B Veitch
Calculus 2 Final Exam Study Guide
This study guide is in no way exhaustive. As stated in class, any type of question from class, quizzes,
exams, and homeworks are fair game. Every problem can have a variation or require multiple techniques.
1. Some Algebra Review
(a) Logarithm Properties
i. y = logb x is equivalent to x = by .
ii. logb b = 1
iii. logb 1 = 0, ln 1 = 0
iv. logb (br ) = r, ln(er ) = r
v. logb (xr ) = r logb (x), ln(xr ) = r ln(x)
vi. logb (M N ) = logb (M ) + logb (N )
M
vii. logb
= logb (M ) − logb (N )
N
viii. The domain of logb (u) is u > 0
ix. All the rules above hold for ln x. Remember that ln x = loge x
x. Change of Base
loga (M ) =
logb M
ln M
=
ln a
logb a
This is useful if you’re asked to find the derivative of y = log5 (x + 1)
(b) Exponential Properties
i. an am = an+m
v. a−n =
ii. (an )m = anm
vi.
iii. (ab)n = an bn
an
1
iv. m = an−m = m−n
a
a
1
an
1
= an
a−n
vii. am/n = (am )1/n = (a1/n )m
(c) Properties of Radicals
i.
ii.
√
n
√
n
a = a1/n
am = am/n
2. Limits
(a) Notation
i. General Limit Notation: lim f (x) = L
x→a
ii. Left Hand Limit: lim− f (x) = L
x→a
iii. Right Hand Limit: lim f (x) = L
x→a+
iv. If lim− f (x) = lim+ f (x) = L then lim f (x) = L
x→a
x→a
x→a
(b) Limits at ±∞. Assume all polynomials are in descending order.
1
B Veitch
Calculus 2 Final Exam Study Guide
i. lim
x→∞
a
= 0, r > 0
xr
axn + ...
a
=
x→∞ bxn + ...
b
iii. lim
axm + ...
=0
x→∞ bxn + ...
axm + ...
= ±∞
x→∞ bxn + ...
ii. If n > m, then lim
iv. If m > n, then lim
(c) Evaluation Techniques
i. If f (x) is continuous, then lim f (x) = f (a)
x→a
lim
x→3
p
p
√
x2 + 4 = 32 + 4 = 13
ii. Factor and Cancel
If you evaluate a rational function and get
0
, then try to factor.
0
x2 − 9
(x − 3)(x + 3)
(x + 3)
= lim
= lim
=6
x→3 x − 3
x→3
x→3
x−3
1
lim
iii. Rationalizing Numerators / Denominators
Try this technique if you have radicals. Multiply top and bottom by the conjugate.
√
√
x − 16
1
x−4
x−4
x+4
√
= lim
·√
= lim
= lim √
x→4 x − 16
x − 16
x + 4 x→4 (x − 16)( x + 4) x→4 x + 4
√
lim
x→4
iv. Combine by Using Common Denominators
Try this when you need to combine fractions within fractions
lim
x→a
2
x+4
−
2
a+4
x−a
= lim
x→a
2
x+4
−
2
a+4
x−a
·
(x + 4)(a + 4)
2(a + 4) − 2(x + 4)
= lim
(x + 4)(a + 4) x→a (x − a)(x + 4)(a + 4)
2a − 2x
−2(x − a)
−2
−2
= lim
= lim
=
x→a (x − a)(x + 4)(a + 4)
x→a (x − a)(x + 4)(a + 4)
x→a (x + 4)(a + 4)
(a + 4)(a + 4)
= lim
(d) Piecewise Functions - Know how to graph and evaluate Piecewise Functions. These are good
ones to test your understanding of left-hand, right-hand, and general limits. They are also used
to test your understanding of continuity.
(e) Definition of Continuity
A function f is continuous at x = a if the following three conditions are satisfied:
i. f (a) must exist
ii. lim f (x) must exist
x→a
iii. lim f (x) = f (a)
x→a
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B Veitch
Calculus 2 Final Exam Study Guide
3. Derivatives
(a) Limit Definition of a Derivative
f (x + h) − f (x)
h→0
h
f 0 (x) = lim
f (x) − f (a)
x→a
x−a
or f 0 (a) = lim
(b) Tangent Lines
When asked to find the equation of a tangent line on f at x = a, you need two things: A point
and a slope.
i. You’re usually given the x value. If they don’t tell you the y value, you must plug the x
value into f (x) to get the y value. Now you have a point (a, f (a))
ii. To find the slope m, you find m = f 0 (a).
iii. The equation of the tangent line is y − f (a) = f 0 (a)(x − a). You may need to write this in
slope-intercept form.
(c) Derivative Formulas
i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
ix.
x.
xi.
xii.
xiii.
xiv.
d
(c) = 0
dx
d
(f ± g) = f 0 ± g 0
dx
d
(x) = 1
dx
d
(kx) = k
dx
d
(xn ) = nxn−1
Power Rule:
dx
d
([f (x)]n ) = n[f (x)]n−1 · f 0 (x)
dx
Product Rule: (f g)0 = f 0 g + f g 0
0
f 0 g − f g0
f
=
Quotient Rule:
g
g2
0
Chain Rule: (f (g(x))) = f 0 (g(x)) · f 0 (x)
d x
(a ) = ax ln a
dx
d x
(e ) = ex
dx
d
1
(ln x) =
dx
x
1
d
(loga x) =
dx
x ln a
d f (x) e
= f 0 (x) · ef (x)
dx
xv.
xvi.
xvii.
xviii.
xix.
xx.
xxi.
xxii.
xxiii.
xxiv.
xxv.
xxvi.
xxvii.
d
1
(ln(f (x))) =
· f 0 (x)
dx
f (x)
d
(sin x) = cos x
dx
d
(cos x) = − sin x
dx
d
(tan x) = sec2 x
dx
d
(sec x) = sec x tan x
dx
d
(csc x) = − csc x cot x
dx
d
(cot x) = − csc2 x
dx
d
1
sin−1 x = √
dx
1 − x2
d
−1
cos−1 x = √
dx
1 − x2
d
1
tan−1 x =
dx
1 + x2
d
−1
cot−1 x =
dx
1 + x2
d
1
sec−1 x = √
dx
x x2 − 1
d
−1
csc−1 x = √
dx
x x2 − 1
REMEMBER: all of these derivatives have a version for the chain rule. For example:
d
1
· 2e2x
tan−1 (ex ) =
dx
1 + (e2x )2
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B Veitch
Calculus 2 Final Exam Study Guide
(d) Critical Points
x = c is a value of f (x) if f 0 (c) = 0 or f 0 (c) does not exist.
(e) Increasing / Decreasing
i. If f 0 (x) > 0 on an interval I, then f (x) is
ii. If f 0 (x) < 0 on an interval I, then f (x) is
increasing.
decreasing.
(f) Concave Up / Concave Down
i. If f 00 (x) > 0 on an interval I, then f (x) is
ii. If f 00 (x) < 0 on an interval I, then f (x) is
concave up.
concave down.
(g) Inflection Points
x = c is an inflection point of f if
i. The point at x = c must exist.
ii. Concavity changes at x = c
(h) First Derivative Test to find Local (Relative) Extrema
i. Find all critical value of f (x) where f 0 (c) = 0
ii. Local Max at x = c if f 0 (x) changes from (+) to (−) at x = c.
iii. Local Min at x = c if f 0 (x) changes from (−) to (+) at x = c.
iv. If f 0 (x) does not change signs, it’s still an important point to plot. It may be a place where
the slope is 0, a corner, an asymptote, a vertical tangent line, etc.
v. Make sure you write your relative max and mins as points (c, f (c))
(i) Second Derivative Test to find Local (Relative) Extrema
i. Find all critical value of f (x) where f 0 (c) = 0
ii. Local Max at x = c if f 00 (c) < 0
iii. Local Min at x = c if f 00 (x) > 0
iv. Make sure you write your relative max and mins as points (c, f (c))
(j) Absolute Extrema
i. (c, f (c)) is an absolute maximum of f (x) if f (c) ≥ f (x) for all x in the domain.
ii. (c, f (c)) is an absolute minimum of f (x) if f (c) ≤ f (x) for all x in the domain.
(k) Finding Absolute Extrema of a continuous f (x) over [a, b]
i. Find all critical values of f (x) on [a, b].
ii. Evaluate f (x) at all critical values.
iii. Evaluate f (x) at the endpoints, f (a) and f (b).
4
B Veitch
Calculus 2 Final Exam Study Guide
iv. The absolute max is the largest function value and the absolute min is the smallest function
value.
(l) L’Hospitals Rule
i. If lim f (x) = 0 and lim g(x) = 0, then
x→a
x→a
f (x)
f 0 (x)
= lim 0
, g 0 (x) 6= 0
x→a g(x)
x→a g (x)
lim
ii. If lim f (x) = ±∞ and lim g(x) = ±∞, then
x→a
x→a
f (x)
f 0 (x)
= lim 0
x→a g(x)
x→a g (x)
lim
iii. Indeterminate Form of Type 0 · ∞
Given lim f (x)g(x), you do the following to put it in the form
x→a
0
∞
or
0
∞
f (x)
1/g(x)
g(x)
B. Rewrite f (x)g(x) as
1/f (x)
iv. Indeterminate Form of 00 , ∞0 , or 1∞
A. Rewrite f (x)g(x) as
Given lim f (x)g(x) , rewrite the limit as
x→a
lim eg(x) ln(f (x))
x→a
then take the limit of the exponent
lim g(x) ln(f (x))
x→a
This should put the limit in the Indeterminate Form of Type 0 · ∞
v. Indeterminate Form of ∞ − ∞
The first thing to try to combine the functions into One Big Fraction.
L’Hospitals Rule.
(m) Inverse Derivative Theorem
Let f (x) be a one-to-one function.
0
f −1 (b) =
5
1
f 0 (f −1 (b))
Then try
B Veitch
Calculus 2 Final Exam Study Guide
4. Summary of Curve Sketching − Always start by noting the domain of f (x)
(a) x and y intercepts
i. x-intercepts occur when f (x) = 0
ii. y-intercept occurs when x = 0
(b) Find any vertical, horizontal asymptotes, or slant asymptotes.
i. Vertical Asymptote: Find all x-values where lim f (x) = ±∞. Usually when the denomix→a
nator is 0 and the numerator is not 0. Rational function MUST be reduced.
ii. Horizontal Aymptotes: Find lim f (x) and lim f (x). There are shortcuts based on the
x→∞
x→−∞
degree of the numerator and denominator.
iii. Slant Asymptotes: Occurs when the degree of the numerator is one larger than the denominator. You must do long division to determine the asymptotes.
(c) Find f 0 (x)
i. Find the critical values, all x-values where f 0 (x) = 0 or when f 0 (x) does not exist.
ii. Plot the critical values on a number line.
iii. Find increasing / decreasing intervals using number line
iv. Use The First Derivative Test to find local maximums / minimums (if any exist).
Remember to write them as points.
A. Local Max at x = c if f 0 (x) changes from (+) to (−) at x = c.
B. Local Min at x = c if f 0 (x) changes from (−) to (+) at x = c.
C. Note: If f 0 (x) does not change signs, it’s still an important point. It may be a place
where the slope is 0, a corner, an asymptote, a vertical tangent line, etc.
(d) Find f 00 (x)
i. Find all x-values where f 00 (x) = 0 or when f 00 (x) does not exist.
ii. Plot these x-values on a number line.
iii. Find intervals of concavity using the number line
iv. Find points of inflection
A. Must be a place where concavity changes
B. The point must exist (i.e, can’t be an asymptote, discontinuity)
(e) Sketch
i. Draw every asymptote
ii. Plot all intercepts
iii. Plot all critical points (even if they are not relative extrema). They were critical points for
a reason.
iv. Plot all inflection points.
v. Connect points on the graph by using information about increasing/decreasing and its
concavity.
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B Veitch
Calculus 2 Final Exam Study Guide
5. Integrals
(a) Definitions
i. Antideriative: An antiderivative of f (x) is a function F (x), where F 0 (x) = f (x).
ii. General Antiderivative: The general antiderivative of f (x) is F (x) + C, where F 0 (x) =
f (x). Also known as the Indefinite Integral
Z
f (x) dx = F (x) + C
iii. Definite Integral:
Z
b
f (x) dx = F (b) − F (a)
a
(b) Approximation Integration Techniques
Z b
b−a
f (x) dx and n (for Simpson’s Rule n must be even), with ∆x =
Given the integral
n
a
and xi = a + i∆x, then
Z b
Left-hand
f (x) dx = ∆x [f (x0 ) + f (x1 ) + f (x2 ) + f (x3 ) + ... + f (xn−1 )]
a
b
Z
Right-hand
f (x) dx = ∆x [f (x1 ) + f (x2 ) + f (x3 ) + f (x4 ) + ... + f (xn )]
a
b
Z
f (x) dx = ∆x [f (x∗1 ) + f (x∗2 ) + f (x∗3 ) + f (x∗4 ) + ... + f (x∗n )] , where x∗i is midpoint in [xi−1 , xi ]
Midpoint
a
b
Z
Trapezoid
f (x) dx =
∆x
[f (x0 ) + 2f (x1 ) + 2f (x2 ) + 2f (x3 ) + ... + 2f (xn−2 ) + 2f (xn−1 ) + f (xn )]
2
f (x) dx =
∆x
[f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + ... + 2f (xn−2 ) + 4f (xn−1 ) + f (xn )]
3
a
Z
b
Simpson’s
a
(c) Common Integrals
Z
i.
Z
k dx = kx + C
Z
ii.
xn dx =
ln x dx = x ln x − x + C
xi.
xn+1
+ C, n 6= −1
n+1
Z
xii.
cos x dx = sin x + C
Z
Z
1
dx = ln |x| + C
x
Z
1
1
iv.
dx = ln |kx + b| + C
kx + b
k
Z
v.
ex dx = ex + C
Z
1
vi.
ekx dx = ekx + C
k
Z
1
vii.
ekx+b dx = ekx+b + C
k
Z
ax
x
+C
viii.
a dx =
ln a
Z
1
ix.
akx dx =
akx + C
k ln a
Z
1
x.
akx+b dx =
akx+b + C
k ln a
sin x dx = − cos x + C
xiii.
iii.
Z
xiv.
Z
xv.
sec2 x dx = tan x + C
csc2 x dx = − cot x + C
Z
xvi.
sec x tan x dx = sec x + C
Z
csc x cot x dx = − csc x + C
xvii.
Z
tan x dx = ln | sec x| + C
xviii.
Z
cot x dx = ln | sin x| + C
xix.
Z
xx.
7
csc dx = ln | csc x − cot x| + C
B Veitch
Calculus 2 Final Exam Study Guide
Z
Z
1
1
−1 x
dx
=
+C
cot
a2 + x2
a
a
Z
1
√
xxv.
dx = sec−1 (x) + C
2−1
x
x
Z
1
xxvi.
− √
dx = csc−1 (x) + C
x x2 − 1
sec x dx = ln | sec x + tan x| + C
Z
x
1
√
xxii.
+C
dx = sin−1
a
a2 − x2
Z
1
1
−1 x
xxiii.
dx
=
+C
tan
a2 + x2
a
a
xxi.
xxiv.
−
(d) Every integral must be written into the proper form in order to use the formulas. For example,
Z
Z
√
x dx = x1/2 dx
Z
Z
3
dx =
5x4
Z
3 −4
x dx
5
8
√
dx =
5
x9
Z
8x−9/5 dx
(e) Finding Area under or between Curves
6. Solids of Revolution
(a) Disk Method - in terms of x
Z
V =
(c) Disk Method - in terms of y
b
Z
2
πR(x) dx
V =
a
Z
πR(y)2 dy
c
(b) Washer Method - in terms of x
V =
d
(d) Washer Method - in terms of y
b
Z
πR(x)2 − πr(x)2 dx
V =
a
c
8
d
πR(y)2 − πr(y)2 dy
B Veitch
Calculus 2 Final Exam Study Guide
You may have to rotate around an axis other than the x or y axis. If you do, you need to adjust the
radius. For example,
Z
V =
d
π(1 + R(y))2 + π(1 + r(y))2 dy
c
(a) Shell - in terms of x
Z
V =
(b) Shell - in terms of y
b
Z
V =
2πxH(x) dx
d
2πyH(y) dx
a
c
7. Arc Length
(b) In terms of y
(a) In terms of x
Z
L=
b
Z
p
1 + (f 0 (x))2 dx
L=
a
c
8. Area of a Surface of Revolution
(a) Rotated around the x-axis
9
d
p
1 + (g 0 (x))2 dy
B Veitch
Calculus 2 Final Exam Study Guide
s
b
Z
SA =
2πy
1+
a
dy
dx
2
Z
dx
s
d
SA =
1+
2πy
c
dx
dy
2
dy
The first integral is in terms of x. That means the y in front of the square root should be in
terms of x. For example, y = x2 + 2. You use x2 + 2 instead of y.
(b) Rotated around the y-axis.
The formula doesn’t change much.
Z
s
b
SA =
2πx 1 +
a
dy
dx
2
Z
dx
SA =
s
d
2πx 1 +
c
dx
dy
2
dy
The second integral is in terms of y. That means the x in front of the square root should be in
√
√
terms of y. For example, if y = x2 , solve for x (x = y) and use y instead of x.
9. Integration Techniques
(a) u Substitution
Z b
Given
f (g(x))g 0 (x) dx,
a
i. Let u = g(x)
ii. Then du = g 0 (x) dx
iii. If there are bounds, you must change them using u = g(b) and u = g(a)
Z b
Z g(b)
f (g(x))g 0 (x) dx =
f (u) du
a
g(a)
(b) Integration By Parts
Z
Z
u dv = uv −
Z
Example:
v du
x2 e−x dx
dv = e−x dx
u = x2
du = 2x dx
v = −e−x
Z
Z
x2 e−x dx = −x2 e−x − −2xe−x dx
You may have to do integration by parts more than once. When trying to figure out what to
choose for u, you can follow this guide: LIATE
10
B Veitch
Calculus 2 Final Exam Study Guide
L Logs
I Inverse Trig Functions
A Algebraic (radicals, rational functions, polynomials)
T Trig Functions (sin x, cos x)
E Exponential Functions
(c) Products of Trig Functions
Z
i.
Z
sinn x cosm x dx
ii.
A. m is odd (power of cos x is odd). Factor out one cos x and place it in front
of dx. Rewrite all remaining cos x as
sin x by using cos2 x = 1−sin2 x. Then
let u = sin x and du = cos x dx
tann x secm x dx
A. m is even (power of sec x is even). Factor out one sec2 x and place it in front
of dx. Rewrite all remaining sec x as
tan x by using sec2 x = 1 + tan2 x.
Then let u = tan x and du = sec2 x dx
B. n is odd (power of sin x is odd). Factor out one sin x and place it in front
of dx. Rewrite all remaining sin x as
cos x by using sin2 x = 1−cos2 x. Then
let u = cos x and du = − sin x dx
B. n is odd (power of tan x is odd). Factor out one sec x tan x and place it
in front of dx.
Rewrite all remain-
ing tan x as sec x by using tan2 x =
sec2 x − 1. Then let u = sec x and
C. If n and m are both odd, you can
du = sec x tan x dx
choose either of the previous methods.
D. If n and m are even, use the following
C. If n odd and m is even, you can use
either of the previous methods.
trig identities
1
(1 − cos(2x))
2
1
cos2 x = (1 + cos(2x))
2
sin(2x) = 2 cos x sin x
D. If n is even and m is odd, the previous
sin2 x =
methods will not work. You can try to
simplify or rewrite the integrals. You
may try other methods.
(d) Partial Fractions
Z
p(x)
dx, the degree of p(x) must be smaller than
q(x)
the degree of q(x). Factor the denominator q(x) into a product of linear and quadratic factors.
Use this method when you are integrating
There are four scenarios.
Unique Linear Factors: If your denominator has unique linear factors
x
A
B
=
+
(2x − 1)(x − 3)
2x − 1 x − 3
To solve for A and B, multiply through by the common denominator to get
x = A(x − 3) + B(2x − 1)
You can find A by plugging in x =
1
. Find B by plugging in x = 3.
2
11
B Veitch
Calculus 2 Final Exam Study Guide
Repeated Linear Factors: Every power of the linear factor gets its own fraction (up to the
highest power).
x
A
D
B
C
=
+
+
+
(x − 3)(3x + 4)3
x − 3 3x + 4 (3x + 4)2
(3x + 4)3
Unique Quadratic Factor:
3x − 1
A
Bx + C
=
+ 2
(x + 4)(x2 + 9)
x+4
x +9
To solve or A, B, and C, multiply through by the common denominator to get
3x − 1 = A(x2 + 9) + (Bx + C)(x + 4)
Repeated Quadratic Factor: Every power of the quadratic factor gets its own fraction (up
to the highest power).
A
Bx + C
Dx + E
Fx + G
2x
=
+ 2
+ 2
+ 2
(3x + 4)(x2 + 9)3
3x + 4
x +9
(x + 9)2
(x + 9)3
(e) Trig Substitution
If you see
√
a2 − x2
√
a2 + x2
√
x2 − a2
Z
x3
√
dx Let
Example:
16 − x2
Z
x3
√
dx =
16 − x2
Substitute
x = a sin(θ)
1 − sin2 (θ) = cos2 (θ)
x = a tan(θ)
1 + tan2 (θ) = sec2 (θ)
x = a sec(θ)
sec2 (θ) − 1 = tan2 (θ)
x = 4 sin θ, dx = 4 cos θ dθ. So x3 = 64 sin3 θ
64 sin3 θ
Z
p
Uses the following Identity
Z
16 − 16 sin2 θ
· 4 cos θ dθ =
64 sin3 θ · 4 cos θdθ
=
4 cos θ
Z
sin3 θ dθ
Now you have an integral containing powers of trig functions. You can refer to that method to
solve the rest of this integral.
Z
sin3 dθ =
cos3 θ
− cos θ + C
3
To get back to x, we need to use a right triangle with the original substitution x = 4 sin x.
√
If cos θ =
16 − x2
, then
4
√
Z
x3
√
dx =
16 − x2
12
16−x2
4
3
3
√
−
16 − x2
+C
4
B Veitch
Calculus 2 Final Exam Study Guide
10. Improper Integrals
(a) Infinity as a bound
∞
Z
i.
t
Z
Z
f (x) dx
f (x) dx = lim
t→∞
a
Z
t→−∞
c
Z
∞
f (x) dx +
−∞
f (x) dx
c
b
Z
f (x) dx
f (x) dx = lim
−∞
Z
f (x) dx =
−∞
a
b
ii.
∞
iii.
t
(b) Discontinuity at a bound
Z b
Z b
i. Discont. at a:
f (x) dx = lim
f (x) dx
t→a+ t
a
Z t
Z b
f (x) dx
f (x) dx = lim−
ii. Discont. at b:
t→b
a
a
Z b
Z c
Z
iii. Discont. at c, a < c < b:
f (x) dx =
f (x) dx +
a
a
b
f (x) dx
c
11. Sequences and Series
(a) Sequences: A list of numbers in a definite order, {a0 , a1 , a2 , a3 , a4 , ...}
i. Sequeeze Theorem If an < bn < cn , for
n ≥ N and lim an = lim cn = L, then
n→∞
n→∞
lim bn = L
n→∞



0,



lim rn = 1,
n→∞




Diverge,
if − 1 < r < 1
if r = 1
elsewhere
iv. A typical heirarchy of sequences
ii. Useful Theorem
If lim |an | = 0, then lim an = 0
n→∞
iii. Geometric Sequence
C < ln n < np < an < n! < nn
n→∞
(b) Series
iii. Geometric Series: If −1 < r < 1, then
i. Definition
∞
X
∞
X
an = a1 + a2 + a3 + a4 + ...
n=1
n=0
∞
X
ii. Partial Sums
arn =
a
1−r
arn−1 =
a
1−r
arn−k =
a
1−r
n=1
SN =
N
X
an = a1 + a2 + a3 + ... + aN
n=1
∞
X
n=k
13
B Veitch
Calculus 2 Final Exam Study Guide
12. Summary of Convergence Tests
Test
Theorem
Comments
Divergence
Given the series
Test
diverges.
P-Test
Integral
Test
X
an , if lim an 6= 0, the series
n→∞
can never be used to test for convergence.
∞
X
1
converges if p > 1. If p ≤ 1,
The series
p
n
n=1
∞
X
1
then
diverges.
p
n
n=1
X
Given the series
an , if you can write an = f (n)
Z ∞
X
f (n) dn converges, then so does
an .
and
Z ∞1
X
If
f (n) dn diverges, so does
an
1
Direct Comparison Test
X
X
an and
bn be series such that 0 ≤ an ≤
X
X
X
bn . If
bn converges, so does
an . If
an
X
diverges, so does
bn .
Let
Can only be used to test for divergence. It
You’re normally using this series as a comparison for Direct or Limit Comparison
Tests
The function f should be easily integrated
using one of our integration techniques
from previous chapters. This test should
not be used when an has factorials.
You need to be able to anticipate the convergence. This allows you to choose an appropriate comparison. You should know
the convergence of one of the series.
Limit Comparison Test
X
X
bn be series where an ≥ 0 and
an
bn ≥ 0. If lim
= L, where L is a positive
n→∞ bn
finite number, then either both series converge or
Let
an and
Alternating
they both diverge.
X
Given the series
(−1)n bn , if bn > 0, decreasing,
Series Test
and bn → 0, then the series converges.
Ratio Test
an+1 .
Let
an be a series. Let L = lim n→∞
an If L < 1, the series converges absolutely.
X
Can be easier than Direct Comparison
Test since getting the required inequalities to work can be time consuming.
You have to check for absolute or
conditional convergence
X
X
|(−1)n bn | =
bn .
by
testing
Use this test when you have exponential
functions or factorials.
If L > 1, the series diverges.
Root Test
If L = 1, the test is inconclusive.
X
p
Let
an be a series. Let L = lim n |an |.
n→∞
Use this test when an = (bn )n .
If L < 1, the series converges absolutely.
If L > 1, the series diverges.
Absolute
If L = 1, the test is inconclusive.
X
If the series
an converges absolutely, then it
If
Conver-
converges.
lutely convergent.
X
|an | converges, when
X
an is abso-
gence
Conditional
Conver-
X
an converges but
|an | diverges,
X
then the series
an converges conditionally.
If the series
X
gence
14
X
You’re usually checking if
|an | conX
verges because
an converged by AST
B Veitch
Calculus 2 Final Exam Study Guide
13. Estimating a Series
(a) Remainder Estimate for Integral Test:
If
X
an is convergent, f (n) is continuous,
n=1
positive, and decreasing, then
Z
∞
RN ≤
f (n) dn
N
If you’re asked to estimate
X
an to 4 decimals (or have an error less than 0.0001), then solve
n=1
integrate and solve for N .
Z
∞
f (n) dn < 0.0001
N
(b) Alternating Series Estimation Theorem: Given a convergent alternating series
X
an =
n=1
X
n=1
then
|RN | ≤ bN +1
14. Power Series
the radius of convergence is R = 0.
(a) A power series centered at 0
X
cn xn = c0 + c1 x + c2 x2 + c3 x3 + ...
This happens when the Ratio Test gives
L > 1.
(b) A power series centered at a
ii. The series converges for all x. The inX
n
2
3
cn (x−a) = c0 +c1 (x−a)+c2 (x−a) +c3 (x−a) +... terval of convergence is (−∞, ∞) and
n=0
n=0
(c) Radius and Interval of Convergence:
Perform the Ratio Test or Root Test (usually Ratio Test) For a given a power series
∞
X
cn (x − a)n , there are only three possibiln=1
the radius of convergence is R = ∞.
This happens when the Ratio Test gives
L = 0.
iii. The series converges on an interval (a −
R, a+R). This happens when the Ratio
ities:
i. The series converges only when x = a.
The interval of convergence is {a} and
Test gives K|x − a|, where you need to
solve K|x − a| < 1.
15. Writing Functions as a Power Series
∞
X
Given f (x) =
cn (x − a)n = c0 + c1 (x − a) + c2 (x − a)2 + c3 (x − a)3 + ... with a radius of convern=0
gence R > 0, then
f 0 (x) =
∞
X
ncn (x − a)n−1 = 1c1 + 2c2 (x − a) + 3c3 (x − a)2 + 4c4 (x − a)3
n=1
Z
f (x) dx =
∞
X
cn (x − a)n+1
1
1
1
= c0 (x − a) + c1 (x − a)2 + c2 (x − a)3 + c3 (x − a)4 + ...
n+1
2
3
4
n=0
1
, ln(1 + x),
(1 + x)2
−1
tan x, etc. Your goal is to differentiate or integrate your function f (x) until it’s in the proper form
You use this technique if you’re asked to find a power series of a function like
A
1−u
15
(−1)n bn ,
B Veitch
Calculus 2 Final Exam Study Guide
A and u can be anything. Once it’s in the proper form, you can write it as a power series.
X
A
=
A(u)n
1 − u n=0
(a) If you had to integrate to get the proper
(b) If you differentiated to get the proper form,
form, then differentiate the power series to
then integrate the power series to get back
get back to the original f (x).
to the original f (x).
16. Taylor Series / Macluarin Series
(b) Maclaurin Series is centered at 0
(a) Taylor Series is centered at a
f (x) =
X
cn (x−a)n = c0 +c1 (x−a)+c2 (x−a)2 +... f (x) =
cn (x)n = c0 + c1 x + c2 x2 + ...
n=0
n=0
with cn =
X
f (n) (a)
n!
with cn =
f (n) (a)
n!
It may helfpul to determine the coefficients c0 , c1 , c2 , ... by using the table
n
f n (x)
f n (a)
cn =
0
f (x)
f (a)
c0 =
0
0
1
f (x)
f (a)
c1 =
2
f 00 (x)
f 00 (a)
c2 =
3
f 000 (x)
f 000 (a)
c3 =
f n (a)
n!
f (a)
0!
f 0 (a)
1!
f 00 (a)
2!
f 000 (a)
3!
(a) If you need to find the Taylor / Maclaurin Series, find a pattern for cn . Then you write the
following series with the appropriate cn
f (x) =
X
cn (x − a)n
n=0
(b) If you need to find the n-th degree Taylor / Maclaurin Polynomial, then just find the coefficients
you need. For example, a 3rd degree Taylor Polynomial requires the coefficients up to c3 .
f (x) = c0 + c1 (x − a) + c2 (x − a)2 + c3 (x − a)3
16