course Mth 272 August 13, 2009 at 9:27 AM Question: `qQuery problem 7.8.6 integrate (x^2+y^2) with respect to y from x^2 to `sqrt(x)
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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: sqrt(1-x^2) dy dx (sqrt(1-x^2)y) =x sqrt(1-x^2) confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). Using a table you will find that an antiderivative with respect to x is arcsin(x)/2 + x `sqrt(1 - x^2) / 2. I don't think your text has dealt with the arcsin function, but arcsin(1) = `pi/2 and arcsin(0) = 0. So substituting limits we get arcsin(1)/2 + 1`sqrt(1 - 1^2)/2 - [arcsin(0)/2 + 0`sqrt(1 - 0^2)/2 ] = `pi/4 (all terms except the first give you zero). Note that y = `sqrt(1-x^2) from x = -1 to 1 is the graph of a circle of radius 1. The integral from x = 0 to x = 1 gives a quarter-circle of radius 1, which has area 1/4 * `pi r^2 = 1/4 * `pi * 1^1 = `pi/4, in agreement with the second integral. ** Note that the region is just 1/4 of the circle of radius 1 centered at the origin. This circle has area pi r^2 = pi * 1^2 = pi, and 1/4 of the circle has area pi/4, in agreement with the integral. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* 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: (4-y^2)-0= (4-y^2) antiderivative: (4y-y^3/3) 8-8/3- (-8-(-8/3)) =16-16/3 =32/3 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. Reversing the order of integration we integrate 1 from -`sqrt(x) to `sqrt(x), with respect to y. Anantiderivative of 1 with respect to y is y; substituting the limits we get `sqrt(x) - (-`sqrt(x) ) = 2 `sqrt(x). We next integrate with respect to x, from 0 to 4. Antiderivative of 2 `sqrt(x) with respect to x is 2 * [ 2/3 x^(3/2) ] = 4/3 x^(3/2). Evaluating between limits 0 and 4 we get [ 4/3 * 4^(3/2) ] - [4/3 * 0^(3/2) ] = 32/3. The two integrals are equal, as must be the case since they both represent the area of the same region &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery problem 7.8.40 Area beneath curve 1/`sqrt(x-1) for 2 <= x <= 5 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Inner (1/sqrt(x-1))-(0)= (1/sqrt(x-1)) outer (1/sqrt(x-1))= 2sqrt(x-1) 2sqrt(5-1) - 2sqrt(2-1) =2 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery problem 7.8.44 area bounded by xy=9, y=x, y=0, x=9 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: inner Y with boundaries from x to 0 x-0= X antiderivative: x^2/2 use x=3 and x=0 3^2/2 - 0= 9/2 outer Y with boundaries from 0 to 9/x 9/x - 0= 9/x antiderivative: 9 ln|x| 9ln|9| - 9ln|3| =9ln|3| Now, the total area is the two answers added together. (9/2) + 9ln|3| Simplified= 9(ln|3|+1/2) 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). 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. 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.