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Completely document your work and your reasoning.
You will be graded on your documentation, your reasoning, and the correctness of your conclusions.
Except where the need for more precision dictates otherwise (e.g., in nuclear physics) all quantities may be rounded to three significant figures. The
generating program works in binary and often generates extraneous digits (e.g., 1.5001 for 1.5, 3.6999 for 3.7).
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function general antiderivative
sin(ax)cos(bx) 1/(b^2-a^2) [ a cos(ax)sin(ax) - b sin(ax)cos(bx) ] + c
cos(ax)cos(bx) 1/(b^2-a^2) [ b cos(ax)sin(bx) - a sin(ax)cos(bx) ] + c
sin(ax)cos(bx) 1/(b^2-a^2) [ b sin(ax)sin(bx) + a cos(ax)cos(bx)] + c
p(x) e^(ax) 1/a p(x)e^(ax) - 1/a INTEGRAL(p'(x)e^(ax),x) + c
p(x) sin(ax) 1/a p(x)cos(ax) + 1/a INTEGRAL(p'(x)cos(ax),x) + c
p(x) cos(ax) 1/a p(x)sin(ax) - 1/a INTEGRAL(p'(x)sin(ax),x) + c
1/(sin(x))^m -1/(m-1) cos(x) / (sin(x))^(m-1) + (m-2)/(m-1) INTEGRAL(1/(sin(x))^(m-2), x) + c
1/sin(x) 1/2 ln | (cos(x)-1) / (cos(x) + 1) | + c
1/(cos(x))^m 1/(m-1) sin(x) / (cos(x))^(m-1) + (m-2)/(m-1) INTEGRAL(1/(cos(x))^(m-2), x) + c
1/cos(x) 1/2 ln | (sin(x)-1) / (sin(x) + 1) | + c
(bx+c)/(x^2+x^2) b/s ln | x^2+x^2 | + c/a arctan(x/a) + c
(cx + d) / [ (x-a)(x-b) ] 1/(a-b) [ (ac + d) ln | x-a | - (bc+d) ln | x-b | ] + c
1 / `sqrt( x^2 +- a^2 ) ln | x + `sqr(x^2 +- a^2 | + c
`sqrt(a^2 +- x^2 ) 1/2 ( x `sqrt(a^2 +- x^2) + a^2 INTEGRAL(1/`sqrt(a^2 +- x^2 ) + c
`sqrt(x^2 - a^2) 1/2 ( x `sqrt(a^2 +- x^2) + a^2 INTEGRAL(1/`sqrt(a^2 +- x^2 ) + c
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Test Problems:
Problem Number 1
Use a Taylor polynomial of appropriate degree to prove that lim{x -> 0} [ (e^x - 1) / x ] = 1.
lim{x>0} e^x-1/x = 1
f(x) = e^x-1/x
f(x) = ln(x-1)/1
f(x) = e^x-1
Calculate the Taylor series of e^x, then substitute into the expression.
The Taylor polynomial is f(x) = f(a) + (x - a) f ' (a) + (x - a)^2 / 2! * f '' ( a) + (x - a)^3 / 3! * f '''(a) + ... .
All the derivatives of e^x are just e^x, and e^0 = 1.
It follows that the Taylor polynomial for e^x about x = 0 is 1 + x + x^2 / 2! + x^3 / 3! + ...
Problem Number 2
Every minute 7.4 grams of salt, along with 4.2 gallons of water, flow into a reservoir containing 37 gallons of water. The salt is immediately and thoroughly
mixed with the water, and 4.2 gallons of the solution flow out of the reservoir. If the reservoir initially contains 10 grams of salt, then what function gives
the concentration of salt as a function of time?
.4.2C gallons of water 4.2C/minute salt leaving system
7.4 g/minute salt entering system S stands for amount of salt in solution so
dS/dt = 7.4-4.2C
The net rate at which water flowing into system is 4.2 4.2 = 0 which is both inflow and outflow
The amount of water in the system at clock time t is 37 gallons + 0 gallons/min * t
So C is S/(37+0t) dS/dt = 7.4 4.2S / (37+0t)
Very good. Now solve your equation
dS/dt = 7.4 4.2S / 37.
Separate your variables to get
dS / (7.4 - 4.2 S) = dt / 37 and integrate both sides. Substitute the initial conditions to evaluate the integration constant.
What do you get?
Problem Number 3
The depth of water in a container is changing at the rate dy / dt = t / 8 + 1 / 4. Find the depth vs. clock time function if it is known that the depth is y = 2.5
when clock time t = 1.1.
dY/dt = t/8 + Ό where y = 2.5 and t = 1.1
must solve this in respect to y so to do so we will end up with the equation
dy = (t/8 + Ό) dt good so far
ln(y) = ln (t/8 + Ό)
This step didn't work. The integral of dy is just y, and the integral of (t / 8 + 1/4) dt is t^2 / 4 + t / 4, so with the integration constant you get
y = t^2 / 4 + t / 4 + c.
Plug your initial conditions into this equation to evaluate c, then use the resulting function to answer the question.
abs(y) = ln (t/8 + Ό)
abs(2.5) = ln((1.1/8) + Ό) = -.9480
.
.
.
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Problem Number 4
Sketch a graph representing the probability distribution of the velocities of the balls in a billiard ball model, assuming that the most likely velocity is 8 .
Using your sketch estimate the expected number of times out of 10000 random velocity observations we will observe a velocity which rounds off to each of
the following values: 3, 8 and 15.
I really do not know how to explain this graph because I am really bad at trying to explain these so any help would be greatly appreciated.
The billiard ball model wasn't used this term so you wouldn't be responsible for this question.
Problem Number 5
Give the first three Fourier approximations to the sawtooth-wave function f(x) = [-x if -`pi <= x <= 0, x if 0 < x < =`pi]. What fraction of the energy of the
function is contained in the constant term and the first three harmonics?
F(x) = -x pi <= x <= 0 x 0< x <= pi
A(0) + a(1)cos(x) + b(1)sin(x) + a(2) cox(2x) + b(2)sin(2x) + a(3)cos(3x) + b(3)sin(3x)
N =1 a(0)+a(1)cos(-x)+b(1)sin(-x) = 1 + pi/squrt(0)cos(x) = 1 + picos(x)
N = 2 a(0) + a(1)cos(x) + b(1)sin(x) + a(2)cos(2x) + b(2)sin(2x) = 1 + picos(x) + pi/sqart(2)cos(2x)
N = 3 a(0) + a(1)cos(x) + b(1)sin(x) + a(2)cos(2x) + b(2)sin(2x) + a(3)cos(2x) + b(3)sin(3x) = 1 + picos(x) + pi/sqart(2)cos(2x) + pi/sqart(3)cos(3x)
.
a(0) = 1 / (2 pi) integral(f(x), x from -pi to pi) = 1 / (2 pi) * pi^2 = pi / 2.
The function f(x) is even, so when multiplied by cos(n x) we obtain an even function. The integral of this even function over the interval from -pi to pi is
double the integral from 0 to pi so we have
a(1) = 2 * int( x cos(x), x, 0, pi) / pi
a(2) = 2 * int( x cos(2x), x, 0, pi) / pi
a(3) = 2 * int( x cos(3x), x, 0, pi) / pi .
f(x) * sin(n x) is, on the other hand, an odd function, so that its integral from -pi to 0 is equal and opposite of its integral from 0 to pi, and the integral from
-pi to pi is 0 so that b(n) = 0 for all n >= 1.
In general an antiderivative of x cos(n x) is found using integration by parts, letting u = x and dv = cos(n x) so that du = dx and v = 1 / n sin(nx). We obtain
u v - integral(v du) = x / n sin(nx) - integral( 1/n sin(nx) dx) = x/n sin(nx) + 1 / n^2 cos(nx).
The change in this antiderivative between x = 0 and x = pi is
pi / n sin(n * pi) + 1 / n^2 cos(n pi) - ( pi / n sin(n*0) + 1 / n^2 cos(n * 0)).
All the sin values are 0, leaving 1 / n^2 cos(n pi) - 1 / n^2 cos(0). If n is even then cos(n pi) = cos(0) = 1 and we obtain a(n) = 0.
If n is odd then cos(n pi) = -1 and we obtain integral -2, so that a(n) = 2 * (-2) / (n^2 pi)= -4 / (n^2 pi).
So the series should be the a(0) term, plus the n = 1 and n = 3 terms:
pi/2 - 4 / pi cos(x) - 4 / (9 pi) cos(3x).
The total energy in the function is
1/pi * integral((f(x))^2, x from -pi to pi) = 2 * integral ( x^2, x from 0 to pi) = 1/ pi * 2 * pi^3 / 3 = 2 pi^2 / 3, approximately 6.7.
The energy in each harmonic is the square of its amplitude, so the energy in the n = 1 harmonic is (4 / pi)^2; in the n = 3 harmonic the energy is (4 / (9
pi))^2.
.
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Problem Number 6
Prove whether the integral of 1 / `sqrt( x^ 1.3 + 1), from x = 1 to infinity, converges or diverges.
This integral is analyzed by looking at its antidervitive which is 10x/2 10(10x^(1.3 +1)) * 1.3
This is not the integral of the function; the integral cannot be expressed in closed form.
If this integral is less than an expression known to converge over this interval then the integral converges
If this integral is greater than an expression known to diverge over this interval then the integral diverges
This integral is less than 1 / (e^3x) = e^-3x and its anti is (1/3)e^-3x so it is easily shown to converge on the interval x to infinity so the integral converges
1 / sqrt(x^1.3 + 1) is not less than e^-(3x) for any x value greater than 1. So this comparison does not work.
For large x, 1 / sqrt(x^1.3 + 1) is close to 1 / sqrt(x^1.3), which is equal to x^-.65. This is x^p with p < 1, so its integral diverges.
However 1 / sqrt(x^1.3 + 1) is less than this divergent integral, not greater, so the comparison doesn't quite work. However if we replace x^-.65 with 1/2
x^-.65, then for any x > 1 we have 1 / sqrt(x^1.3 + 1) > 1/2 x^-.65, and the latter does diverge. This proves that the original integral diverges.
Problem Number 7
Prove the Mean Value Inequality for Integrals: that if f(x) has upper bound M and lower bound m on [a,b], and if f(x) is differentiable on [a, b], then the
integral of f from a to b lies between m(b-a) and M(b-a) (hint: consider single-interval upper and lower sums).
.i really dont know what is going on here so any help would be greatly appreciated here and just help me out please.
.
For any Riemann sum of the form
sum ( f(c_i) `dx)
we know that f(c_i) is greater than the lower bound m, so that f(c_i) `dx > = m `dx and
sum(f(c_i) `dx) >= sum ( m `dx ) = m sum (`dx) = m * (b - a).
Similarly f(c_i) `dx < M `dx, and a similar arrangement shows that the sum is therefore less that M * ( b - a ).
Since the integral is the limiting value of the Riemann sum, the integral must also lie between m * (b - a) and M * (b - a).
.
Problem Number 8
Integrate arctan(x) with respect to x, from x = 1.2 to x = 2.7.
F(x) = arctan(x) from x = 1.2 to x = 2.7
F(x) = arctan(u) u =x du = dx
F(x) = 1 / (1+u^2)
F(x) = 1/ (1+u^2) from 1.2 to 2.7
F(x) = 1 / (1+(2.7)^2) 1 / (1+(1.2)^2) = -.2892
The derivative of arcTan(x) is 1 / (1 + x^2).
To integral arcTan(x), use integration by parts with u = arcTan(x) and dv = dx, so that du = 1 / (1 + x^2) and v = x. Then the integral becomes
u v - integral ( v du) = arcTan(x) * x - integral ( x / ( 1 + x^2) ). The latter integral is easily integrated by the substitution u = x^2.
The result is
x arcTan(x) - 1/2 ln(1 + x^2).
.
Problem Number 9
Explain whether the midpoint rule underestimates the integral between two points for a graph which is concave up or concave down. Explain the same for the
trapezoidal rule.
.The midpoint rule underestimates the integral between two points for a graph when it is both concave up and concave down because it is only accurate up to a
certain point. The trapezoidal rule is very accurate and does not underestimate the integral at all.
A graph which is concave up tends to 'dip down' in the middle and give a low estimate of the function's average value on the interval, resulting in a low
estimate.
If the graph is concave down then the straight line of the trapezoidal approximation lies above the curve and therefore tends to give a high estimate.
Problem Number 10
Antidifferentiate 8 / (x^2 + 5 x + 11.25) without the use of tables.
Must complete the square on x^2 + 5x on the bottom
When done x^2 + 5x + 25/4 25/4 = (x-5/2)^2 25/4
Denominator (x^2 + 5x + 11.25) becomes (x+5/2)^2 + 25/4 + 11.25) = (x + 5/2^2 7) = (x+5/2)^2 + 7
The integrand is -8 / (x+5/2)^2 + 7 u = x + 5/2
-8 / u^2 + a^2 can be integrated by trigonometric substitution in table of integral given in test
Good.
Problem Number 11
Sketch a graph of a smooth curve which is asymptotic to y = 5 for negative values of x and to y = - 5 for positive values of x, and which passes through the
origin. Assuming that this graph represents the function f '', sketch graphs of f ' and f, assuming that both graphs pass through the origin.
f is an antiderivative of f so to get a graph of f . It isnt expected to calculate all the exact values of the Areas, etc
.. but you start the graph of f at
x=0 with a slope of 5 your slope decreases to 0 then to -5 by the time y=5, then should increase first to 0 and to 5 by the time x=5 and should then decrease
to 0 by the time x=5
sketch a reasonable graph of f. The slopes of your f graph should equal the slopes of your f graph.
.
Problem Number 12
Give the differential equation for a mass m experiencing a restoring force F = - k x, and with maximum displacement A. What is the period of the motion?
.i do not know what is going on here so any help would be greatly appreciated here and please just help me out
F = m a (force = mass * acceleration) and acceleration is the second derivative of position. Thus a = x '' and
F = - k x becomes
m * x '' = - k x.
This is a second-order equation which can be written
m x '' + k x = 0. Substituting x = e^(r t) you get characteristic equation
m r^2 + k = 0 so that
r^2 = - k / m and
r = +-sqrt(k / m) * i
leading to solution
x = c1 e^(sqrt(k/m) t i) + c2 e^(-sqrt(k/m) i).
Using Euler's identity the real part of this solution is equivalent to
x = A cos(sqrt(k/m) t + phi), where A and phi are arbitrary constants.
The period of this function is determined by the fact that a cycle corresponds to a change in of 2 pi in the value of sqrt(k/m) * t; this change occurs when t =
2 pi / (sqrt(k/m)) so this is the period of motion.
Typically we use omega to stand for sqrt(k/m) so the function would be expressed as
x = A cos(omega t + phi) and the period would be 2 pi / omega.
Problem Number 13
Find the integral of 7 x ^ 5 * e^( .9 x ^ 6), between x = 1.6 and x = 2.6 , in two ways.
First find an antiderivative of the function, in terms of the original variable x, and apply the First Fundamental Theorem.
Then use an appropriate u = ... substitution and rewrite the integral in terms of u. Don't convert the antiderivative back to the original variable, but
simply apply the First Fundamental Theorem to an antiderivative expressed in terms of u.
Have no idea what I need to do here I just am really clueless when it comes to these kinds of problems. Its probably very simple so please help me.
If you let u = x^6 then du = 6 x^5 so that 7 x^5 = (7/6) du and your integrand becomes
7/6 e^(.9 u) du, which is easily integrated to yield
7 / (.9 * 6) e^(.9 u).
In terms of x this is
7 / (.9 * 6) e^(.9 * x^6).
Over the limits of integration this yields
7 / (.9 * 6) e^(.9 * (2.6)^6) - 7 / (.9 * 6) e^(.9 * (1.6)^6).
You can easily evaluate this using your calculator.
Now if u = x^6, then the limits on u would be 1.6^6 and 2.6^6 so the form 7 / (.9 * 6) e^(.9 u) yields
7 / (.9 * 6) e^(.9 * 2.6^6) - 7 / (.9 * 6) e^(.9 * 1.6^6), yielding the same result.
Problem Number 14
Use Riemann Sums to obtain the integral required to solve the following, and evaluate the integral: How much work is required to pump a fluid with weight
density 13800 Newtons / m^3 out of a rectangular pool with dimensions 270 m by 17 m by depth 8 m, to a height of 10 meters above the bottom of the
pool?
.density = 13800 V= (length*width*height)
W = F = density(height) / Volume = (13800)(10) / 36720 = 3.76 Joules
When the fluid depth is y the height to which it must be raised is (10 meters - y) above the water level. A layer of thickness `dy would have volume `dy *
270 m * 17 m and mass 13800 N/m^3 times this. Summing all such contributions we would have
work = integral( 13800 N/m^3 * 270 m * 17 m * (10 - y) , y from 0 to 8 meters).
The integral is easily enough computed.
.
Problem Number 15
Find the volume of the solid obtained by rotating the first arch of the curve y = sin( 5 x) about the line y = -1.201.
The first arch of the curve extends from the point where 5 x = 0 to the point where 5 x = pi, so it extends from x = 0 to x = pi/5.
The distance from y = -1.2 to y = sin(5x) is sin(5x) - 1.2, so this would be the radius of the circle. The area of the circle would be pi * (sin(5x) - 1.2)^2 and
the volume of a slice of width `dx would be pi * (sin(5x) - 1.2)^2 * `dx, so the volume integral would be
integral ( pi * (sin(5x) - 1.2)^2 dx, x from 0 to pi/5).
The integral is evaluated by standard techniques of integration.
Rotated about y = -1.201 circular disk of radius x^-2
A = pir^2 = pi(x^-2)^2 = pi(x)^-4
V = pix^-4dx
Integral (pi(x)^-4 dx at y = -1.201)
Y = sin (5x) x = y^5(cos5x)
The remainder of the questions are assessment questions, which are optional but which can help, but not hurt, your score on the exam.
However I have no way of knowing what problems were generated for the test you printed out. You would need to type out a copy of each problem in order
for me to respond. You are of course welcome to do so.
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