course Mth 272 8/1 8:45 pm 032.
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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: To solve this problem, we must first take the integral with respect to x. Using our integration rules, we find that our first integral is sqrt(1-x^2). Now we can plug in our bounds, 0 and x. We find that our expression remains sqrt(1-x^2) because the lower bound is 0. Now we need to take the integral of this expression again. From the table, and as stated in the given solution, our integral is arcsin x/2 + xsqrt(1-x^2)/2. We can now substitute in our bounds, so arcsin(1)/2 + 1?sqrt(1 - 1^2)/2 - [arcsin(0)/2 + 0?sqrt(1 - 0^2)/2 ] = `pi/4. 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). 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: To solve this problem, we can first integrate with respect to x, finding that we end up with just x. We can now substitute our bounds, ending up with 4-y^2 because our lower bound is 0. We can now take the integral with respect to y, and we get 4y ?1/3y^3 according to our integration rules. Finally we can substitute in our bounds to get the area. So Area = (4(2) -1/3(2^3)) ?(4(-2) -1/3(-2)^3) = 16/3 + 16/3 = 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. 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: To find the area under the curve, we need to first integrate and then plug in our bounds to solve. When we graph the function, we see that we have to integrate it in two parts. The first is from 1/sqrt(x-1) to 1, and then from [2,5]. The first we take with respect to y, finding that our integral is y/sqrt(x-1). When we plug in our bounds, we find that the area for this section is 1/sqrt(x-1). The next integral we take with respect to x. So according to our integration rules, this expression? integral is 2(x-1)^(1/2). Now we can plug in the bounds to find our area. So Area = 2(5-1)^(1/2) ?2(2-1)^(1/2) = 2(4)^(1/2) ?2(1)^(1/2) = 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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery problem 7.8.44 area bounded by xy=9, y=x, y=0, x=9 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can first graph these equations to see where they lay and what integrals we need to take. We find that the lines intersect at the point(3,3). From 0<=x<=3, the equation y=x is on top of y=0. From 3<=x<=9, y=9/x is on top of y=0. So we need to take two integrals according to what function is the top function. We can start with the [0,3] part of the problem, finding that the integral of y=x with regard to y is y. So the area for that is equal to Area = x -0 = x. Now we can take the integral with respect to x, finding that it is equal to 1/2x^2. We can now plug in the bounds of [0,3], finding that Area = ?3^2) -1/2(0^2) = 9/2 -0 = 9/2. Next we can take the integral of y = 9/x, finding that it is 9ln(abs(x)). We can now find the area under this segment. So Area = 9ln(abs(9)) ?9ln(abs(3)) = 9.888. Now we can find the total area by adding the two areas together, so Area = 9/2 + 9.888 = 14.388. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 ).