#$&* course MTH 173 11/10 22:15
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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'.** STUDENT COMMENT I'm not sure but I believe I understand. INSTRUCTOR RESPONSE: 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. This idea might be clearer if expressed in terms of a trapezoidal approximation of the graph of f(x): Just like any subinterval, the chosen increment would define a thin trapezoid for the function f(x). The trapezoid would have width `dx, and would lie somewhere between x = a and x = b. x would be any point on the x axis between the left and right boundaries of the increment (i.e., any point of the x axis which lies on the boundary of the trapezoid), and f(x) would be the 'graph altitude' at x. If we sum up the areas of all the increments we get a result which is close to the actual area under the graph of f(x). The smaller the increment `dx the closer the sum of the incremental areas will tend to be to the actual area, and as `dx approaches zero, the sum of the incremental areas will approach the actual area. As long as the trapezoid is thin (i.e., as long as `dx is small), the two altitudes of the trapezoid will both be close to f(x). So f(x) will be close to the average altitude and f(x) * `dx will be close to the area of the trapezoid, which is ave. altitude * width. On the interval `dx, the function f(x) - g(x) would similarly form a trapezoid whose 'average altitude' is close to f(x) - g(x) and whose width is `dx. The area of the trapezoid would therefore be close to (f(x) - g(x)) * `dx. In the given solution the distributive law is be used to show that the sum of the (f(x) - g(x) ) * `dx contributions is equal to the sum of the f(x) * `dx minus the sum of the g(x) * `dx, so that the integral of f(x) - g(x) is equal to the integral of f(x) minus the integral of g(x). It's worth summarizing we have done here: • we began with a partition of the interval [a, b] (i.e., a subdivision into small increment), then • we considered one increment of the partition (i.e., one subinterval), • we summed the contributions of all increments, and • we drew conclusions about the integral. This procedure, where we partition some interval the consider what happens on a typical subinterval, or increment, is a crucial step in finding the integrals necessary to solve a wide variety of problems. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: According to me this is not always right. I mean this will be correct only when the function f(x) tends to m from values of y more than m, that is the values of f(x) is always greater than m and decreases at a decreasing rate and thus tending to m. If f(x) is below y = m. and is increasing at a decreasing rate, then too the function f(x) will tend to m for all x on [a,b]. If the function is above f(x) = m, every value of f(x) will be more than m. When f(x) is divided into a number of rectangles for either a left hand or the right hand sum the area of each rectangle will be greater than m(dx). When the area of the rectangles is summed over the entire interval of [a,b] the integral will greater than the m(b-a). If the function tends m from below the line y = m , each value will be less than m and thus the integral will be less than m(b-a). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: Integral of f(x) is the area under the curve f(x) over an interval and integral of g(x) will be the area under the curve over the same interval. Thus the ratio of integral of f(x)/integral of g(x) will just give the ratio of the 2 areas. Whereas f(x) / g(x) will give a completely new function and the integral of which will be the area under the curve for the new function over the same interval. The two values will be different and have no relation with each other. Thus the integral of f(x)/g(x) will not be equal to the value of the integral of f(x) / integral of g(x). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: It is known that f(x) is the rate function and F(x) is its integral function. Thus the slope of the function F(x) at a particular x value will be equal the value of the function f(x). Thus starting the graph of F(x) from origin we see what the value of f(x) is and maintain the slope of the F(x) curve accordingly. Higher the value of f(x) steeper is the F(x) curve. As soon as f(x) crosses the x axis and takes negative value the function F(x) starts decreasing and smaller the value f(x) faster will F(x) be decreasing. We know that F(x) passes through the origin and thus F(0) = 0. Since F’(x) = f(x), the integral of the function f(x) will given F(x). Suppose we calculate the integral from x = 0 to x = a. This value of the integral will be equal to the area under the curve which can be calculated by trapezoidal approximation graph. This value of the integral will also be equal to F(a) - F(0). Since F(0) = 0, the value of the integral of f(x) over the interval x = 0 to x = a will be equal to F(a). Thus the function F(x) can be plotted by integrating f(x) over the various intervals and obtaining the value of the function F(x) at a particular x value. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** STUDENT COMMENT So in general using trapezoid approx.. is another way to see what is going on? INSTRUCTOR RESPONSE It's very similar, in that if you first think of the trapezoids you see how the area is close to the sum of the trapezoidal areas. In each increment of the partition we choose a single value of x, and evaluate f(x) at that value. Using that value as the average 'graph altitude' we approximate the area of the trapezoid as f(x) * `dx. This has the same effect as replacing the trapezoid by a rectangle of altitude f(x) and width `dx. As long as `dx is small, f(x) is close to the actual average 'graph altitude' for the interval, and the error in the approximation is small. As `dx approaches zero, the error in the approximation approaches zero. The limiting value of our approximations is the integral, and it represents the area beneath the graph. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhat is the derivative of the given function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given function y = x^(1/4) Using the power theorem the derivative of the function is y’ = (1/4)*(x^(-3/4)) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe derivative y=x^(1/4), which is of the form y = x^n with n = 1/4, is y ' = n x^(n-1) = 1/4 x^(1/4 - 1) = 1/4 x^(-3/4). *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery problem 3.1.39 was 3.1.27 (formerly 4.1.24) derivative of (`theta-1)/`sqrt(`theta) What is the derivative of the given function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given function f(theta) = (theta - 1)/sqrt(theta) = theta^(1/2) - theta^(-1/2) Thus the derivative of the function will be f’(theta) = (1/2)(1/theta^(1/2)) - (-1/2)(theta^(-3/2)). confidence rating #$&*: 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: Given expression x^7 + 5x^5 - 4x^3 - 7 The expression as x^7 as its highest power. When the expression is differentiated for 7 times x^5 will have been reduced to 0, so will x^3 have been and 7 too. x^7 will has been differentiated to a factor having x^0 as the x coefficient leaving the expression as only a constant. Thus on the eight derivative the value of the derivative becomes 0. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The first derivative is 7 x^6 + 25 x^4 - 12 x^3. The second derivative is the derivative of this expression. We get 42 x^5 + 200 x^3 - 36 x^2. It isn't necessary to keep taking derivatives if we notice the pattern that's emerging here. If we keep going the highest power will keep shrinking but its coefficient will keep increasing until we have just 5040 x^0 = 5050 for the seventh derivative. The next derivative, the eighth, is the derivative of a constant and is therefore zero. The main idea here is that the highest power is 7, and since the power of the derivative is always 1 less than the power of the function, the 7th derivative of the 7th power must be a multiple of the 0th power, which is constant. Then the 8th derivative is the derivative of a constant and hence zero. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery problem 3.2.4 (3d edition 3.2.6) (formerly 4.2.6) derivative of 12 e^x + 11^x. What is the derivative of the given function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given function y = 12 e^x + 11^x The derivative of the given function will be y’ = 12e^x + 11^x(ln(11)) confidence rating #$&*: 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: Given function pi^2 + pi^x, we know that pi^2 is a constant, thus the value of differential will be equal to pi*x*ln(pi) 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)^6, in $1000, t in years since purchase. What are the value and meaning of V(4) and ov V ' (4)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Problem submitted through text_17 assignment under the name 3.2.40 confidence rating #$&*: 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 ********************************************* Question: `qQuery problem 3.2.43 was 3.2.40 (3d edition 3.2.30) (formerly 4.2.34) value V(t) = 25(.85)^6, in $1000, t in years since purchase. What are the value and meaning of V(4) and ov V ' (4)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Problem submitted through text_17 assignment under the name 3.2.40 confidence rating #$&*: 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 #*&!