course Mth 174
There were a few questions that i did not understand so if you could helpe me i would greatly appreciate it. Thanks
Calculus II Test 2
Problem Number 1
Prove whether the integral of 1 / ( .7 + e^( 3 x)), from x = 1 to infinity, converges or diverges.
= (.7 + e^3x)^(-1/2) dx = 2(.7 + e^3x)^(1/2) * (1/3)e^3x ?(1/3) from 1 to b
= (7/5) + 2e^(3x) * (1/3)e^3x ?(1/3) = 2e(3x) * (1/3)^3x ?(1/3) = 2e^3b * (1/3)e^3b ?(1/3)
b goes to infinity and the area under curve is not finite so the integral diverges
An antiderivative of this function is 10·x/7 - 10·LN(10 e^(3·x) + 7)/21. However you don't want to think about the details of that integration. All you want to know is whether the integral converges or diverges.
The way to answer that question is to compare the expression with an expression that is known to converge or to diverge.
If 1 / ( .7 + e^( 3 x)) is less than an expression known to converge over this interval then the integral converges.
If 1 / ( .7 + e^( 3 x)) is greater than an expression known to diverge over this interval then the integral diverges.
An obvious comparison is that 1 / (.7 + e^(3x) ) is less than 1 / (e^(3x)) = e^(-3x). This function has antiderivative -1/3 e^(-3x) and is easily shown to converge on the interval from 1 to infinity.
So the given integral must also converge.
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.Problem Number 2
Antidifferentiate sin^ 8( 7 x) cos^ 6( 7 x) with or without the use of tables.
= (1/ 7^2 ?7^2)[7sin^8(7x) * sin^6(7x) + 7 cos^8(7x) * cos^6(7x)] + c
Let z = 7 x so that dz = 7 dx and dx = dz / 7.
Since sin^2(z) = 1 - cos^2(z) you can write sin^8(z) cos^6(z) as
(sin^2(z))^4(cos^6(z)) =
(1 - cos^2(z))^4 (cos^6(z)).
In general cos^n(z) can be integrated by parts, letting dv = cos(z) so that v = sin(z), with u = cos^(n-1)(z) so that du = (n - 1) sin(z) cos^(n - 2)(z). This leads to a reduction formula, reducing the integration of cos^n(z) to that of cos^(n-2)(z).
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.Problem Number 3
Find the volume of the solid obtained by rotating the curve y = x^- 2, between x = 0 and x = .15, about the x axis. Find the volume of the solid obtained by rotating the same curve about the y axis.
The volume of the slice = pi(x^2)deltay = pi(e^-y)deltay
Total volume = sum pi(x^2)deltay = pi(e^-y)deltay
Total volume = pi(e^-y)^2dx from 0 to .15 = pi from 0 to .15 e^-2ydx = pi (-1/2)e^-2y from 0 to .15
= pi (-1/2)(e^-2 ?e^0) = pi/2 (1-e^-2) = 1.36
If the curve is rotated about the x axis then a slice is a circular disk of radius x^(-2). The area of the disk is pi r^2 = pi ( x^-2)^2 = pi x^-4; the volume of a slice is therefore pi x^(-4) `dx and the integral is
integral(pi x^(-4) dx, x from 0 to .15).
It turns out that this integral diverges so the volume is infinite.
If the solid is rotated about the y axis then the radius of a disk at y is equal to the distance from the y axis to the curve. This distance will be the x coordinate of the point. Since y = x^-2, we have x = y^(-1/2), so the radius is y^(-1/2) and the area of the disk is pi r^2 = pi * (y^(-1/2))^2 = pi y^-1 = pi / y. The volume of a disk of thickness `dy is therefore pi / y * `dy. When x = .15 we have y = .15^-2 = 400/9 (decimal expansion 44.444...); as x -> 0 we see that y -> infinity so the integral is
integral(pi / y dy, y from .15 to infinity).
This integral also diverges.
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.Problem Number 4
Antidifferentiate 8 / (-x^2 + 5 x - 13.25) without the use of tables.
= 8x / (-1/3x^3 + (5/2)x^2 ?13.25x) = 8x (-(1/3)x^3 + (5/2)x^2 ?13.25x)^-(1/2)
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If you complete the square on x^2 - 5 x you get x^2 - 5 x + 25/4 - 25/4 = (x - 5/2)^2 - 25/4. So the denominator (-x^2 + 5 x - 13.25) becomes ( - (x + 5/2)^2 + 25/4 - 13.25) = (-(x+5/2)^2 - 7) = - ( (x+5/2)^2 + 7). So the integrand is
- 8 / ( (x+5/2)^2 + 7 ).
Substituting u = x + 5/2 you get the form
-8 / (u^2 + a^2),
which can be integrated by trigonometric substitution, and which is in the table of integrals given with the test.
.Problem Number 5
Use a Riemann sum to represent the approximate mass of a rod of length 25 cm whose density at a point x cm from its end is 6 + 20 x / 25 grams/cm. Find the exact mass of the rod.
Not sure of how to start this problem so if you have any suggestions just let me know I am really confused about this one.
.Partition the interval from 0 to 25 cm using x_0 = 0, x_1, x_2, ..., x_n = 25. Sample point c_i is in the ith interval, interval width is `dx (25 cm - 0 cm) / n = 25 / n, suppressing the units.
The density of the ith interval is 6 + 20 c_i / 25, and the mass of the interval is density * interval width = (6 + 20 c_i / 25) * `dx.
The total mass is therefore approximated by the Riemann sum
sum((6 + 20 c_i / 25) * `dx), which as n -> infinity approaches the integral
integral((6 + 20 c_i / 25) dx, x from 0 to 25).
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.Problem Number 6
Find the future value after 12 years of income stream function $ 50,000 * e^( .071 t) per year, where t is in years from the present, provided we expect money to grow at a constant annual rate of 6%, compounded continuously.
B = P(1 + r)^t
= 50,000e^(.071 t) * (1 + (.06))^12
= 235862.71
The income during a typical interval containing clock time t is
income stream * `dt = $ 50,000 * e^( .071 t) `dt.
This income occurs at clock time t so it still has time (12 year - t) to grow. If t is in years, then this income will grow by factor (1 + r)^(12 - t) = 1.06^(12 - t) and will end up as
( 50,000 * e^( .071 t) `dt) * 1.06^(12-t).
Summing up all such contributions and integrating from t = 0 to t = 12 we have
integral( 50,000 * e^( .071 t) * 1.06^(12-t) dt, t from 0 to 12).
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Problem Number 7
Suppose that for people over age 60 the relative likelihood of death is given by a probability density function p(t) = a e^(-a t). For what value of a would 99% of the people alive at age 60 be dead by age 100? What would then be the average age of people over 60?
P(t) = ae^(-at)
= ae^(-atdt from 0 to 100 = t(0.99)dt from 0 to 60 + t(.99) from 60 to 100dt
= 75
The probability of death during a time interval `dt at age t is p(t) * `dt.
The probability of death between t = a and t = b is therefore
integral(p(t) dt, t from a to b).
The probability of death between age 60 and 100 is therefore
integral(p(t) dt, t from 60 to 100),
and this probability is said to be .99. Thus
integral(p(t) dt, t from 60 to 100) = .99; substituting we have
integral ( a e^(-a t) dt, t from 60 to 100) = .99.
You can perform the integration on the left-hand side, and solve the resulting equation for a.
The mean value of t is then the 'first moment' integral
integral ( t * p(t), t from 60 to infinity).
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Problem Number 8
If p(x) = k x e^(- 1.19 x), 0 <= x <= 1, represents a probability distribution, then
?What is the value of k?
?What is the mean value of x?
This problem again is really confusing me a little so could you help me out of how to start the problem and go from there
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The integral of a probability distribution function over its entire range of definition must be 1. So
integral ( k x e^(-1.19 x) dx, x from 0 to 1) = 1.
Perform the integration then solve for k.
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.Problem Number 9
Find the integral of the function f(x) = e^x, evaluated between x = 0 and x = `pi/2. Then calculate the error obtained by TRAP(3), MID(3) and SIMP(3). Estimate the error you would expect from TRAP(30), MID(30) and SIMP(30).
(3) Midpoint = 3.767 Trapezoid = 6.535 Simpson = 2.345
(30) Midpoint = 3.810 Trapezoid = 70.99 Simpson = 26.20
Find the integral of the function. Do the integration so you can compare the results.
Then include your steps in your solution. Your results are incorrect. They will not differ by that much.
You are asked to estimate the error in each. The errors for these 30-interval approximations will be small; some will be much smaller than others. Be sure you investigate how the error of each method changes with the number of intervals.
Problem Number 10
If we increase the number of intervals in a trapezoidal approximation by 5, by what factor do we expect our error to change? By what factor do we expect error to change if we are using Simpson's Rule?
If the trapezoidal approx is increased by 5 then it would be around 350 and the others would go up by a factor of as well but the simpson would go up by more because you have to use the trapezoidal answer to figure the simpson one out.
The trapezoidal and midpoint rules have errors that decrease be factor n^2; Simpson by factor n^4.
This was received a few days ago. For some reason it did not post at that time. I just discovered the error.
Be sure to send me your revised solutions and/or self-critiques based on my notes. Try to include much more detail, as well as questions.