6.1
• Graph y = x and y = x3
1. Find the area between y = x and y = x3 from x = 0 to x = 1
(a)
(b)
(c)
(d)
Shade in the region between the two graphs from x = 0 to x = 1.
Which function is “on top”?
Write down the integral representing the area between the two graphs.
Now calculate the area by computing the integral.
2. Find the area between y = x and y = x3 from x = 0 to x = 2.
(a) Shade in the region between the two graphs from x = 0 to x = 2.
(b) Which function is “on top”? When does it change?
(c) Write down the integral representing the area between the two graphs using
the absolute value formula.
(d) Now rewrite that as a sums of integrals based on your answer to part b).
(e) Now calculate the area by computing the integrals.
3. Find the area between y = x and y = x3 .
(a) We have not been given bounds on x. What could the question be asking
for?
(b) Shade in the regions between the two graphs that are bounded. What are
the x values for the bounded regions between the two graphs?
(c) Which function is “on top”? When does it change?
(d) Write down the integral representing the area between the two graphs using
the absolute value formula.
(e) Now rewrite that as a sums of integrals based on your answer to parts b)
and c).
(f) Now calculate the area by computing the integrals.
• Find the area between sin x and cos x from x = 0 to x = π.
1. Graph both functions on the interval [0, π].
2. Shade in the region between the two graphs from x = 0 to x = π.
3. Which function is “on top”?When does it change?
4. Write down the integral representing the area between the two graphs using the
absolute value formula.
5. Now rewrite that as a sums of integrals based on your answer to part b).
6. Now calculate the area by computing the integrals.
• Find the area between y = tan−1 x and y = π4 x2 .
1. Graph both functions. Note that these two graphs intersect at (0, 0) and
(1, π/4).
2. Shade in the bounded region. Which function is on top?
3. Write an integral expressing the area between these two graphs.
Instructor: Students should get the integral
Z 1
π
tan−1 x − x2 dx
4
0
R
After some time, they will notice that they can not calculate tan−1 x dx. Ask
for suggestions, and if no one suggests it, then explain that the graphs should be
expressed as functions of y. This will lead to the following question:
√
4. Find the area between x = tan y and x = √2π y.
Instructor: Now is a good time to discuss why we choose the positive square
root when solving for x in the equation y = π4 x2 .
5. Graph both functions. Note that these two graphs intersect at (0, 0) and
(π/4, 1).
6. Shade in the bounded region. Which function is on top?
Instructor: This is a good time to point out that “on top” for functions of y
means to the right.
7. Write an integral expressing the area between these two graphs.
8. Now calculate the area by computing the integral.
5.5
•
R
tan x dx
Instructor: Ask for suggestions for this integral and eventually students will ask to
rewrite it.
Z
Z
tan x dx =
sin x
dx
cos x
Now the instructor should attempt the u substitution u = sin x, and fail. Most likely,
a student will suggest the substitution u = cos x. Ask the students to try this
substitution on their own. Once they have
Z
tan x dx = − ln | cos x | + C,
ask for another way this can be written. I usually ask “what will Quest do?”
Z
tan x dx = − ln | cos x | + C = ln | (cos x)−1 | + C = ln | sec x | + C
• Moving a constant over to du
R
2
1. xex dx
Instructor: This example is easy enough, just point out that when choosing u,
we don’t need to see the entire du in the integral. Just up to multiplication by a
constant. In this case, u = x2 , but du = 2x dx is not in the integral. However,
1
du = x dx is there.
R2 5x
2. e dx
Instructor: Again, this example is easy, but notice that whenever u is linear,
then du is just a number, and since we can change du by a constant, we never
need to see du. We can always do a linear substitution: u = mx + b.
Z
x
3.
dx
5 − 2x
Instructor: Here the students will do a linear substitution (they’ve been led to
this by the previous problems) in order to simplify the denominator, but this
will complicate the numerator. This leads them into more difficult substitutions
that require “solving for x” after having chosen u. Or even more difficult
substitutions like the next one.
•
x3
√
dx
x2 + 1
Instructor: I would consider letting the students try for a few minutes, then trying
to elicit the suggestion of u = x2 + 1. Note that it is not easy to see du now (up to a
constant, we’re looking for x dx), then give them the hint to write
Z
Z
x3
x2
√
√
dx =
· x dx
x2 + 1
x2 + 1
Z
Now
1
du = x dx
2
At this point I say “let’s try to swap out everything we can” and write
Z
Z
x2
1
√
√ · du
· x dx =
2
u 2
x +1
u = x2 + 1
and point out that we don’t know the numerator still. Invariably, a student will
produce the solution that we should replace x2 with u − 1.
•
Z
tan θ sec2 θ dθ
Note: This example will also be included in section 7.3, Trigonometric Integrals,
which is where I prefer to address it.
Instructor: I would begin this with the class, and ask for suggestions. Some student
will come up with the substitution u = tan θ. Then have the students work the
problem that way to get
Z
Z
1
1
2
tan θ sec θ dθ = u du = u2 + C = tan2 θ + C
2
2
Then ask for another way to solve this integral. If no one has any suggestions, give
them the hint that it’s another substitution. If no one has any suggestions, rewrite
Z
Z
2
tan θ sec θ dθ = sec θ tan θ sec θ dθ
and someone will eventually suggest the substitution u = sec θ, giving
Z
Z
1
1
2
tan θ sec θ dθ = u du = u2 + C = sec2 θ + C
2
2
At this point I always wait, leaving both solutions on the board. I just wait for as
long as it takes until someone notices there might be an issue. At this point, the
students really need to be engaged enough to see for themselves that they’ve gotten
two different answers to the same problem, and to be concerned about that. Then I
talk about how these answers only differ by a constant, which is wrapped up in the
“+C”. I also bring up the example that ln(2x) + C is the same as ln(x) + C for the
same reason
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?