#$&* course Mth 272 08/04 9 pm 032.
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Given Solution: `a An antiderivative would be x^2 y + y^3 / 3. Evaluating this antiderivative at endpoints y = x^2 and y = `sqrt(x) we get [ x^2 * `sqrt(x) + (`sqrt(x) ) ^ 3 / 3 ] - [ x^2 * x^2 + (x^2)^3 / 3 ] = x^(5/2) + x^(3/2) / 3 - ( x^4 + x^6 / 3) = x^(5/2) + x^(3/2) / 3 - x^4 - x^6 / 3. This can be simplified in various ways, but the most standard form is just decreasing powers of x: - x^6/3 - x^4 + x^(5/2) + x^(3/2)/3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: Ok ********************************************* Question: `qQuery problem 7.8.14 integrate `sqrt(1-x^2) from 0 to x wrt y then from 0 to 1 wrt x YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x(1-x^2)^1/2 u= 1- x^2 du= 2x -2/3 (1-x^2)^3/2 from 0 to 1= 2/3 confidence rating #$&*: 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): ------------------------------------------------ Self-critique Rating: Ok ********************************************* 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 integrate 4y- y^3/3 from -2 to 2= 32/3 The vertex is (4,0) and the parabola intercepts at (0,2), (0, -2) 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. 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): Needed to describe graph more. Self-critique Rating: Ok ********************************************* Question: `qQuery problem 7.8.40 Area beneath curve 1/`sqrt(x-1) for 2 <= x <= 5 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: inside integral: 1/sqrt(x-1) outside: 2 sqrt( x-1) b/w 2 & 5= 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: Ok ********************************************* Question: `qQuery problem 7.8.44 area bounded by xy=9, y=x, y=0, x=9 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: outer: x= 0 to 3 x^2/2 = 9/2 inner: y= 0 to x area: 9ln|3|+9/2 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). 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: Ok ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. "