course Mth 151 When I submitted q_a and querry 11 I asked you if you had received my orientation number eight. I am very sorry if you took the time to look that up because it is listed on my access page as understanding first assignmets.
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17:43:41 5.4.4 (was 5.4.4 integrate `sqrt(9-x^2) from -3 to 3
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RESPONSE --> This graph is of a half-circle, so in order to find the area under the curve, you can use the formula for the area of a circle, A=pi*r^2, and divide by two, giving you: A= 1/2 pi r*2. In this case, since we are going from -3 to 3, r= 3, and the area equals 1/2 pi*3^2, or 9pi/2
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17:46:43 ** 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 --> I got the correct answer, but I wasn't sure of the reasoning behind it. The only reason I used the area of a circle formula is because I couldn't figure out the integral of (9-x^2)^1/2, since du=2x, I wasn't sure how to make that work out since there was no other term in the integral to which I could multiply or divide any constant to come up with 2x. I'm still unsure of how to take the integral of this function.
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17:51:54 5.4.13 (was 5.4.10 (x^2+4)/x from 1 to 4
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RESPONSE --> The first thing I did was split this integral up into 2 parts: the int from 1 to 4 of (x^2)/x and the int from 1 to 4 of 4/x. This left me with integrals of x and 4x^(-1). Solving these integrals, I got 1/2 x^2 + 4 lnx. Then plugging in 4 and 1 for x, I got: (1/2 (16) + 5.545) - (1/2 (1) +0) =13.545 - 0.5 = 13.045
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17:53:53 ** 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 --> I integrated correctly, but I got a slightly different answer for the definite integral. I got 13.045, but I'm hoping this difference is attributed only to rounding errors, and not a more serious error I may have made in my calculations.
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17:58:29 5.4.20 (was 5.4.16 3x^2+x-2 from 0 to 3
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RESPONSE --> As long as I am interpreting the grouping correctly, this seemed to be a fairly straightforward integral to solve. The integral of 3x^2 +x -2 = x^3 +.5x^2 -2x. Solving for the definite integral: ((3^3) + .5(3^2) -2(3)) - ((0^3) + .5(0^2) -2(0)) = 27 + .5(9) - 6 - 0 = 25.5
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17:58:37 ** 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 --> ok
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20:00:48 ** 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 --> I accidentally clicked ahead. My answer: The integral of the square root of (2/x) = int of sqrt 2 / int of sqrt x =int (sqrt2)*(x^-1/2) =2*(sqrt2)(x^1/2) =2(sqrt2)(sqrtx) =2*(sqrt(2x)) From 1 to 4: (2*sqrt(2*4)) - (2*sqrt(2*1)) = (2*sqrt(8)) - (2*sqrt(2)) = 5.657 - 2.828 = 2.828 Critique: My response was correct, but it took me awhile to realize I could separate in the integrand into two integrals in order to make it easier to solve. Even after I did that, I found it hard to treat sqrt(2) as any other constant. I kept trying different things with it before I realized I should just leave it alone, the same as any other constant.
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20:08:03 Extra Problem (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 --> To find the average value over an interval [a,b], you must multiply the regular integral you would find by 1/(b-a). 1/(10-0) int (5e^(.2x-2)) u= .2x-2 du=.2 we must multiply and divide by .04, or 1/25. (1/10)*(25/1) (e^(.2x-2)) 25/10 (e^(.2x-2)) 2.5(e^(.2x-2)) [0,10]: (2.5e^((.2*10)-2)) - (2.5e^((.2*0)-2) 2.5e^0 - 2.5e^-2 2.5 - 0.338 2.16 = the average value
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20:10:12 ** 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 --> I think I got the correct answer, because what I got was close to 2.2, but I must have gone about solving it in a completely different way.
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20:16:41 Extra Problem (was 5.4.56 ave val of 1/(x-3)^2 from 0 to 2
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RESPONSE --> 1/b-a = 1/2-0 1/2 int (x-3)^(-2) (1/2)*(-1)((x-3)^-1) (-1/2)*((x-3)^(-1)) [0,2] (-.5((-1)^(-1))) - (-.5((-3)^(-1))) (1/2) - (1/6) 1/3 = average value
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20:17:32 ** 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 --> I got the same answer again, but I still can't really follow the steps given here, and so I know I'm doing something different, but I can't really tell what it is.
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20:23:04 Does the average value make sense in terms of the graph?
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RESPONSE --> I really don't know how to do this problem. Am I supposed to graph the original function or what I got after I integrated? I graphed both on my calculator, and there really wasn't too much difference between them. However, I'm guessing that this average value does make sense considering the interval we're dealing with is only from [0,2], but I know that's not a sufficient answer.
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20:24:14 ** 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 --> Okay, this makes more sense now. I can follow the steps and so from now on, I'll know what to do when asked this question, but I never would've figured that out on my own.
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