Math 121 Fall 2009, Final ExamPRACTICE

Instructions

Closed book, closed notes, except for one 8.5”-by-11” (or A4) sheet of paper, okay to use both sides. You may be required to turn in your note sheet with the exam, so write your name on it.

90 minutes are allowed for this exam.

Clearly indicate your answer.

You must show all relevant work and justify your answers appropriately.

Partial credit will be given, but not without sufficient support.

No calculators that have a QWERTY-type keyboard are allowed, unless previously approved. The proctor's discretion is final.

Answers that violate simple upper or lower bounds may result in more point deductions.

There is a shortcut we learned about the average value of a sine or cosine; you may use that shortcut (but say you’re using it).

THIS PRACTICE EXAM DOES NOT COVER EVERYTHING THAT MIGHT BE ON THE FINAL!

The final is comprehensive, but more emphasis will be placed on the material from the last few weeks of the course.

Study old tests and homeworks.

  1. For each integration method listed, give an example integral that is best solved with that method. You do not have to solve the integrals.
  2. By partsSee Ch 7.1
  3. By u-substitutionSee Ch 5.5
  4. By partial fractionsSee Ch. 7.4
  5. By trig identitiesSee Ch 7.2
  6. By trig substitutionSee Ch 7.3
  7. By numerical methods (Trapezoid, Simpson’s, etc.) See Ch 7.7 or make up the ugliest integral you can imagine.
  1. Suppose a person (over 21 years old) has a blood-alcohol content (BAC) of 0.05%. What is their BAC after 5 hours? Their liver gets rid of alcohol at the rate of 0.01 percentage points per hour until their BAC reaches 0.02%, after which it gets rid of alcohol at this rate: f(t) = 0.01% * exp(-t/3), where t is time since hitting a BAC of 0.02%.

After 3 hours, their BAC is down to 0.02 and the exponential-decrease takes over.

How much is eliminated in the next 2 hours? Integral from t=0 to t=2 of 0.01%*exp(-t/3) dt = 0.01%*(-3)*[exp(-t/3)]_0^2 = 0.01%*(-3)*0.4865=0.014597%

Subtract that from the 0.02% remaining to get 0.0054025% as your final answer.

  1. Let f(x) = sin(x) from 0 to pi.
  2. Write an integral formula that will compute the average value of f(x)

(1/(pi-0)) * integral from x=0 to x=pi of sin(x) dx

  1. Find a numerical value for your answer to part (a) by doing the integration.

First, find an LB and UB and guess. An LB of 0 and a UB of 1 are obvious; a guess might be 1/2 or 2/3

(1/pi) * [-cos(x)]_0^pi = (1/pi) [-cos(pi)- -cos(0)] = (1/pi)[--1 - -1] = 2/pi = 0.6366ish, which meets our bounds and is close to our guess.

  1. Find the sum of this series, if it converges:

It looks like a Geometric series, but starts at 0 instead of 1. Re-index to get sum_{n=1} e^(-n+1), and then flip the inside over to get sum_{n=1} (1/e)^(n-1); now it is in the form sum_{n=1} a*r^(n-1), with a=1 and r=1/e, which is roughly 0.367. Since abs(r)<1, the sum will converge, and it converges to a/(1-r) = 1/(1-1/e) = 1.5819767

  1. a) What is the Maclaurin series for sin(a*x) ?

The Maclaurin series for sin(x) is x-x^3/3! + x^5/5! – x^7/7! + etc

Substitute (ax) for x to get (ax)-(ax)^3/3! + (ax)^5/5! – (ax)^7/7! + etc

You could stop here or you could split out the “a” terms if you like.

b) show that the derivative of sin(a*x) is a*cos(a*x) using your series from part (a).

Take the derivative of (ax)-(ax)^3/3! + (ax)^5/5! – (ax)^7/7! + etc term-by-term to get

a – a^3 * 3x^2/3! + a^5* 5x^4/5! – a^7 * 7x^6 / 7! + etc

Cancel the coefficients 3, 5, 7, etc. with the first part of the factorials to get

a – a^3 * x^2/2! + a^5 * x^4/4! – a^7 *x^6 / 6! + etc

Now, change a^3 into a * a^2, and a^5 into a*a^4, etc. and group with the x’s:

a*1 – a(ax)^2/2! + a(ax)^4/4! – a(ax)^6 / 6! + etc

and factor an “a” out front:

a(1 – (ax)^2/2! + (ax)^4/4! – (ax)^6 / 6! + etc)

The part in the parentheses is now cos(ax), so we have

a*cos(ax)

  1. The following integral is important in applied probability ; evaluate the integral using symbolic methods to start with (that is, don’t just use the midpoint rule or a similar tactic). Boil your answer down to a final number in decimal form.

The first temptation is to try u-substitution, with u=-x/2, but then du=-(1/2) dx, and the just-plain-x in the integrand doesn’t go away. So, that leaves integration by parts, with u=x, dv=e^(-x/2), etc; or perhaps (I hope!) you have a formula for the integral of x*e^(ax) in your notes (Integral Formula #96 on the very last page of the book). It is (1/a^2)(ax-1)e^(ax) + C. We set a=-0.5, and we want to evaluate at infinity and zero; we get

(1/.25) (-0.5x-1)e^(-0.5x) from 0 to infinity, which is

(1/.25)[ (-0.5(inf)-1)e^(-0.5inf) - (-0.5*0-1)e^(-0.5*0)], which is

(1/.25)[ 0 - (-1)e^(0)], which is

(1/.25)[ 0 +(+1)1] = 4

Whoops! Forgot the 0.5 in the original integral; throw it in to get 0.5*4 = 2

  1. Suppose f ’’(0) > 0. If you take the Maclaurin series up to first order, will you usually get an LB or UB or neither on f(x) for x values close to 0? Explain. Assume that f is a nice function; that is, we’re talking about what will usually happen, ignoring weird cases.

The Maclaurin series up to first order is f(0) + f '(0) * x , that is, the tangent line at x=0. Since f "(0) is positive, the function is concave up near 0, so the tangent line will fall below the function. This means that the Maclaurin series (to first order) gives a Lower Bound on the function. Notice that it’s irrelevant whether the function is increasing or decreasing.

  1. A) Use the Trapezoid method with n=4 to integrate the function f(x) = sin(pi x) from x=0 to x=1. Also, say whether you should get an LB, UB, or neither by using the Trapezoid method in this case.

The first steps are, as usual: sketch and find a quick LB and UB and guess of the area. LB=0 as usual, UB=1*1 (the curve fits in the unit square), and a guess might be 1/2 or 2/3.

The first step of actually doing the numerical integration is to break up the interval from 0 to 1 into n=4 pieces; your boundaries are at 0, 1/4, 1/2, 3/4 , and 1. Then, set up a table like this (we include Simpson’s method here for part (B) as well)

x / f(x)=sin(pi x) / This trapezoid / running sum / this Simpson's area / running sum
0 / 0 / 0 / 0 / 0 / 0
0.25 / 0.707107 / 0.088388 / 0.088388 / 0 / 0
0.5 / 1 / 0.213388 / 0.301777 / 0.319036 / 0.319036
0.75 / 0.707107 / 0.213388 / 0.515165 / 0 / 0.319036
1 / 1.23E-16 / 0.088388 / 0.603553 / 0.319036 / 0.638071
Trapz answer / Simpson's answer

The two answers are between our bounds and near our guess, so we are probably doing okay.

The Trapezoid method would draw straight lines underneath the sine curve, since it’s concave-down from 0 to 1. This means that the resulting area estimate will be a LB on the true area.

The question didn’t ask you to find the answer exactly, but it’s easy enough that you could. You get (1/pi)*(-1)[cos(pi*x)-cos(pi*0)] = 2/pi=0.6366ish. Note the similarity to the question about average value of a function, up above. My tests often (but not always) have some kind of sneaky connection between problems like that, that you can also use to help check your work.

B) Use Simpson’s method with n=4 to integrate the function f(x) = sin(pi x) from x=0 to x=1.

See above. Here are the formulas used in the table, not that you would have used Excel on the exam. You could also use the 1,2,2,1 pattern for the Trapezoid method, and the 1 4 2 4 1 pattern for Simpson’s method, but I like this way of doing it.

A / B / C / D / E / F / Row#
x / f(x)=sin(pi x) / this trapezoid / running sum / this Simpson's area / running sum / 2
0 / =SIN(PI()*A3) / 0 / 0 / 0 / 0 / 3
0.25 / =SIN(PI()*A4) / =(B4+B3)/2*(A4-A3) / =D3+C4 / 0 / =F3+E4 / 4
0.5 / =SIN(PI()*A5) / =(B5+B4)/2*(A5-A4) / =D4+C5 / =(1/6)*(B5+4*B4+B3)*(A5-A4)*2 / =F4+E5 / 5
0.75 / =SIN(PI()*A6) / =(B6+B5)/2*(A6-A5) / =D5+C6 / 0 / =F5+E6 / 6
1 / =SIN(PI()*A7) / =(B7+B6)/2*(A7-A6) / =D6+C7 / =(1/6)*(B7+4*B6+B5)*(A7-A6)*2 / =F6+E7 / 7
trapz answer / Simpson's answer
  1. Give the Maclaurin series for f(t) = t * e^t, up to 3rd order. Then give the general n’th order term.

You could do this by taking derivatives:

f(t) = t * e^t

f’(t) = 1*e^t + t*e^t

f’’(t) = e^t + 1*e^t + t*e^t

f’’’(t) = e^t + e^t + 1*e^t + t*e^t

then plug in t=0 to get f=0,f’=1, f’’=2, f’’’=3, then write your polynomial: 0 + 1*t/1! + 2*t^2/2! + 3*t^3/3! + …; this simplifies to 0+t+t^2/1! + t^3/2!. The general term is then t^n/(n-1)!, for n starting at 1.

Or, you could start with the Maclaurin series for e^t, then multiply by t:

t*e^t = t*(Maclaurin series for e^t) = t*(1+t+t^2/2! + t^3/3! + …) = t + t^2 + t^3/2! + t^4/3!; again the general term seems to be t^n/(n-1)! (n starting at 1)

  1. Find a general formula for the volume of a paraboloid: f(x) = sqrt(a*x), spun around the x-axis, for x values between 0 and b.

We will use disks (instead of washers or shells) to do this. We start with: the integral from x=0 to x=b of pi * (sqrt(a*x))^2 dx

This becomes: pi*a*integral from 0 to b of x dx, which is pi*a*(0.5 x^2) from 0 to b, which is pi*a*0.5*b^2.