7.1
• What is
Z
x · cos x dx ?
1. Fill in the right hand side of the following equation by taking the derivative:
(x · sin x)0 =
2. Integrate both sides of the equation.
Instructor: When instructing students to integrate both sides of the equation,
I tell not to worry about the “+C”. They may need to be encouraged to use the
Fundamental Theorem of Calculus on the left hand side. The students should,
of course, get stuck on integrating x · cos x, and hopefully they recognize that
this is the original question. They’re equations should look like:
Z
3. Now solve for
R
(x · sin x)0 = sin x + x · cos x
Z
Z
0
(x · sin x) dx =
sin x dx + x · cos x dx
Z
x · sin x = − cos x + x · cos x dx
x · cos x dx.
•
• If we think of substitution (u substitution, section 5.3) as the inverse to the chain
rule in differentiation, we may think of integration by parts as the inverse to the
product rule.
1. Fill in the right hand side of the following equation by taking the derivative
(using the product rule):
(f (x) · g(x))0 =
2. Now integrate both sides of the resulting equation.
Instructor: When instructing students to integrate both sides of the equation,
I tell not to worry about the “+C”. They may need to be encouraged to use the
Fundamental Theorem of Calculus on the left hand side. They’re equations
should look like:
(f (x) · g(x))0 = f 0 (x)g(x) + f (x)g 0 (x)
Z
3. Now solve for
(f (x) · g(x))0 dx = f 0 (x)g(x) + f (x)g 0 (x)
Z
Z
0
f (x) · g(x) =
f (x)g(x) dx + f (x)g 0 (x) dx
R
f 0 (x)g(x) dx. This is the formula for integration by parts!
• L.I.P.E.T.
When attempting integration by parts, the first step is to choose u and dv in the left
hand side of the formula
Z
Z
u dv = u · v − v du
The mnemonic LIPET can help with this choice. LIPET helps us remember the
order for the five kinds of functions:
L: log
I: inverse trig P: power
E: exponential
T: trig
√
2x , ex , ( 14 )x
cos x, csc x
ln x, log x tan−1 x, sin−1 x x2 , 3 x, x1
When
attempting integration by parts, if the left hand side of the above equation,
R
u dv, is a product of two types of functions from the LIPET list, then choose the
function of the type farthest to the left as u, and the function (along with the
differential) of the type farthest to the right as dv.
R
1. Try LIPET on the following integral: x · cos x dx.
Instructor: I especially like students to work this example after having done
the first activity listed her.
R
2. x ln x dx
Instructor: If there is time, I like to give the students the following list:
R ln x
R
R ln x
dx
dx
x ln x dx
2
x
x
1
It helps them remember that 2 should be interpreted as a power function and
x
Z
ln x
that
dx is better done with substitution. Generally, don’t forget to look
x
for substitutions first, as they usually provide the best method for solving an
integral.
R
3. t2 et dt
Instructor: This is a gentle introduction into using LIPET multiple times in a
single problem.
R
4. ln x dx
Instructor: Here, I give them the hint to rewrite the integral as
Z
Z
ln x dx = ln x · 1 dx
and to think of 1 as the power function x0 , then do LIPET.
R
5. ex cos x dx
(a) Do LIPET once, then stop.
Instructor: Equation will be
Z
Z
x
x
e cos x dx = e sin x − ex sin x dx
(b) Separately, do LIPET on
R
ex sin x dx.
(c) Plug this into your original computation.
Instructor: Equation should be
Z
Z
x
x
e cos x dx = e sin x − ex sin x dx
Z
x
x
x
= e sin x − −e cos x + e cos x dx
Z
x
x
= e sin x + e cos x − ex cos x dx
This is a great opportunity to make sure that students are using parentheses
when they do integration by parts twice. Give them time to articulate
amongst themselves that they have come up against a wall, the same
integral they started with. If possible, one of the students should make the
leap that they can bring that integral to the other side of the equation.
R
(d) Solve for ex cos x dx.
Instructor: If students are still stuck, hint that they can solve
R
algebraically. You may even suggest replacing all factors of ex cos x dx
with the variable I, then solve for I.
R 3 x2
6. x e dx
Instructor: I like to give students a chance to try LIPET on this integral, and
fail. Then I ask why LIPET isn’t working. If the students do not reach this
2
conclusion themselves, I make a strong point that ex is NOT an exponential
function (a fact that comes up again in series).
(a) Rewrite
Z
3 x2
x e dx =
Z
2
x · x2 ex dx
(b) Perform the substitution t = x2 .
(c) Perform LIPET on the resulting integral.
Instructor: If you are using the u-dv notation, this is a convenient place to
point out that substituting with the variable u and then doing integration
by parts can be confusing. Just switch variables, then do LIPET.
R
√
7. sin x dx
√
(a) Perform the substitution t = x.
Instructor: This is a very difficult problem for students. It breaks the
formal rules set up for substitutions up to this point. It can be approached
as a backwards substitution (x = t2 ), but I never treat it this way. That is,
of course, up to you, and will avoid some of the issues that come up when
having students do this activity.
√
If you are treating the substitution t = x, students will almost certainly
get stuck in the following spot:
√
t =
x
1
dt = √ dx
2 x
and find that there is no matching dt in the original integral. I usually tell
them that I would really like to do the following cheat
√
2 x dt = dx
but that this mixes x’s and t’s. Then I ask what I can do to get around that
problem. Usually
with that phrasing, a student will eventually suggest
√
replacing x with t once it has crossed over to join the dt differential,
yielding
2t dt = dx
then I have the students complete the substitution, giving
Z
Z
√
sin x dx = 2t sin t dt
R
(b) Perform LIPET on 2t sin t dt
√
(c) Complete the integral by replace t with x
6.2
• A new way to see cylinders.
1. Graph the region R bounded by x = 2, x = 3, y = 0, and y = 1.
2. Rotate the region R about the x-axis.
(a) Graph this 3-d object. What geometric object is it?
(b) What is the volume formula for this object? Use this formula to calculate
the volume of this object.
Instructor: The idea here is to get students using the volume formula for a
cylinder, and seeing how the area formula for the disc is used. This should
help them remember to use the formula for the disc in the integral formula
coming up.
(c) What is the formula for calculating the volume of this object using
integrals? Note that the area of a disc is used in this formula, also.
3. Rotate the region R about the y-axis.
(a) Graph this 3-d object. What geometric object is it?
Instructor: This may not be something they’ve seen before, but students
should articulate that the object is basically a difference of cylinders, and
the class should treat it this way after that.
(b) What is the volume formula for this object? Use this formula to calculate
the volume of this object.
Instructor: Students should use this idea of a difference of cylinders to get
to the following equation:
V = πR2 · h − πr2 · h = πR2 − πr2 · h
We want them to interpret the volume as the area of the base times the
height, and the area of the base is the area of an annulus:
A = πR2 − πr2
which will be used in the integral formula as well.
(c) What is the formula for calculating the volume of this object using
integrals? Note that the area of an annulus is used in this formula, also.
• A new way to see cones.
1. Graph the region R bounded by y = x, x = 1, and y = 0.
2. Rotate the region R about the x-axis.
(a) Graph this 3-d object. What geometric object is it?
(b) What is the volume formula for this object? Use this formula to calculate
the volume of this object.
Instructor: Don’t focus on the formula for the volume of a cone as much
as we focused on the volume of the cylinder or the difference of cylinders.
The cone volume isn’t used in the integral computation, we’re just allowing
the student to see that the integral computation matches up with other
geometric formulas where they are available. Then, in a following problem,
we discuss when the solid of revolution is not one of our familiar geometric,
we are left only with the integral formula.
(c) What is the formula for calculating the volume of this object using
integrals? Note that we still use the formula for the area of a disc, since
“slices” of this object are 2-dimensional discs.
3. Rotate the region R about the y-axis.
(a) Graph this 3-d object. What geometric object is it?
Instructor: At this point, students should collectively come to the
conclusion that the object is a cylinder minus a cone.
(b) What is a volume formula for this object? Use this formula to calculate the
volume of this object.
Instructor: If there is time, you might use this opportunity to discuss how
the volume of a cone is 1/3 the volume of a full cylinder, and so for this
object - a cylinder minus a cone - the volume will be 2/3 the volume of a
full cylinder. Again, we don’t want to spend too much time on this formula
as it will not be used in the integral equations.
(c) What is the formula for calculating the volume of this object using
integrals? Note that we still use the area of an annulus, since “slices” of this
object are 2-dimensional annuli.
• Seeing solids of revolution as deformations of cylinders.
1. What is the volume of the solid obtained by rotating the region R bounded by
y = sin x + 2, y = 0, x = 0, and x = 2π about the x-axis.
Instructor: The emphasis here is not on the actual functions being used, but
that the object they graph will look like a deformation of a cylinder. This allows
them to see how we are generalizing the idea of the volume of a cylinder, and
also why the area for a disc still comes into play in these equations.
(a) Graph the region R bounded by y = sin x + 2, y = 0, x = 0, and x = 2π.
(b) Imagine rotating this region about the x-axis. Graph this 3-d object as best
you can. Can you see it as a familiar geometric object that has been warped
in some way?
(c) What are the “slices” of this solid? Are they the same as the slices of the
geometric object you thought about it in part b)? Note that since this
object has been “warped” slightly, or “deformed” slightly, there are no
geometric formulas we know to calculate its volume. But the “slices” are
still familiar geometric objects (in this case, discs), and we’re exploiting that
fact to calculate the volume of this more complicated solid.
(d) What is the formula for calculating the volume of this object using integrals?
Instructor: We’re not interested in the actual volume of this object, or
integrating the functions involved (in this case, a sin2 x which involves a trig
identity and may be too involved at this point.) So if you like, just have the
students set up the integral without trying to compute its numerical value.
2. What is the volume of the solid obtained by rotating the region R bounded by
y = sin x + 4, y = sin x + 2, x = 0, and x = 2π about the x-axis.
Instructor: This is clearly similar to the first part of this activity, but we are
now exploring the idea of a deformation of a “cylinder minus a cylinder”, which
will still use the formula for an annulus.
3. Imagine rotating this region about the x-axis. Graph this 3-d object as best you
can. Can you see it as a familiar geometric object that has been warped in some
way?
4. What are the “slices” of this solid? Are they the same as the slices of the
geometric object you thought about it in part b)? Note that since this object
has been “warped” slightly, or “deformed” slightly, there are no geometric
formulas we know to calculate its volume. But the “slices” are still familiar
geometric objects (in this case, annuli), and we’re exploiting that fact to
calculate the volume of this more complicated solid.
5. What is the formula for calculating the volume of this object using integrals?
Instructor: We’re not interested in the actual volume of this object, or
integrating the functions involved (in this case, a sin2 x which involves a trig
identity and may be too involved at this point.) So if you like, just have the
students set up the integral without trying to compute its numerical value.
5.3
• Exploring
x
Z
g(x) =
f (t) dt
0
1.
Z
x
t dt, x ≥ 0
g(x) =
0
(a) Let f (t) = t, t ≥ 0. Graph f (t). Make sure to use a t-axis and a y-axis.
(b) We’re going to compute
Z x
t dt, x ≥ 0.
g(x) =
0
(c)
(d)
(e)
(f)
First, plot an x > 0 on the t-axis. This is some positive real number, but
we’re not choosing its value. We’re treating the variable x like a constant
here, usually reserved for symbols like a, b, or especially c.
Now graph f (x), which is just x again in this case, but on the y-axis.
Let’s use the geometric interpretation of g(x). Thus g(x) is the area under
the graph of f (t) and above the t-axis from t = 0 to t = x. What shape does
this give us?
Use the geometry formula for the area of a triangle to determine g(x) (i.e.
A = 21 · b · h.)
We see that g(x) = 12 x2 , an antiderivative of f (x). (Note that the variables
have switched. g(x) is an antiderivative of f (x), not f (t).)
2.
Z
x
cos t dt, 0 ≤ x ≤ 2π
g(x) =
0
(a) Let f (t) = cos t, 0 ≤ t ≤ 2π. Graph f (t). Make sure to use a t-axis and a
y-axis.
(b) We’re going to try to graph g(x). Let’s start by filling in the following table
as best as we can:
x
0
g(x)
π
2
π
3π
2
2π
Instructor: Students should reach the following conclusions:
R0
i. g(0) = 0 cos t dt = 0 using a basic property of definite integrals.
R π/2
ii. g(π/2) = 0 cos t dt: students will need more help here to see that
they can not find an exact value, but the class will reach a consensus
that it is some positive value α, and you want to make sure they
identify α with the area of that geometric shape, the “first hill” of f (t).
Rπ
R π/2
Rπ
iii. g(π) = 0 cos t dt = 0 cos t dt + π/2 cos t dt = 0: students should now
see that the “first hill” of f (t) will
R π cancel with the “first valley” of f (t).
Point out that this means that π/2 cos t dt = −α. Make sure they
identify this in the graph of f (t).
iv. g(3π/2) = −α: using the above logic about the graph of f (t), students
should be able to see that the total integral here will be −α.
v. g(2π) = 0 follows similarly.
The table will look like this now:
x
0
π
2
π
3π
2
2π
g(x)
0
α
0
−α
0
(c) Plot the points in the chart. What curve fits these points?
(d) What do you guess g(x) is?
Instructor: It should be clear now that g(x) = sin x. Point out that this is
an antiderivative of f (x). Again, note the change in variables.
•
d
dx
Z
ln x
√
sin t dt
x
1. First, use the property that
Z b
Z c
Z b
f (x) dx =
f (x) dx +
f (x) dx
a
a
c
Instructor: Students should get
Z ln x
Z c
Z ln x
d
d
sin t dt
=
sin t dt +
sin t dt
√
√
dx
dx
x
x
c
Z c
Z ln x
d
d
=
sin t dt +
sin t dt
√
dx
dx
x
c
2. Second, use the property that
Z b
Z
f (x) dx = −
a
a
f (x) dx
b
Instructor: Students should get
d
dx
Z ln x
d
d
sin t dt
=
sin t dt +
√
dx
dx
c
x
Z
√
c
d
= −
dx
Z
!
x
−
sin t dt
c
Z √
x
!
sin t dt
c
d
+
dx
Z
d
+
dx
Z
ln x
sin t dt
ln x
sin t dt
c
c
3. Last, use the Chain Rule with the First Part of the Fundamental Theorem of
Calculus:
Z u
Z u
d
d
du
f (t) dt
=
f (t) dt ·
dx
du a
dx
a
du
= f (u) ·
dx
Instructor: Students should get
!
Z ln x
Z √x
√ d√
d
d
d
−
x + sin (ln x)
sin t dt +
sin t dt
= − sin x
ln x
dx
dx
dx
dx
c
c
√ 1
1
= − sin x · √ + sin (ln x) ·
x
2 x
NAME:
EID:
Final Exam M408S Spring 2015
To receive credit, write your name on every sheet, show all work, give reasons for answers.
1. (10 points)
Z
Estimate
0
1
x · tan−1 (x2 ) dx using the Taylor polynomial T3 (x) for x · tan−1 (x2 )
centered at 0.
2. (10 points)
What is the radius of convergence for
f (x) =
∞
X
(2n)!
n=0
(n!)2
xn ?
NAME:
EID:
Final Exam M408S Spring 2015
To receive credit, write your name on every sheet, show all work, give reasons for answers.
3. (10 points)
Compute T3 (x) centered at π/3 for f (x) = cos x.
4. (10 points)
The power series
f (x) =
∞
X
n=1
n2
n
xn
+n+1
has radius of convergence R = 1. What is the interval of convergence?