course mth 173 assignment #017£l±Ý‘ñئ‰’íÀUxÍ´x‚óq ê‹T{©àŹ§
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15:25:51 Explain in terms of the contribution to the integral from a small increment `dx why the integral of f(x) - g(x) over an interval [a, b] is equal to the integral of f(x) over the interval minus the integral of g(x) over the same integral.
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RESPONSE --> The integral of f(x) - g(x) over the period of 'dx is equal to the average f(avg) 'dx - g(avg)'dx or 'dx(f(avg) - g(avg)) This is stands for the area or the difference between the two finctions on a graph.
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15:31:10 ** In terms of the contribution from a single increment, the rectangle has 'altitude' f(x) - g(x) so its area is [ f(x) - g(x) ] `dx. Adding up over all increments we get sum { [ f(x) - g(x) ] `dx, which by the distributive law of addition over multiplication is sum [ f(x) `dx - g(x) `ds ], which is not just a series of additions and can be rearrange to give sum (f(x) `dx) - sum (g(x) `dx). As the interval `dx shrinks to zero, the sums approach the definite integrals and we get int( f(x), x, a, b ) - int(g(x), x, a, b), where int ( function, variable, left limit, right limit) is the definite integral of 'function' with respect to 'variable' from 'left limit' to 'right limit'.**
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RESPONSE --> What I had written my answer essentially got you to the rectangle that you talk about in the answer. And as 'dx gets smaller and smaller then the rieman sums of the integral approch the integral itself, as you have written in the answer window: int( f(x), x, a, b ) - int(g(x), x, a, b),
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16:00:42 Explain why the if f(x) > m for all x on [a,b], the integral of f(x) over this interval is greater than m (b-a).
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RESPONSE --> Given that f(x)>m and 'dx= (b-a) then f(b) - f(a) must be greater than m('dx) and therfore greater than m(b-a).
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16:01:44 ** This is also in the text, so look there for an alternative explanation and full rigor. The idea is that if f(x) > m for all x, then for any interval the contribution to the Riemann sum will be greater than m * `dx. So when all the contributions are added up the result is greater than the product of m and the sum of all `dx's. The sum of all `dx's is equal to the length b-a of the entire interval. So the Riemann sum must be greater than the product of m and this sum--i.e., greater than m * ( b - a ). **
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RESPONSE --> okay
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16:13:38 Explain why the integral of f(x) / g(x) is not generally equal to the integral of f(x) divided by the integral of g(x).
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RESPONSE --> for example if we have (x^2 - 5x) / (x^4) we cannot say that (2x -5) / (4x^3). We have to say (2x / 4x^3) - (5x / 4x^3) because there must be a common denominator.
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16:19:15 ** For now assume that f and g are both positive functions. The integral of f(x) represents the area beneath the curve between the two limits, and the integral of g(x) represents the area beneath its curve between the same two limits. So the integral of f(x) divided by the integral of g(x) represents the first area divided by the second. The function f(x) / g(x) represents the result of dividing the value of f(x) by the value of g(x) at every x value. This gives a different curve, and the area beneath this curve has nothing to do with the quotient of the areas under the original two curves. It would for example be possible for g(x) to always be less than 1, so that f(x) / g(x) would always be greater than f(x) so that the integral of f(x) / g(x) would be greater than the integral of f(x), while the length of the interval is very long so that the area under the g(x) curve would be greater than 1. In this case the integral of f(x) divided by the integral of g(x) would be less than the integral of f(x), and would hence be less than the integral of f(x) / g(x). **
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RESPONSE --> My explination was similar to the one in the notes, which had to do with derivatives instead of integrals but when reading your explination it makes sense to me.
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16:31:05 Given a graph of f(x) and the fact that F(x) = 0, explain how to construct a graph of F(x) such that F'(x) = f(x). Then explain how, if f(x) is the rate at which some quantity changes with respect to x, this construction gives us a function representing how much that quantity has changed since x = 0.
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RESPONSE --> We can say that the graph of f(x) is representative to the change in F(x) because f(x) = F'(x) so if we find the area under the graph of f(x) then we get the change in F(x).
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16:35:49 ** To construct the graph you could think of finding areas. You could for example subdivide the graph into small trapezoids, and add the area of each trapezoid to the areas of the ones preceding it. This would give you the approximate total area up to that point. You could graph total area up to x vs. x. This would be equivalent to starting at (0,0) and drawing a graph whose slope is always equal to the value of f(x). The 'higher' the graph of f(x), the steeper the graph of F(x). If f(x) falls below the x axis, F(x) will decrease with a steepness that depends on how far f(x) is below the axis. If you can see why the two approaches described here are equivalent, and why if you could find F(x) these approaches would be equivalent to what you suggest, you will have excellent insight into the First Fundamental Theorem. **
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RESPONSE --> when finding the area as described you would use the trapazoidal aproximation the area of trapazoids would reprsent a point on the graph of F(x). ex. (trapazoidal area from t=0->1 is 5units. So the first of the F(x) points would be (1,5).
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16:41:39 Query problem 3.1.16 (3d edition 3.1.12 ) (formerly 4.1.13) derivative of fourth root of x. What is the derivative of the given function?
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RESPONSE --> f(x)= '4sqrt (x) = x^(.25) since(x^n)' = nx^(n-1) f'(x)= .25x ^ (-.75)
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16:41:58 The derivative y=x^(1/4), which is of the form y = x^n with n = 1/4, is y ' = n x^(n-1) = 1/4 x^(1/4 - 1) = 1/4 x^(-3/4). *&*&
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RESPONSE --> okay
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16:47:59 Query problem 3.1.27 (formerly 4.1.24) derivative of (`theta-1)/`sqrt(`theta) What is the derivative of the given function?
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RESPONSE --> f(x)=('theta -1) / 'theta^(.5) f(x)= ('theta / 'theta^(.5)) - (1/ 'theta^(.5)) f'(x) = ('theta / .5'theta^(-.5)) - (1/ .5'theta^(.5))
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16:53:40 ** (`theta-1) / `sqrt(`theta) = `theta / `sqrt(`theta) - 1 / `sqrt(`theta) = `sqrt(`theta) - 1 / `sqrt(`theta) = `theta^(1/2) - `theta^(-1/2). The derivative is therefore found as derivative of the sum of two power functions: you get 1/2 `theta^(-1/2) - (-1/2)`theta^(-3/2), which simplifies to 1/2 [ `theta^(-1/2) + `theta^(-3/2) ]. . **
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RESPONSE --> I messed up by not eleminating theta from my first derivative and beyond that I didn't simplify 1 / theta^(.5) to theta^(-.5) But everything you have done here makes sense to me when I see it.
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17:05:32 Query problem 3.1.60 (3d edition 3.1.48) (formerly 4.1.40) function x^7 + 5x^5 - 4x^3 - 7 What is the eighth derivative of the given function?
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RESPONSE --> f(x)= x^7 + 5x^5 - 4x^3 - 7 since the n dirivative of f is f^n(x) f^8(x)= (x^7 + 5x^5 - 4x^3 -7) ^ 8 f^8(x)= (7x^6 + 25x^4 - 12x^2) ^8 (-7 is eliminated becasue it is constant) = 8 (7x^6 + 25x^4 - 12x^2) ^7 =(56x^6 + 200x^4 - 96x^2) ^7
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17:11:28 ** The first derivative is 7 x^6 + 25 x^4 - 12 x^3. The second derivative is the derivative of this expression. We get 42 x^5 + 200 x^3 - 36 x^2. It isn't necessary to keep taking derivatives if we notice the pattern that's emerging here. If we keep going the highest power will keep shrinking but its coefficient will keep increasing until we have just 5040 x^0 = 5050 for the seventh derivative. The next derivative, the eighth, is the derivative of a constant and is therefore zero. The main idea here is that the highest power is 7, and since the power of the derivative is always 1 less than the power of the function, the 7th derivative of the 7th power must be a multiple of the 0th power, which is constant. Then the 8th derivative is the derivative of a constant and hence zero. **
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RESPONSE --> in the book it said that n dirivative was f'^n(x) i misinterperted this to mean it was f to the n power but it means finding the dirivative n times, so what you have done in the answer window. Which makes perfect sense to me.
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17:14:46 Query problem 3.2.4 (3d edition 3.2.6) (formerly 4.2.6) derivative of 12 e^x + 11^x. What is the derivative of the given function?
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RESPONSE --> f(x) = 12 e^x +11 ^x f'(x)= (12e^x) + (ln(11) * 11^x)
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17:14:59 ** The derivative of a^x is ln(a) * a^x. So the derivative of 11^x is ln(11) * 11^x. The derivative of the given function is therefore 12 e^x + ln(11) * 11^x. **
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RESPONSE --> okay
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17:16:27 Query problem 3.2.10 (3d edition 3.2.19) (formerly 4.2.20) derivative of `pi^2+`pi^x. What is the derivative of the given function?
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RESPONSE --> f(x) = 'pi^2 + 'pi^x f'(x)= 2'pi + ln('pi) * 'pi^x
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17:20:43 ** `pi is a constant and so therefore is `pi^2. Its derivative is therefore zero. `pi^x is of the form a^x, which has derivative ln(a) * a^x. The derivative of `pi^x is thus ln(`pi) * `pi^x. **
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RESPONSE --> I failed to evaluate 'pi^2 for a constant, which it is so the derivative overall is wrong but, the part that it is right (ln('pi) * 'pi^x) is right.
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17:29:32 Query problem 3.2.40 (3d edition 3.2.30) (formerly 4.2.34) value V(t) = 25(.85)^6, in $1000, t in years since purchase. What are the value and meaning of V(4) and ov V ' (4)?
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RESPONSE --> V(t) = 25(.85) ^t V'(t)= 25 (ln(.85) .85^t) V(4)= 13.05 * $1000 = $13,050 is the change in $ when compared to time V'(4) = -2.12 * $1000 = -$2,120 is the rate of change at t=4
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17:33:25 ** V(4) is the value of the automobile when it is 4 years old. V'(t) = 25 * ln(.85) * .85^t. You can easily calculate V'(4). This value represents the rate at which the value of the automobile is changing, in dollars per year, at the end of the 4th year. **
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RESPONSE --> okay
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17:34:55 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I think that I understand finding the dirivative and the integral concepts that we have gone over in this query.
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