course Mth 272
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RESPONSE --> The graph is a half circle with a radius of 3. Area = 1/2 pi r^2 = 1/2 pi(3^2) = 9/2 pi The integral is (9-x)^3/2 confidence assessment: 2
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12:07:43 The graph of y = `sqrt( 9-x^2) is a half-circle of radius 3 centered at the origin. We can tell this because any point (x, `sqrt(9-x^2) ) lies at a distance of `sqrt( x^2 + (`sqrt (9-x^2))^2 ) = `sqrt(x^2 + 9-x^2) = `sqrt(9) = 3 from the origin. The area of the entire circle is 9 `pi square units. The region beneath the graph is a half-circle is half this, 9/2 `pi square units, which is about 14.1 square units. This area is the integral of the function from x=-3 to x=3. **SERIOUS STUDENT ERROR: Take the int and get 9x -1/3(x^3) INSTRUCTOR COMMENT: The integral of `sqrt( 9 - x^2) is not 9x - 1/3 x^3. The derivative of 9x - 1/3 x^3 function is 9 - x^2, not `sqrt(9-x^2). **
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RESPONSE --> self critique assessment: 2
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12:07:56 5.4.17 (was 5.4.13) (was 5.4.10 (x^2+4)/x from 1 to 4
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RESPONSE --> antiderivative is x+4(1/x)dx = [x^2/2 + 4ln(x)] 1 to 4 [4^2/2 + 4ln(4)] - [1^2/2 + 4ln(1)] = 13.0452 confidence assessment: 2
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12:07:58 The correct integral is not too difficult to find once you see that (x^2 + 4 ) / x = x + 4/x. There is an addition rule for integration, so you can integrate x and 4/x separately and recombine the results to get x^2/2 + 4 ln(x) + c. The definite integral is found by evaluating this expression at 4 and at 1 and subtracting to get (4^2 / 2 + 4 ln(4) ) - (1^2 / 2 + 4 ln(1) ) = 8 + 4 * 1.2 - (1/2 + 0) = 12 (approx). As usual check my mental calculations. ** STUDENT ERROR: The int is((x^3)/3 + 4x)(ln x) + C INSTRUCTOR CORRECTION: ** That does not work. You can't integrate the factors of the function then recombine them to get a correct integral. The error is made clear by taking the derivative of your expression. The derivative of (x^3/3 + 4x) ln(x) is (x^2 + 4) ln(x) + (x^2/3 + 4). Your approach does not work because it violates the product rule. **
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RESPONSE --> self critique assessment: 2
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12:08:11 Extra Problem (formerly 5.4.20) (was 5.4.16 Integrate 3x^2+x-2 from x = 0 to x = 3
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RESPONSE --> anitderivative is x^3+ 1/2(x^2) - 2x [3^3+ 1/2(3^2) - 2(3)] - [0^3+ 1/2(0^2) - 2(0)] = 25.5 - 0 = 25.5 confidence assessment: 2
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12:08:13 an antiderivative of f(x) = 3 x^2 + x - 2 is F(x) = x^3 + x^2/2 - 2x. Evaluating at 3 we get F(3) = 25.5. At 0 we have F(0) = 0. So the integral is the change in the antiderivative function: F(3) - F(0) = 25.5 - 0 = 25.5. **
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RESPONSE --> self critique assessment: 2
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12:08:26 5.4.28 (was 5.4.24) (was 5.4.20 Integrate sqrt(2/x) from 1 to 4
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RESPONSE --> Antiderivative of sqrt(2)(x^-.5)dx is 2 sqrt(2x) [2 sqrt(2(4))] - [2 sqrt(2(1))] = 2.828 confidence assessment: 2
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12:08:29 The function can be written as `sqrt(2) / `sqrt(x) = `sqrt(2) * x^-.5. An antiderivative is 2 `sqrt(2) x^.5 = 2 `sqrt(2x). Evaluating at 4 and 1 we get 2 `sqrt(2*4) = 4 `sqrt(2) and 2 `sqrt(2) so the definite integral is 4 `sqrt(2) - 2 `sqrt(2) = 2 `sqrt(2), or approximately 2.8. **
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RESPONSE --> self critique assessment: 2
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12:08:45 5.4.63 (was 5.4.52 What is the average value of 5e^(.2(x-10)) from x = 0 to x = 10?
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RESPONSE --> 1/10-0 int 0 to 10 5 e^(.2x-2)dx 1/10(5/.2 e^.2x-2) [5/2(e^(.2(10)-2)] - [5/2(e^(.2(0)-2)] = 2.1617 2.1617/10 = .2162 2.1617 = 5 e^(.2x-2) x = 5.8073 confidence assessment: 2
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12:08:47 The area under a curve is the product of its average 'height' and its 'width'. The average 'height' is the average value of the function, the area is the definite integral and the 'width' is the length of the interval. It follows that average value = definite integral / interval width. To integrate 5 e^(.2 ( x - 10) ): If you let u = .2x - 2 you get du/dx = .2 so dx = du / .2. You therefore have the integral of 5 e^u du / .2 = (5 / .2) e^u du. The integral of e^u du is e^u. So an antiderivative is 5 / .2 e^u = 5 / .2 e^(.2x - 2). Using the antiderivative 25 e^(.2(x-10)) at 0 and 10 we get about 22 for the definite integral (i.e., the antiderivative function 25 e^(.2(x-10)) changes by 22 between x = 0 and x = 10). The average value (obtained by dividing the integral by the length of the interval) is thus about 22 / 10 = 2.2. ** ERRONEOUS STUDENT SOLUTION: The average value is .4323. INSTRUCTOR COMMENT: This average value doesn't make sense. The function itself has value between 0 and 1 (closer to 1) when x=0 and value 5 when x=10 so its average value is probably greater than .4323. Unless the graph has a serious dip between the point where its value is 1 and the point where its value is 5, its average value would be between 1 and 5 and wouldn't be less than 1. **
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RESPONSE --> self critique assessment: 2
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12:09:00 5.4.66 (was 5.4.56 ave val of 1/(x-3)^2 from 0 to 2
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RESPONSE --> antiderivative is -1/x-3 [-1/(2-3)] - [-1/(0-3)] = 2/3 (2/3)/2 = 1/3 confidence assessment: 2
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12:09:02 An antiderivative of 1 / (x-3)^2 is -1 / (x-3). At 0 and 2 this antiderivative takes values 1/3 and 1 so the integral is 1 - 1/3 = 2/3, the change in the value of the antiderivative. The average value of the function is therefore ave value = integral / interval width = 2/3 / (2-0) = 2/3 / 2 = 1/3. **
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RESPONSE --> self critique assessment: 2
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12:09:12 Does the average value make sense in terms of the graph?
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RESPONSE --> confidence assessment:
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12:09:14 When x = 1, f(x) = .25 1 / (x-3)^2 = 1/3. Solve for x. Inverting both sides you get (x-3)^2 = 3 so x-3 = +-`sqrt(3) so x = 3 + `sqrt(3) or x = 3 - `sqrt(3), or approximately x = 4.732 or x = 1.268. The 1.268 makes sense for this interval; 4.732 isn't even in the interval. **
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RESPONSE --> self critique assessment:
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