#$&* course MTH 174 6/29 The relative accuracy of the trapezoidal and midpoint rules depends on the nature of the function. As shown in my previous notes for a positive function which is concave down the midpoint rule is more accurate. A brief numerical example is y = x^2 on the interval (-1, 0), where trap gives 1/2, actual is 1/3 and mid is 1/4. 1/3 is closer to 1/4 than to 1/2. The same is true if the function is positive and concave up, for the same reasons. In fact I think that the argument can be extended to show that if concavity doesn't change on an interval, mid has to beat trap. I'd have to draw a picture or two to be sure, but it seems to be fairly obvious so I'll leave that to you. Use my previous notes as a guide.
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Optional preliminary questions: ********************************************* Question: `q001. The graph of f(x) = 1 / sqrt(x) has a vertical asymptote at x = 0. The function is not defined at x = 0, since 1 / 0 is undefined. So the limiting value of f(x), as x approaches infinity, is infinite. What is the integral of 1 / sqrt(x) on each of the following intervals? The interval from 1/4 to 1 The interval from 1/16 to 1 The interval from 1/64 to 1 The interval from 1/216 to 1 The interval from 1/1024 to 1 The numbers 1/4, 1/16, 1/64, 1/216 and 1/1024 get closer and closer to 0, each being 4 times as close to zero as the preceding, so the the last of these numbers is 216 times closer to zero than the first. Do you think the value of the integral will change much if we continue letting the interval approach zero? The above integrals could be expressed as follows: integral(1 / sqrt(x), x from c to 1), for c values 1/4, 1/16, 1/64, 1/216, 1/1024. Why would it make sense to define the integral of f(x) on the interval from 0 to 1 to be the limiting value of these integrals, as c values continue to approach zero in the same manner? Why does it therefore make sense to make the following definition: integral ( 1 / sqrt(x), x from 0 to 1) = limit { c -> 0 } (integral( 1 / sqrt(x), x from c to 1)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. The graph of f(x) = 1 / sqrt(x) has the x axis as a horizontal asymptote, which it approaches as x approaches infinity. What is the integral of 1 / sqrt(x) on each of the following intervals? The interval from 1 to 4 The interval from 1 to 16 The interval from 1 to 64 The interval from 1 to 256 The interval from 1 to 1024 The numbers 4, 16, 64, 256, 1024 move away from x = 1 more and more rapidly. It is clear that if the pattern continues, these numbers will eventually exceed any bound, so that they will approach infinity. Do you think the value of the integral will change much if we continue letting the interval increase in this manner? What do you think will happen to the value of the integral? The above integrals could be expressed as follows: integral(1 / sqrt(x), x from 1 to c), for c values 4, 16, 64, 256, 1024. Why would it make sense to define the integral of f(x) on the interval from 1 to infinity to be the limiting value of these integrals, as c values continue to approach infinity in the same manner? Why does it therefore make sense to make the following definition: integral ( 1 / sqrt(x), x from 1 to infinity) = limit { c -> 0 } (integral( 1 / sqrt(x), x from 1 to infinity)? Why does it make sense to say that integral(1 / sqrt(x), x from 1 to infinity) approaches infinity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. Using the above as a guide, determine whether each of the integrals integral ( 1 / x^2, x from 0 to 1) and integral ( 1 / x^2, x from 1 to infinity) is finite or infinite. In the finite case, find the value of the integral. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: Questions related to text assignment: ********************************************* Question: Section 7.6 Problem 6 approx using n=10 is 2.346; exact is 4.0. What is n = 30 approximation if original approx used LEFT, TRAP, SIMP?
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21:44:07 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Find the magnitude from n = 10 to n =30 It is 3 Divide the error by the magnitude for LEFT Divide the error by the magnitude squared for TRAP Divde the error by the magnitude ^4 for SIMP confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: LEFT and RIGHT approach the exact value in proportion to the number of steps used. MID and TRAP approach the exact value in proportion to the square of the number of steps used. SIMP approachs the exact value in proportion to the fourth power of the number of steps used. Using these principles we can work out this problem as follows: ** The original 10-step estimate is 2.346, which differs from the actual value 4.000 by -1.654. If the original estimate was done by LEFT then the error is inversely proportional to the number of steps and the n = 30 error is (10/30) * -1.654 = -.551, approximately. So the estimate for n = 30 would be -.551 + 4.000 = 3.449. If the original estimate was done by TRAP then the error is inversely proportional to the square of the number of steps and the n = 30 error is (10/30)^2 * -1.654 = -.184, approximately. So the estimate for n = 30 would be -.184 + 4.000 = 3.816. If the original estimate was done by SIMP then the error is inversely proportional to the fourth power of the number of steps and the n = 30 error is (10/30)^4 * -1.654 = -.020, approximately. So the estimate for n = 30 would be -.02 + 4.000 = 3.98. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand what the flaw in my reasoning was. I was too focused on the error as opposed to the approximation. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: Section 7.6 Problem 5 problem 7.6.5 (previously problem 7.6.10) If TRAP(10) = 12.676 and TRAP(30) = 10.420, estimate the actual value of the integral. **** What is your estimate of the actual value and how did you get it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** You need to use the inverse square proportionality of the error with the number of steps. Trap(30) is approximately (10/30)^2 = 1/9 of TRAP(10). So the difference 10.420 - 12.676 = -2.256 between TRAP(10) and TRAP(30) is approximately 8/9 of the error of TRAP(10). It follows that the error of TRAP(10) is 9/8 * -2.256 = -2.538. Our best estimate of the integral is therefore -2.538 + 12.676 = 10.138. ** Ok I notice that I multiplied by 8/9 and you multiplied by 9/8 where did this come from?????? And why???
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Given Solution: 1 / (u^2-16) = 1 / [(u+4)(u-4)] . Since for 0 < x < 4 we have 1/8 < 1 / (u+4) < 1/4, the integrand is at most 1/4 times 1/(u-4) and at least 1/8 of this quantity, so the original integral is at most 1/4 as great as the integral of 1 / (u-4) and at least 1/8 as great. That is, 1/8 int(1 / (u-4), u, 0, 4) < int(1 / (u^2-4), u, 0, 4) < 1/4 int(1 / (u-4), u, 0, 4). Thus if the integral of 1 / (u-4) converges or diverges, the original integral does the same. An antiderivative of 1 / (u-4) is ln | u-4 |, which is just ln(4) at the limit u=0 of the integral but which is undefined at the limit u = 4. We must therefore take the limit of the integral of 1/(u-4) from u=0 to u=x, as x -> 4. The integral of 1 / (u-4) from 0 to x is equal to ln(x-4) - ln(4). As x -> 4, ln(x - 4) approaches -infinity, Thus the integral diverges. STUDENT QUESTION I am lost from the start of this problem I see where the integrand is (1/4) at its most but how can it be 1/8 at its least. Can you show me a step by step as to how we should have found this. INSTRUCTOR RESPONSE Since 1 / ( (u + 4) ( u - 4) ) = (1 / (u + 4) ) * ( 1 / (u - 4) ), and since on this interval 1/8 < 1 / (u + 4) < 1/4 it follows that 1/8 | 1 / (u - 4) | < |1 / (u + 4) ) * ( 1 / (u - 4) | < 1/4 | 1 / (u - 4) | and therefore that on this interval 1/8 integral | 1 / (u - 4) du | < integral | 1 / ( (u + 4) ( u - 4) ) du | < 1/4 integral | 1 / (u - 4) du |. Since | 1 / (u - 4) | -> infinity as u -> 4, we proceed as indicated to show that this integral approaches -infinity. Our previous inequality 1/8 integral | 1 / (u - 4) du | < integral | 1 / ( (u + 4) ( u - 4) ) du | < 1/4 integral | 1 / (u - 4) du | thereby 'sandwiches' the magnitude of the desired integral between two values, both of which approach infinity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: Section 7.7 Problem 7 problem 7.7.7 (previously 7.7.30) rate of infection r = 1000 t e^(-.5t) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** First graph the function using standard graphing techniques from first-semester calculus: This graph increases at first as you move to the right from t = 0. However e^(-.5 t) eventually approaches zero much faster than t increases so the graph has an asymptote at the positive t axis. So it increases for small positive t but eventually returns almost to the t axis, and it can't be strictly increasing. Its concavity changes from downward (negative) for small positive t to upward for larger t; the point at which the concavity changes is important. We use the standard technique from first-semester calculus to find the point at which this function maximizes. The first derivative is dr/dt = 1000 e^(-.5 t) - 500 t e^(-.5 t). Setting this derivative equal to 0 we get • 1000 e^(-.5 t) - 500 t e^(-.5 t) = 0; dividing through by e^-.5 t we get the equation 1000 - 500 t = 0, which is easily solved to obtain t = 2. A first-or second-derivative test confirms that the t = 2 graph point is a relative maximum. Concavity is determined by the second derivative r'' = e^(-.5 t) [ -1000 + 250 t ], which is 0 when t = 4. This is a point of inflection because the second derivative changes from negative to positive at this point. So the function is concave downward on the interval (-infinity, 4) and concave upward on (4, infinity). The first derivative has a critical point where the second derivative is zero. This occurs at x = 4, which was identified in the preceding paragraph as the point of inflection for the original function. Since the second derivative goes from negative to positive, this point is a minimum of the first derivative. The first derivative is a decreasing function from t = 0 to t = 4 (2d derivative is negative) and is then an increasing function with asymptote y = 0, the x axis, which it approaches through negative values. Its maximum value for t >= 0 is therefore at t = 0. ** People are getting sick the fastest when the rate of infection is highest. This occurs at the relative maximum of the rate function, which was found above to occur at t = 2. Thus people are getting sick the fastest 2 days after the epidemic begins. To find how many people get sick during a time interval, you integrate the rate function over that interval. In this case the interval doesn't end; so you need to integrate the rate function r = 1000 t e^(-.5t) from t = 0 until forever, i.e., from t = 0 to t = infinity. An antiderivative of the function is F(t) = 1000 int ( t e^(-.5 t)) = 1000 [ -2 t e^(-.5t) - int ( e^(-.5 t) ) ] = 1000 [ -2 t e^(-.5 t) - 4 e^(-.5 t) ]. Integrating from 0 to x gives F(x) - F(0) = 1000 [ -2 t e^(-.5 x) - 4 e^(-.5 x) ] - 1000 [ -2 * 0 e^(-.5 *0 ) - 4 e^(-.5 * 0 ) ] = 1000 e^-(.5 x) [ -2 t - 4 ] - (-4000). As x -> infinity, e^-(.5 x) [ -2 t - 4 ] -> 0 since the exponential will go to 0 very much faster than (-2 x - 4) will approach -infinity. This leaves only the -(-4000) = 4000. ** The calculator is fine for checking yourself, but you need to use the techniques of calculus to determine inflection points, maxima, minima etc.. The careful use the calculator to enhance rather than replace mathematical understanding. I get a lot of students in these courses who are now at 4-year institutions and who have taken courses based on the graphing calculator, or even TI-92, and many of them tend to have a very difficult time in courses that don't permit them, and in courses were mathematical understanding is required. ** ** You have to use the techniques of calculus to determine these behaviors. Plugging values in won't show you the exact location of intercepts, maxima, minima, etc.. ** STUDENT QUESTION I didn’t know where to go with the antiderivative but I think I understand your conclusion on that as well. INSTRUCTOR RESPONSE The infection is the rate-of-change function, so the antiderivative is the change-in-amount function. Specificly we have the rate of change of the number of people who are or have been sick, with respect to clock time. The 'amount' is the number of people, so the antiderivative function is the change in the number of people (i.e., in the number who have been or are sick). The definite integral between two clock times therefore tells you how many people are or have been sick between those clock times. If we integrate from some clock time from the initial instant to infinity, we get the total number of people who will get sick. STUDENT COMMENT I used the derivatives because this is a rate problem and the antiderivative is just a change in quantity formula. INSTRUCTOR RESPONSE You are given the rate. If you know the average rate of change of change of a quantity with respect to time on an interval, you multiply it by the time interval in order to get the change in the quantity. This is as opposed to subtracting two quantities and dividing by the time interval, which corresponds to the derivative. Summing up average rate * time interval corresponds to the integral, which is what should have been used here. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand it now. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q004. Determine whether each of the integrals below converges or diverges: • integral ( 1/x, x from 0 to 1 ) • integral ( 1/x, x from 1 to infinity ) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Ln(1) - ln(0) = diverges Ln(inf) - (ln(1) = diverges confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. Determine whether each of the integrals below converges or diverges: • integral ( 1/x, x from 0 to 1 ) • integral ( 1/x, x from 1 to infinity ) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Ln(1) - ln(0) = diverges Ln(inf) - (ln(1) = diverges confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: #*&!