Exam 1 - Emory Math/CS Department

Exam 1: Things to Know
Indefinite Integrals
Z
xn dx =
1
xn+1 + C
n+1
Z
1
dx = ln |x| + C
x
Z
ex dx = ex + C
(n 6= −1)
Z
sin x dx = − cos x + C
Z
cos x dx = sin x + C
Z
sec2 x dx = tan x + C
Z
sec x tan x dx = sec x + C
Z
1
√
= sin−1 x + C
2
1
−
x
Z
1
= tan−1 x + C
1 + x2
Indefinite Integrals to Know How to Derive
1.
Z
Z
tan x dx =
sin x
dx = −
cos x
Z
1
du = − ln |u| + C = − ln | cos x| + C
u
where we use
u = cos x =⇒ du = − sin x dx
2.
Z
Z
sec x dx =
sec x(sec x + tan x)
dx =
sec x + tan x
Z
sec2 x + sec x tan x
dx =
sec x + tan x
Z
= ln | sec x + tan x| + C
where
u = sec x + tan x =⇒ du = (sec x tan x + sec2 x) dx
3.
Z
Z
ln x dx = x ln x −
dx = x ln x − x + C
u = ln x =⇒ du =
v = x ⇐ dv = dx
Note: Use similar method for inverse trig functions
dx
x
1
du = ln |u| + C
u
Trig Identities
Pythagorean Identities:
1. (sin and cos)
sin2 x + cos2 x = 1
2. (tan and sec)
sin2 x cos2 x
1
+
=
=⇒ tan2 x + 1 = sec2 x
2
2
cos x cos x
cos2 x
Half-Angle Formulas:
1. sin2 x =
1−cos 2x
2
2. cos2 x =
1+cos 2x
2
Integration Techniques
1. u-substitution
2. Integration by parts:
R
u dv = uv −
R
v du
3. Trig Substitution: evaluate integrals in the form
Z
Z
m
n
sin x cos x dx
and
tanm x secn x dx
4. Partial Fractions: denominators contain linear factors (including repeated) and irreducible
quadratic factors of the form x2 − a (no repeats), polynomial division
5. Approximate Integration: approximate definite integrals using midpoint rectangles and
trapezoids, use formula (will be given) to find bound on error
6. Improper Integrals: If f (x) is continuous on (−∞, ∞) and g(x) is continuous on [a, c) and
(c, b], then
Z ∞
Z t
f (x) dx = lim
f (x) dx
Z
a
t→∞ a
b
Z
f (x) dx = lim
t→−∞ t
−∞
Z
∞
Z
f (x) dx = lim
t→−∞ t
−∞
Z
t→∞ a
g(x) dx
t→c−
b
a
Z
g(x) dx = lim
t→c+
Z
b
Z
g(x) dx = lim
a
t→c−
b
g(x) dx
t
t
Z
g(x) dx + lim
a
f (x) dx
t
Z
c
t
Z
f (x) dx + lim
g(x) dx = lim
Z
f (x) dx
a
c
a
b
t→c+
b
g(x) dx
t
if the limits on the right exist as finite numbers. Otherwise, we say that the improper
integrals diverge.
Arc Length and Surface Area
Let
s
ds =
1+
s
ds =
dy
dx
2
dx
dy
2
1+
dx
if y = f (x), a ≤ x ≤ b
dy
if x = g(y), a ≤ y ≤ b
1. Arc Length:
Z
b
ds
s=
a
2. Surface Area:
Z
b
S=
2πy ds
(rotation about x-axis)
2πx ds
(rotation about y-axis)
a
Z
S=
a
b