#$&* course MTH 173 8/7/11noon
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Given Solution: `a** In terms of the contribution from a single increment, the rectangle has 'altitude' f(x) - g (x) so its area is [ f(x) - g(x) ] `dx. Adding up over all increments we get sum { [ f(x) - g(x) ] `dx, which by the distributive law of addition over multiplication is sum [ f(x) `dx - g(x) `ds ], which is not just a series of additions and can be rearrange to give sum (f(x) `dx) - sum (g(x) `dx). As the interval `dx shrinks to zero, the sums approach the definite integrals and we get int( f(x), x, a, b ) - int(g(x), x, a, b), where int ( function, variable, left limit, right limit) is the definite integral of 'function' with respect to 'variable' from 'left limit' to 'right limit'.** The question asks you to focus on a single increment. This means that we first divide the interval [a, b] into a large number of subintervals (i.e., we 'partition' the interval [a, b]), and consider what happens in a typical subinterval, or increment. The typical increment will have some width, which we represent by `dx, and we consider a typical x value within the increment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qExplain why the if f(x) > m for all x on [a,b], the integral of f(x) over this interval is greater than m (b-a). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&&&&&&& ************************ the integral of the function between two x values will be greater than the product of m and the length of the interval, and the average value of the function must be less than M and greater than m &&&&&&&&&&&&&&&&&&&&&&&& ************************ confidence rating #$&*:2.5 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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 ). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qExplain 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). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&&&&&&& ************************ The area underneath the curve for f(x) divided by the area underneath the curve for g(x) will probably not be the same as the area underneath the curve of the new graph f(x)/g(x). say you have y = x and y = 3x^2 + 1 for the interval x= 0 to x= 3. the integral for y = x is 4.5, and the integral for y = 3 is 30. 4.5/30 =~ .15 then the integral for the same interval for x/(3x^2 + 1) is approx .56 It's just a different graph. &&&&&&&&&&&&&&&&&&&&&&&& ************************ confidence rating #$&*:2.5 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qGiven 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&&&&&&& ************************ Okay, so you want to graph F(x) in which F'(x) will look a certain way. So if you want to make a graph in which the derivative looks a certain way, you just have to think about your rate of change. The slope of F(x) is the value of F'(x). If F'(x), the graph you're given, has, for example, a straight line, the rate will be constant, and you'll have a linear graph for F(x). If F'(x) is linear, that means your rate will change uniformly, and also your F(x) will probably be a parabola, its location/exact shape depending on where the F'(x) is negative or positive and its magnitude. That sort of thing. &&&&&&&&&&&&&&&&&&&&&&&& ************************ confidence rating #$&*:2.5 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Gotcha ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery problem 3.1.16 (3d edition 3.1.12 ) (formerly 4.1.13) derivative of fourth root of x. `qWhat is the derivative of the given function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&&&&&&& ************************ y' = 1/4x^(-3/4) &&&&&&&&&&&&&&&&&&&&&&&& ************************ confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe derivative y=x^(1/4), which is of the form y = x^n with n = 1/4, is y ' = n x^(n-1) = 1/4 x^(1/4 - 1) = 1/4 x^(-3/4). *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery problem 3.1.39 was 3.1.27 (formerly 4.1.24) derivative of (`theta-1)/`sqrt (`theta) What is the derivative of the given function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&&&&&&& ************************ .5(theta)^-.5 + .5theta^(-1.5) = .5(theta^-.5 + theta^-1.5) &&&&&&&&&&&&&&&&&&&&&&&& ************************ confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** (`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) ]. . ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery problem 3.1.61 was 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&&&&&&& ************************ equation is of degree 7, so the derivative should end up being 0. &&&&&&&&&&&&&&&&&&&&&&&& ************************ confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The first derivative is 7 x^6 + 25 x^4 - 12 x^3. The second derivative is the derivative of this expression. We get 42 x^5 + 200 x^3 - 36 x^2. It isn't necessary to keep taking derivatives if we notice the pattern that's emerging here. If we keep going the highest power will keep shrinking but its coefficient will keep increasing until we have just 5040 x^0 = 5050 for the seventh derivative. The next derivative, the eighth, is the derivative of a constant and is therefore zero. The main idea here is that the highest power is 7, and since the power of the derivative is always 1 less than the power of the function, the 7th derivative of the 7th power must be a multiple of the 0th power, which is constant. Then the 8th derivative is the derivative of a constant and hence zero. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery problem 3.2.4 (3d edition 3.2.6) (formerly 4.2.6) derivative of 12 e^x + 11^x. What is the derivative of the given function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&&&&&&& ************************ 12e^x + ln(11)*11^x &&&&&&&&&&&&&&&&&&&&&&&& ************************ confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery problem 3.2.18 was 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&&&&&&& ************************ ln(pi)*pi^x &&&&&&&&&&&&&&&&&&&&&&&& ************************ confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** `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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery problem 3.2.43 was 3.2.40 (3d edition 3.2.30) (formerly 4.2.34) value V(t) = 25(.85)^t, in $1000, t in years since purchase. What are the value and meaning of V (4) and ov V ' (4)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&&&&&&& ************************ V(4) = 13.05, and is the value of the car at time = 4 years in $1000. v'(t) = 25*ln(.85)*.85^t V'(4) = -2.12/yr and is the rate of change of the value over the car over time at 4yrs in $1000. &&&&&&&&&&&&&&&&&&&&&&&& ************************ confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!