1220-006 CALCULUS II
Review sheet for Midterm 1
Alan M. Watson
September 26, 2015
Abstract
Below I’m summarizing very briefly what you should know from each section and I’m
providing a few sample exercises from each section from chapter 6. Chapter 7 consists mainly
of integrals, so I’m not including sample exercises (you have enough of them in the textbook).
1
Section 6.1: The natural logarithm function
Things you should know:
(i) We defined the natural logarithm as
Z
ln(x) =
1
x
1
dt
t
in order to fix the gap we had in the power rule and we saw that this logarithm has the
same properties as the one you learned when studying algebra.
(ii) We saw that
d
dx
ln(x) = x1 .
Sample questions:
• Compute the derivative of ln(x +
• Compute
R1
• Compute
R
t+1
0 2t2 +4t+3
x2 +x
2x−1
√
x2 + 1)
dt
dx (use long division first).
√
2
• Find the length of the curve y = x4 − ln x, 1 ≤ x ≤ 2. Remember that the length of a
Rbp
curve y = f (x) between x = a and x = b is given by a 1 + [f 0 (x)]2 dx
• Find f 0 (1) if f (x) =
R ln(x2 +x−1)
1
cos2 (t) dt.
1
2
Sections 6.2/6.3/6.4: Inverse functions and their derivatives;
the natural exponential function
Things you should know:
(i) If a function f is strictly monotonic in its domain, then it has an inverse (remember, the
monotonicity condition ensures that any horizontal line meets the graph of your function
at one point at most.)
(ii) Inverse function theorem: Suppose that a function f is differentiable and strictly
monotonic on an interval [a, b]. If f 0 (x) 6= 0 at a certain point x in [a, b], then f −1 is
differentiable at the corresponding point y = f (x) and the derivative is given by
0
f −1 (y) =
1
f 0 (x)
(iii) In section 6.3 we defined ex as the inverse function of ln(x) (which made sense because
ln(x) is monotonically increasing) and we compited its derivatives and integrals
Z
d x
e = ex ,
dx
ex dx = ex + C
(iv) In section 6.4 we learned how to extend the results of 6.3 to arbitrary exponential functions
and we found
Z
d x
1 x
x
a = ln(a) · a ,
ax dx =
a +C
dx
ln(a)
Sample questions:
• Show that the function f (x) =
Rx√
0
t4 + t2 + 10 dt has an inverse.
0
• Find f −1 (2) if f (x) = x5 + 5x − 4.
0
• Find f −1 (2) if f (x) = 2 tan x, for − π2 < x < π2 .
Rx
2
• Sketch the graph of the function f (x) = 0 e−t dt (in the exam I would ask you to find
the domain, determine where the function is increasing/decreasing and so forth.)
• Find the maximum and minimum values of f (x) =
• Compute
R
e3/x
x2
ln x
.
1+(ln x)2
dx
• Consider the function
x
Z
log10 (t2 + 1) dt
f (x) =
0
Where is it concave up? Does it have any inflection points.
2
3
Section 6.5: Exponential growth and decay
Things you should know:
(i) In this section your are given a time-dependent quantity y(t) (population, amount of
bacteria, etc) and you are allowed to assume that the rate of variation at any given time
d
(i.e. dt
y) is proportional to the current quantity, namely
d
y(t) = ky(t)
dt
where k is some constant.
(ii)
d
dt y(t)
= ky(t) is an example of a differential equation (an equation involving a function
y(t) and one or more of its derivatives) and it can be solved by separating variables, namely
by writing
dy
= kdt
y
and by integrating both sides (with respect to y on the left and t on the right)
Z
Z
1
dy = k dt =⇒ ln(y) = kt + C =⇒ y(t) = ekt+C = |{z}
eC ekt = y0 ekt
y
y0
(iii) At this point you have y(t) = y0 ekt , where y0 and k are constants that need to be determined and all you need to do is use the information that you are given in the problem in
order to solve for those.
(iv) We finally learned how to compute limits arising as variations of
lim (1 + h)1/h = e
h→0
Sample questions:
• Find the limit limn→∞
3n−1 5n
.
3n
• Pick any word problem from section 6.5.
4
Section 6.6: First-order Linear differential equations
Things you should know:
(i) We learned how to solve differential equations of the form
dy
+ P (x)y = Q(x)
dx
where P (x), Q(x) are known functions of x and where the function y(x) is unkown.
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(ii) The trick was to multiply both sides of the equation by the factor e
use the product rule to conclude that
Z
R
R
y(x) = e− P (x) dx Q(x)e P (x) dx dx
R
P (x) dx
and then to
(iii) Since the solution involves integration, y(x) will be only determined up to a constant.
Hence, any differential equation generally comes with an initial condition (i.e. the value of
y at some point is prescribed - see examples below).
(iv) I won’t be asking you to solve any word problem from this section, but an equation like
this is likely to appear.
Sample questions:
dy
• Solve sin x dx
+ 2y cos x = sin(2x)y when y( π6 ) = 2.
• Solve
5
dy
dx
+ 2x(y − 1) = 0 when y(0) = 3.
Section 6.7: Approximations for differential equations
We skipped this section. I’ll come back to it in the future, time permitting.
6
Section 6.8: The inverse trigonometric functions and their
inverses
Things you should know:
(i) Trigonometric functions are periodic, so they are not invertible over their full domain. The
first thing we did was to restrict them to the largest possible interval over which they are
invertible.
(ii) We then defined the inverse trigonometric functions by requiring:
• x = sin−1 y ⇐⇒ y = sin x,
− π2 ≤ x ≤
• x = cos−1 y ⇐⇒ y = cos x,
0≤x≤π
• x=
tan−1 y
⇐⇒ y = tan x,
• x = sec−1 y ⇐⇒ y = sec x,
π
2
− π2 ≤ x ≤
π
2
0 ≤ x ≤ π,
x 6=
π
2
(iii) We next used these defining expression in order to compute the derivatives of the inverse
trigonometric functions using the chain rule.
For instance, take y = sin−1 (x) let’s compute
d/dx
x = sin y =⇒ 1 = cos y ·
dy
dx .
Starting with the definition we have:
dy
dy
1
1
1
[∗]
= √
=⇒
=
=
−1
dx
dx
cos y
cos(sin x)
1 − x2
where in [∗] we used that the cosine of the angle whose sine is x is precisely
4
√
1 − x2 .
Sample questions:
2x
• Show that tan(2 tan−1 (x)) = 1−x
2.
R x+1
dx.
• Compute the integral √4−9x
2
• Show (as I did above for the sine) that
7
tan−1 (x) =
d
dx
1
.
1+x2
Section 6.9: The hyperbolic functions and their inverses
Things you should know: Hyperbolic functions are defined in terms of exponentials, so they
have no further mystery.
1. We defined cosh x =
ex +e−x
2
and sinh x =
ex −e−x
2
and we noted that
cosh2 x − sinh2 x = 1
2. In analogy with the trigonometric setting, we defined
tanh x =
sinh x
,
cosh x
sechx =
1
,
cosh x
cshx =
1
sinh x
3. You should know how to compute the derivatives of all these functions.
4. Unlike the case of inverse trigonometric functions, there are closed expressions for the
inverse hyperbolic functions (which you don’t need to memorize)
p
p
sinh−1 x = ln(x + x2 + 1), cosh−1 x = ln(x + x2 − 1)
!
√
2
1
1
+
1
+
x
1
−
x
tanh−1 x = ln
, sech−1 x = ln
2
1−x
x
5. Using these closed expressions, one can compute the derivatives of the inverse hyperbolic
functions, but one can also do that faster using the definition of an inverse function together
with the chain rule.
For instance, if y = sinh−1 x, we can compute
d/dx
y = sinh−1 x =⇒ x = sinh y =⇒ 1 = cosh y·
dy
dx
as follows:
dy
1
dy
1 [∗]
1
=⇒
=
= p
=√
2
dx
dx
cosh y
1 + x2
1 + sinh y
where in [∗] we used that cosh2 x − sinh2 x = 1.
Sample questions:
• Find the length of the curve y = a cosh xa between x = −a and x = a, where a is any real
number. Hint: Remember that the length of the curve y = f (x) over an interval [a, b] is
Rbp
given by a 1 + [f 0 (x)]2 dx.
5
• Compute the integral
R
sinh(2x1/4 )
√
4 3
x
dx.
• Show that the function f (x) = tan−1 (sinh x) is odd, increasing, and has an inflection point
at the origin.
• Show that the area under the curve y = cosh t over 0 ≤ t ≤ x is numerically equal to its
arc-length.
8
Section 7.1: Basic integration rules
This section has no new content, but provides good practice.
You are basically given a list of integrals that can be computed using techniques from Calculus
I and chapter 6, once an appropriate change of variable has been performed.
R
R
For instance the integrals sin(x) dx and x6 sin(x7 ) dx are essentially the same after you
use the change of variables u = x7 .
R
R tan(3x)
The integrals ex dx and e1+9x2 dx are essentially the same after you use the change of
variables u = tan(3x).
Tip: If you are given an integral to compute, and you have no idea about how to proceed,
a good strategy is to look for a composition of functions within your integrand, call the inner
function u and perform the change of variables. This might not be enough to solve the problem
completely, but it is likely to transform your integral into a less messy one (i.e. one that you
can find on your tables, or one that is closer to one of those than the integral you started with.)
We did an example like this in class
R
√ tan(x)
2
. In the denominator you have a composition
sec (x)−4
p
sec2 (x) − 4. You certainly don’t have any integral on your table that looks like that, but you
have plenty of examples with sec x replaced by x. If you set u = sec x, then du = tan(x) sec(x)dx
and your integral becomes
Z
1
√
du
u u2 − 4
This integral is not in your tables either, but it looks a lot more like a sec−1 -type.
9
Section 7.2: Integration by parts
Things you should know:
(i) This technique is useful when you have to integrate products of functions, or complicated
functions (like ln x or tan−1 x whose derivative is simple.
(ii) The formula to remember is
Z
Z
u(x)v 0 (x) dx = u(x)v(x)− u0 (x)v(x) dx,
Z
a
b
Z b
u(x)v 0 (x) dx = u(x)v(x)|ba −
u0 (x)v(x) dx
a
(iii) Given a product of functions, identify one with u(x) and the other with v 0 (x) and use the
integration by parts formula to reduce your original integral to a simpler one.
6
(iv) Given a complicated function f (x) (e.g. f (x) = ln x or f (x) = tan−1 x, apply the integration by parts formula with u(x) = f (x) and v 0 (x) = 1.
10
Section 7.3: Some trigonometric integrals
Things you should know: I’m just sketching the procedure you should follow in each case:
for examples, please check your notes or the textbook.
R
R
(i) Integrals of the type sinn (x) dx and cosn (x) dx
• If n is odd, factor one sin(x) or cos(x) out and use sin2 (x) + cos2 (x)
R = 1 to ?re-write
even power. You’ll be left with a bunch of integrals sin x cos x dx or
Rthe remaining
cos x sin? x dx that can be solved with the power rule plus an integral of a smaller
power of sin or cos, so that you can iterate the process.
• If n is even, use the half-angle formulas cos2 (x) =
R
(ii) Integrals of type sinn (x) cosm (x) dx
1+cos(2x)
, sin2 (x)
2
=
1−cos(2x)
2
• If either n or m are odd, say n, factor one sin x out, so you’ll end up with
Z
sin x sineven x cos? x dx
?
2
?
Write the even power of sin x in terms of cos x using sin2 x = (1 − cos
R x) , expand
if necessary, and you’ll end up with a bunch of integrals of the form sin x cos? x dx
to which you can apply the power rule.
• If both n and m are even, use the half-angle formulas as above.
R
(iii) Integrals of type sin(mx) cos(nx) dx
• Use suitable trig identities to re-write
the products ofR sines and cosines as sums
R
thereof. The resulting integrals sin [(m ± n)x] dx or cos [(m ± n)x] dx are then
straightforward (right?).
R
R
(iv) Integrals of type tanm (x) dx and cotm (x) dx
• Factor one tan2 x out and re-write it in terms of sec2 x using
tan2 x = sec2 x − 1. Since
R
d
2
2
sec x tan? x dx that you can
dx tan x = sec x, you are left with a bunch of integrals
solve using the power rule and possibly the integral of a smaller power of tan x to
which you can apply the same procedure.
Sample questions:
R
R
• Compute cos5 (x) dx and sin4 (x) dx
R
R
• Compute tan5 (x) dx and tan4 (x) dx
R
• Compute sin(3t) sin(t) dt
R
• Compute x cos2 x sin x dx (use integration by parts first).
7
11
Section 7.4: Rationalizing substitutions
Things you should know: The goal here is to transform integrals involving roots into integrals
of rational functions (products of polynomials) to which you can apply other techniques.
√
√
(i) For integrals containing n ax + b (no powers of x), use the change u = n ax + b.
√
(ii) For integrals containing a2 − x2 , use the change x = a sin θ, with − π2 ≤ θ ≤ π2 . The root
will disappear since
p
p
p
a2 − x2 = a2 − a2 sin2 x = |a| 1 − sin2 x = |a| cos(x)
√
(iii) For integrals containing a2 + x2 , use the change x = a tan θ, with − π2 ≤ θ ≤ π2 . The root
will disappear since
p
p
p
a2 + x2 = a2 + a2 tan2 x = |a| 1 + tan2 x = |a| sec(x)
√
(iv) For integrals containing x2 − a2 , use the change x = a sec θ, with 0 ≤ θ ≤ π (different
from π2 ). The root will disappear since
p
p
p
x2 − a2 = a2 sec2 −a2 = |a| sec2 x − 1 = ±|a| tan(x)
√
(v) For integrals containing ax2 + bx + c, complete the square under the square-root sign
and you’ll be reduced to one of the previous cases.
(vi) There are times in which it might not be necessary to do any of this. For instance:
Z
Z
1
x
√
√
dx,
dx
2
3−x
3 − x2
Here you can either use the changes of variable above, or identify these integrals in the
tables (the first one is of type sin−1 , and the second one follows by the power rule.
12
Section 7.5: Integration of rational functions using partial
fractions
Things you should know: The goal here is to re-write an arbitrary quotient of polynomials
into a sum of simpler functions to which you can apply other procedures.
(i) If the degree of the numerator is greater than or equal to the degree of the denominator,
start by doing long division. You’ll end up with a polynomial plus another rational function
whose denominator has strictly larger degree.
(ii) Factor the denominator completely: in the most general setting, the denominator will look
like
(x + a)n (x2 + bx + c)m
where x2 + bx + c is irreducible (e.g. x2 + x + 1 or x2 + 3 - you’ll get complex roots when
you apply the quadratic formula).
8
• Each factor (x + a)n in the denominator will contribute n terms:
A1
A2
An
+
+ ··· +
x + a (x + a)2
(x + a)n
where A1 , . . . , An are constants to be determined
• Each factor (x2 + bx + c)n in the denominator will contribute n terms:
A1 x + B 1
An x + Bn
A2 x + B 2
+ ··· + 2
+ 2
2
2
x + bx + c (x + bx + c)
(x + bx + c)n
where A1 , B1 . . . , An , Bn are constants to be determined
(iii) Once you have written down your decomposition, clear the denominators and evaluate
the resulting equation in as many x-values as unknown constants you have. Use those
equations to solve for the unknown constants.
(iv) You are now reduced to computing the integrals of simpler functions of the following types:
•
1
x−a : ln-type.
1
(x−a)m : power-rule
•
2x+b
:
(x2 +bx+c)m
•
1
: tan−1 -type.
x2 +bx+c
1
: this is possibly
(x2 +bx+c)m
•
•
type.
ln-type (if m = 1) or power rule (if m > 1).
the toughest.
√
√
1
c tan(θ) or x = c sec(θ)
If b = 0 you have (x2 +c)
m and you should use x =
√
(depending on whether c is positive or negative). If c is positive, using x = c tan(θ)
√
and dx = c sec2 θdθ the integral becomes
√
√ Z
Z
Z
Z
1
c sec2 θ
c
sec2 θ
[∗]
dx
=
dθ
=
dθ
=
cos2m−2 θ dθ
(x2 + c)m
cm
sec2m θ
(c tan2 θ + c)m
where in [∗] I used tan2 θ + 1 = sec2 θ. The last integral is an even power of cos θ,
which you should know how to compute (half-angle formula...)
If b 6= 0 you should start by completing the square, so you’ll end up with something
that looks like ((x+α)12 +β)m and if you set u = x + α you are reduced to the previous
case.
13
Section 7.6: Integration strategies
This section has no new content: it basically contains a collection of integrals that can be
computed using techniques from previous chapters, once an appropriate change of variable has
been performed. Check the tips I wrote above within section 7.1 and COMPUTE AS MANY
INTEGRALS AS POSSIBLE.
9