 7.8.40 Area beneath curve 1/`sqrt(x-1) for 2 <= x <= 5 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Setting up the integral you first integrate 1 with respect to y from 0 to 1/(sqrt(x-1)) then integrate that result with respect to x from 2 to 5. So you get y and plug in the limits you get (1/sqrt(x-1))-0=(1/sqrt(x-1)) Integrating again you get 2*(sqrt(x-1)). Plugging in the limits you get 2*sqrt(4)-2*sqrt(1)=4-2=2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: `a The region goes from x = 2 to x = 5, and from the x axis to the curve y = 1 / `sqrt(x-1). To find the area you integrate 1 over the region. The integral would be from y = 0 to y = 1/`sqrt(x-1) with respect to y (this is the inside integral), then from x = 2 to x = 5 with respect to x (the outside integral). Inside integral: antiderivative of 1 is y; substituting limits we have [1/`sqrt(x-1)] - 0 = 1 / `sqrt(x-1). Outside integral is now of 1/`sqrt(x-1) with respect to x, from x = 2 to x = 5. Antiderivative is 2 `sqrt(x-1) which can be found thru the u substitution u = x - 1. Evaluating at limits we have 2 `sqrt(5-1) - 2 `sqrt(2-1) = 2 `sqrt(4) - 2 `sqrt(1) = 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery problem 7.8.44 area bounded by xy=9, y=x, y=0, x=9 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y=x is a line running through all point y=x so(1,1)(2,2)etc. Xy=9 solved for y is y=9/x. This is similar to y=1/x, it is asymptotic to the x and y axis and passes through the point (3,3). So we are looking at the region beneath these two functions to the x axis and from x=0 to x=9. We have 0<=x<=3 and 0<=y<=x for the area beneath y=x. For the area beneath y=9/x out through x=9 we have 3<=x<=p and 0<=y<=9/x. This means we will have to add two different double integrands. The first integrand will be:inner integrand 1 with respect to y from0 to x, outer integrand of the result with respect to x from 0 to 3. We get int.(x,x,0,3) from the first integral. Then we get (x^2)/2 from the second, evaluated from 0 to 3 you get 9/2 - 0 = 9/2. The second integrand will be: inner integrand 1 with respect to y from 0 to 9/x, outer integrand of the result with respect to x from 3 to 9. You get 9/x from the first integral. Then we get 9*ln|x| from the second, evaluated from 3 to 9 you get 9*ln(9) - 9*ln(3)=9*ln(3). Add these two double integrand results together to get the area. (9/2) + 9*ln(3)=9((1/2)+ln(3)) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: `a xy = 9 means y = 9/x; graph decreases at decreasing rate from asymptote at y axis toward asymptote at x axis. y = x is straight line at 45 deg to x axis. y = x intersects xy = 9 when x^2 = 9 or x = 3 (x^2 = 9 found by substituting y = x into xy = 9). The graph is bounded by y = 0, which is the x axis. Also by x = 9. So the region lies above the x axis, below the y = x curve from x = 0 to x = 3 ( where y = x is lower than y = 9 / x) and below the xy = 9 curve from x = 3 to x = 9. The description: 0 <= x <= 3, 0 <= y <= x plus 3 <= x <= 9, 0 <= y <= 9/x. We are integrating area so for our first region: Our 'inner' integral is the integral of 1 with respect to y, between the limits y = 0 and y = x. Our antiderivative is therefore y, which changes between the limits from 0 to x, a change of x. So our 'inner' integral gives us x. Integrating this result x with respect to x, between limits x = 0 and x = 3, our antiderivative is x^2 / 2. Between the limits x = 0 and x = 3 our antiderivative changes from 0^2 / 2 = 0 to 3^2 / 2 = 9/2, so the change in the antiderivative is 9/2. For our second region: Our 'inner' integral is the integral of 1 with respect to y, between the limits y = 0 and y = 9 / x. Our antiderivative is y, which changes between the limits from 0 to 9 / x, a change of 9/x. So our 'inner' integral gives us 9 / x. Integrating this result 9 / x with respect to x, between limits x = 0 and x = 3, our antiderivative is 9 ln | x |. Between the limits x = 3 and x = 9 our antiderivative changes from 9 ln ( 3) to 9 ln ( 9 ), a change of 9 ln(9) - 9 ln(3) = 9 ln(9/3) = 9 ln(3). The area of our region is therefore 9/2 + 9 ln(3), which we simplify to 9 ( ln(3) + 1/2 ). So we first integrate 1 from x = 0 to x = 3 (outer integral) and from y = 0 to y = x (inner integral). Inner integral gives antiderivative y; substituting limits we get x - 0 = x. Integrating x from x = 0 to x = 3 we get antiderivative x^2 / 2; substituting limits yields 3^2 / 2 - 0^2 / 2 = 9/2 (which is the area of the triangle formed between x = 0 and x = 3). Then we integrate 1 from x = 3 to x = 9 (outer integral) and from y = 0 to y = 9 / x (inner integral). Inner integral gives antiderivative y; substituting limits we get 9/x - 0 = 9/x. Integrating x from x = 3 to x = 9 we get antiderivative 9 ln | x | ; substituting limits yields 9 ln | 9 | - 9 ln | 3 | = 9 ln | 9/3 | = 9 ln | 3 |. Total area is therefore 9 ln | 3 | + 9/2 = 9 ( ln | 3 | + 1/2 ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. ***What a relief that I am done. That was a lot of assignments in such a short amount of time.***

Assignment 32

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&#This looks good. Let me know if you have any questions. &#