course MTH 174 ̳Gassignment #007
......!!!!!!!!...................................
20:02:27 query problem 7.6.6 approx using n=10 is 2.346; exact is 4.0. What is n = 30 approximation if original approx used LEFT, TRAP, SIMP?
......!!!!!!!!...................................
RESPONSE --> This is 7.6.11 in the second edition. Remembering that for LEFT and RIGHT the factor in ratio of errors is n, TRAP and MID is n^2 and for SIMP it is n^4 When n =10, the approx for LEFT(10) is 2.346 For LEFT(30), n increased by a factor of 3. The ratio of errors is also n for LEFT, so That means that 1.654/(4.0 - x) = 3 Solving for x: 1.654 = 3(4.0 - x) 1.654 = 12.0 - 3x 1.654 - 12.0 = -3x x = (1.654 - 12.0)/-3 = approx. 3.449 For TRAP(30), the factor for errors is n^2, so 1.654 = 9(4.0 - x) 1.654 = 36.0 - 9x 1.654 - 36.0 = -9x x = (1.654 - 36.0)/-9 = approx. 3.816 For SIMP(30), the factor for errors is n^4, so 1.654 = 81(4.0 - x) 1.654 = 324.0 - 81x 1.654 - 324.0 = -81x x = (1.654 - 324.0)/-81 = approx. 3.980
.................................................
......!!!!!!!!...................................
20:03:38 If the approximation used LEFT then what is your estimate of the n = 30 approximation and how did you get it?
......!!!!!!!!...................................
RESPONSE --> The ratio of errors is n for LEFT, so That means that 1.654/(4.0 - x) = 3 Solving for x: 1.654 = 3(4.0 - x) 1.654 = 12.0 - 3x 1.654 - 12.0 = -3x x = (1.654 - 12.0)/-3 = approx. 3.449
.................................................
......!!!!!!!!...................................
20:04:03 If the approximation used TRAP then what is your estimate of the n = 30 approximation and how did you get it?
......!!!!!!!!...................................
RESPONSE --> For TRAP(30), the factor for errors is n^2, so 1.654 = 9(4.0 - x) 1.654 = 36.0 - 9x 1.654 - 36.0 = -9x x = (1.654 - 36.0)/-9 = approx. 3.816
.................................................
......!!!!!!!!...................................
20:04:13 If the approximation used SIMP then what is your estimate of the n = 30 approximation and how did you get it?
......!!!!!!!!...................................
RESPONSE --> For SIMP(30), the factor for errors is n^4, so 1.654 = 81(4.0 - x) 1.654 = 324.0 - 81x 1.654 - 324.0 = -81x x = (1.654 - 324.0)/-81 = approx. 3.980
.................................................
......!!!!!!!!...................................
20:09:19 This problem has been omitted from the present edition and may be skipped: query problem 7.6.10 If TRAP(10) = 12.676 and TRAP(30) = 10.420, estimate the actual value of the integral.
......!!!!!!!!...................................
RESPONSE --> This problem is 7.7.16 in the second edition, I will show how I worked it. Again, the factor change in n is 3, so the factor change in the error ratio is 3^2 = 9 Using x = Actual value estimation: (x - 12.676) / (x - 10.420) = 9 x - 12.676 = 9(x - 10.420) x - 12.676 = 9x - 93.78 - 12.676 + 93.78 = 9x - x 81.113 = 8x x = 10.139 The estimated actual value of the integral is 10.139
.................................................
......!!!!!!!!...................................
20:09:45 What is your estimate of the actual value and how did you get it?
......!!!!!!!!...................................
RESPONSE --> Using x = Actual value estimation: (x - 12.676) / (x - 10.420) = 9 x - 12.676 = 9(x - 10.420) x - 12.676 = 9x - 93.78 - 12.676 + 93.78 = 9x - x 81.113 = 8x x = 10.139 The estimated actual value of the integral is 10.139
.................................................
......!!!!!!!!...................................
20:12:53 By what factor should the error in the second approximation be less than that in the first, and how does this allow you to estimate the integral based on the difference in the two approximations?
......!!!!!!!!...................................
RESPONSE --> The error should decrease by a factor of 3^2 for a TRAP approximation with a factor of 3 difference in n. These ratios allow solving the equation that I set up to find an approximate value for the integral.
.................................................
......!!!!!!!!...................................
21:50:04 a < b, m = (a+b)/2. If f quadratic then int(f(x),x,a,b) = h/3 ( f(a) / 2 + 2 f(m) + f(b) / 2).
......!!!!!!!!...................................
RESPONSE --> When f(x) = 1 the integral will be integral 1 dx from a to b The integral is x from a to b = b - a Substituting in for f(a) = 1, f(m) = 1, and f(b) = 1 b - a = (b -a)/3 (1/2 + 2 + 1/2) b - a = (b -a)/3 (3) b - a = b - a When f(x) = x the integral will be integral x dx from a to b The integral is x^2/2 from a to b = b^2/2 - a^2/2 Substituting in for f(a) = a, f(m) = m = (b + a)/2, and f(b) = b b^2/2 - a^2/2 = (b -a)/3 [a/2 + 2(a+b)/2 + b/2] b^2/2 - a^2/2 = (b -a)/3 [(a + b )/2 + 2 (a+b)/2] b^2/2 - a^2/2 = (b - a)/3 * 3 (a + b)/2 b^2/2 - a^2/2 = (b - a) * (a + b)/2 b^2/2 - a^2/2 = (ba - ab + b^2 -a^2)/2 b^2/2 - a^2/2 = (b^2 -a^2)/2 b^2/2 - a^2/2 = b^2/2 - a^2/2 When f(x) = x^2 the integral will be integral x^2 dx from a to b The integral is x^3/3 from a to b = b^3/3 - b^3/3 Substituting in for f(a) = a^2, f(m) = m^2 = (a+b)^2/4, and f(b) = b^2 b^3/3 - a^3/3 = (b - a)/3 [a^2/2 + 2(a+b)^2/4 + b^2/2] b^3/3 - a^3/3 = (b - a)/3 [(a^2 + b^2)/2 + 2(a^2 + 2ab + b^2)/4] b^3/3 - a^3/3 = (b - a)/3 [2(a^2 + b^2)/4 + 2a^2 + 4ab + 2b^2)/4] b^3/3 - a^3/3 = (b - a)/3 [(2a^2 + 2b^2)/4 + 2a^2 + 4ab + 2b^2)/4] b^3/3 - a^3/3 = (b - a)/3 [(2a^2 + 2b^2 + 2a^2 + 4ab + 2b^2)/4] b^3/3 - a^3/3 = (b - a)/3 [(2a^2 + 2b^2 + 2a^2 + 4ab + 2b^2)/4] b^3/3 - a^3/3 = (b - a)/3 (4a^2 + 4b^2 + 4ab)/4 b^3/3 - a^3/3 = (4ba^2 + 4b^3 + 4ab^2 - 4a^3 - 4ab^2 - 4ba^2)/12 b^3/3 - a^3/3 = (4b^3 - 4a^3)/12 b^3/3 - a^3/3 = (b^3 - a^3)/3 b^3/3 - a^3/3 = b^3/3 - a^3/3 Using the properties about Sums and Constant Multiples of the Integrand lets us use these three together to show the integral of Ax^2 + Bx + C dx from a to b is equal to integral Ax^2 dx from a to b + integral Bx dx from a to b + integral C dx from a to b The multiple A of x^2 can be brought outside the integrand as can the multiple B of x and the multiple C of 1. We can add all the integrals together to arrive at an integral of Ax^2 + Bx + C dx from a to b which is A(b^3/3 - a^3/3) + B(b^2/2 - a^2/2) + C(b - a)
.................................................
......!!!!!!!!...................................
21:50:35 How did you show that if f(x) = 1, the equation holds?
......!!!!!!!!...................................
RESPONSE --> When f(x) = 1 the integral will be integral 1 dx from a to b The integral is x from a to b = b - a Substituting in for f(a) = 1, f(m) = 1, and f(b) = 1 b - a = (b -a)/3 (1/2 + 2 + 1/2) b - a = (b -a)/3 (3) b - a = b - a
.................................................
......!!!!!!!!...................................
21:50:48 How did you show that if f(x) = x, the equation holds?
......!!!!!!!!...................................
RESPONSE --> When f(x) = x the integral will be integral x dx from a to b The integral is x^2/2 from a to b = b^2/2 - a^2/2 Substituting in for f(a) = a, f(m) = m = (b + a)/2, and f(b) = b b^2/2 - a^2/2 = (b -a)/3 [a/2 + 2(a+b)/2 + b/2] b^2/2 - a^2/2 = (b -a)/3 [(a + b )/2 + 2 (a+b)/2] b^2/2 - a^2/2 = (b - a)/3 * 3 (a + b)/2 b^2/2 - a^2/2 = (b - a) * (a + b)/2 b^2/2 - a^2/2 = (ba - ab + b^2 -a^2)/2 b^2/2 - a^2/2 = (b^2 -a^2)/2 b^2/2 - a^2/2 = b^2/2 - a^2/2
.................................................
......!!!!!!!!...................................
21:51:02 How did you show that if f(x) = x^2, the equation holds?
......!!!!!!!!...................................
RESPONSE --> When f(x) = x^2 the integral will be integral x^2 dx from a to b The integral is x^3/3 from a to b = b^3/3 - b^3/3 Substituting in for f(a) = a^2, f(m) = m^2 = (a+b)^2/4, and f(b) = b^2 b^3/3 - a^3/3 = (b - a)/3 [a^2/2 + 2(a+b)^2/4 + b^2/2] b^3/3 - a^3/3 = (b - a)/3 [(a^2 + b^2)/2 + 2(a^2 + 2ab + b^2)/4] b^3/3 - a^3/3 = (b - a)/3 [2(a^2 + b^2)/4 + 2a^2 + 4ab + 2b^2)/4] b^3/3 - a^3/3 = (b - a)/3 [(2a^2 + 2b^2)/4 + 2a^2 + 4ab + 2b^2)/4] b^3/3 - a^3/3 = (b - a)/3 [(2a^2 + 2b^2 + 2a^2 + 4ab + 2b^2)/4] b^3/3 - a^3/3 = (b - a)/3 [(2a^2 + 2b^2 + 2a^2 + 4ab + 2b^2)/4] b^3/3 - a^3/3 = (b - a)/3 (4a^2 + 4b^2 + 4ab)/4 b^3/3 - a^3/3 = (4ba^2 + 4b^3 + 4ab^2 - 4a^3 - 4ab^2 - 4ba^2)/12 b^3/3 - a^3/3 = (4b^3 - 4a^3)/12 b^3/3 - a^3/3 = (b^3 - a^3)/3 b^3/3 - a^3/3 = b^3/3 - a^3/3
.................................................
......!!!!!!!!...................................
21:51:14 How did you use your preceding results to show that if f(x) = A x^2 + B x + c, the equation must therefore hold?
......!!!!!!!!...................................
RESPONSE --> Using the properties about Sums and Constant Multiples of the Integrand lets us use these three together to show the integral of Ax^2 + Bx + C dx from a to b is equal to integral Ax^2 dx from a to b + integral Bx dx from a to b + integral C dx from a to b The multiple A of x^2 can be brought outside the integrand as can the multiple B of x and the multiple C of 1. We can add all the integrals together to arrive at an integral of Ax^2 + Bx + C dx from a to b which is A(b^3/3 - a^3/3) + B(b^2/2 - a^2/2) + C(b - a)
.................................................
......!!!!!!!!...................................
22:10:46 query problem 7.7.19 integrate 1 / (u^2-16) from 0 to 4 if convergent
......!!!!!!!!...................................
RESPONSE --> integral of 1 / (u^2-16) from 0 to 4 Factoring the denominator 1 / [(u + 4) (u - 4)] Splitting the integrand into parts A / (u + 4) + B / (u - 4)] = 1/ [(u + 4) (u - 4)] A(u - 4) + B(u + 4) = 1 Au - 4A + Bu + 4B = 1 Au + Bu = 0 therfore A = -B -4A + 4B = 1 -4(-B) + 4B = 1 4B + 4B = 1 8B = 1 B = 1/8 since A = -B, A = -1/8 We now have two integrals -1/8 integral 1/(u + 4) from 0 to 4 and 1/8 integral 1/(u - 4) from 0 to 4 and The integrals evaluate to: lim as b -> 4- [-1/8 ln |b + 4| + 1/8 ln |b - 4|] ln |b - 4| gets more and more negative as b gets close to 4 from the left so the integral diverges.
.................................................
......!!!!!!!!...................................
22:11:34 does your integral converge, and why or why not?
......!!!!!!!!...................................
RESPONSE --> ln |b - 4| gets more and more negative as b gets close to 4 from the left so the integral diverges.
.................................................
......!!!!!!!!...................................
22:11:46 If convergent what is your result?
......!!!!!!!!...................................
RESPONSE --> It does not converge.
.................................................
......!!!!!!!!...................................
22:12:54 Why is there a question as to whether the integral does in fact converge?
......!!!!!!!!...................................
RESPONSE --> There is an asymptote at x=4 (and also at x = -4, but the integral only goes from 0 to 4)
.................................................
......!!!!!!!!...................................
22:13:08 Give the steps in your solution.
......!!!!!!!!...................................
RESPONSE --> I gave the steps previously.
.................................................
......!!!!!!!!...................................
22:13:24 If you didn't give it, give the expression whose limit showed whether the integral was convergent or divergent.
......!!!!!!!!...................................
RESPONSE --> I gave this previously.
.................................................
......!!!!!!!!...................................
22:50:21 query problem 7.7.44 (was #39) rate of infection r = 1000 t e^(-.5t)
......!!!!!!!!...................................
RESPONSE --> a) The graph goes from 0 increasing at decreasing rate until t = 2 and then decreasing at a decreasing rate.
.................................................
......!!!!!!!!...................................
22:51:31 describe your graph, including asymptotes, concavity, increasing and decreasing behavior, zeros and intercepts
......!!!!!!!!...................................
RESPONSE --> The graph goes from 0,0 increasing at decreasing rate until t = 2 where the slope is 0 and then decreasing at a decreasing rate.
.................................................
......!!!!!!!!...................................
22:52:01 when our people getting sick fastest and how did you obtain this result?
......!!!!!!!!...................................
RESPONSE --> People are getting sick the fastest when the rate of infection tops off. In this case it reaches a maximum. This maximum can be found by finding the derivative of 1000 t e^(-.5t). d/dt = 1000 e^(-.5t) *(-.5) + e(-.5t) * 1000 d/dt = 1000 e ^(-.5t) * ( 1 - .5t) Setting this derivative equal to 0 and solving for t will give us the time that the rate of change in the rate of infection is 0. 1000 e ^(-.5t) * ( 1 - .5t) = 0 t = 2 will make this equation true, so the fastest rate of infection is at day 2
.................................................
......!!!!!!!!...................................
22:52:11 How many people get sick and how did you obtain this result?
......!!!!!!!!...................................
RESPONSE --> The total number of people infected is found from the integral of 1000 t e^(-.5t) from 0 to infinity dt. Using integration by parts with u = 1000 t and v' = e^(-.5t): du/dt = 1000 v = -2 e^(-.5t) uv - integral u'v 1000t * -2 e^(-.5t) - integral -2 * 1000 * e^(-.5t) dt 1000t * -2 e^(-.5t) + 2 * 1000 integral e^(-.5t) dt 1000t * -2 e^(-.5t) + 2 * 1000 * -2 e^(-.5t) 1000[-2t * e^(-.5t) + -4 e^(-.5t)] 1000[e^(-.5t)(-2t - 4)] from 0 to b 1000[e^(-.5t)(-2b - 4)] - 1000[e^(-.5t)(-2(0) - 4)] 1000[e^(-.5t)(-2b - 4)] - 1000[e^(-.5(0))* (-2(0) - 4)] 1000[e^(-.5t)(-2b - 4)] - 1000(1 * - 4) 1000[e^(-.5b)(-2b - 4)] + 4000 Taking the limit as b -> infinity: 1000[e^(-.5b)(-2b - 4)] + 4000 As b gets larger [e^(-.5b) gets closer to 0 leaving the total number of infected people at 4000.
.................................................
......!!!!!!!!...................................
22:53:15 What improper integral arose in your solution and, if you have not already explained it, explain in detail how you evaluated the integral.
......!!!!!!!!...................................
RESPONSE --> integral of 1000 t e^(-.5t) dt from 0 to infinity. I have already explained how I evaluated it.
.................................................
......!!!!!!!!...................................
22:53:21 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE -->
.................................................
"