Download Review Handout For Math 1220

Review Handout For Math 1220
Logarithms
Suppose u and b are positive numbers and b 6= 1.
logb u is the power of b which gives u: p = logb u ⇔ u = bp .
Suppose u, v, a and b are positive numbers with a 6= 1
and b 6= 1, and p is any real number.
logb u
logb a
log u = log10 u , ln u = loge u
loga u =
logb 1 = 0 , logb b = 1
logb u = logb v =⇒ u = v
logb up = p logb u
blogb u = u
logb uv = logb u + logb v
logb
u
v
= logb u − logb v
Inverse Trigonometric Functions
For |x| ≤ 1, sin−1 x is the angle θ such that − π2 ≤ θ ≤
π
2
and sin θ = x.
For |x| ≤ 1, cos−1 x is the angle θ such that 0 ≤ θ ≤ π and cos θ = x.
For any real number x, tan−1 x is the angle θ such that − π2 < θ <
π
2
and tan θ = x.
For any real number x, cot−1 x is the angle θ such that 0 < θ < π and cot θ = x.
For |x| ≥ 1, sec−1 x is the angle θ such that 0 ≤ θ <
π
2
or π ≤ θ <
3π
2
and sec θ = x.
For |x| ≥ 1, csc−1 x is the angle θ such that 0 < θ ≤
π
2
or π < θ ≤
3π
2
and csc θ = x.
For − 1 ≤ x ≤ 1 , sin(sin−1 x) = x
For −
For − 1 ≤ x ≤ 1 , cos(cos−1 x) = x
For 0 ≤ x ≤ π , cos−1 (cos x) = x
For any real number x, tan(tan−1 x) = x
For −
For any real number x, cot(cot−1 x) = x
For 0 < x < π , cot−1 (cot x) = x
For x ≤ −1 or x ≥ 1, sec(sec−1 x) = x
For 0 ≤ x <
π
2
or π ≤ x <
3π
,
2
sec−1 (sec x) = x
For x ≤ −1 or x ≥ 1, csc(csc−1 x) = x
For 0 < x ≤
π
2
or π < x ≤
3π
,
2
csc−1 (csc x) = x
π
2
π
2
≤x≤
<x<
π
2
π
2
, sin−1 (sin x) = x
, tan−1 (tan x) = x
Rectangular and Polar Coordinates
Rectangular Coordinate
Polar Coordinate
√
P = (r, θ) where r = a2 + b2 and
P = (x, y) =⇒
θ=



tan−1 ( ab )


π + tan−1 ( ab ) , if a < 0
, if a > 0
⇐= P = (r, θ)
P = (x, y) where x = r cos θ
and y = r sin θ
Series
n
X
ak = a1 + · · · + an
k=1
n
X
(ak ± bk ) =
k=1
n
X
n
X
ak ±
k=1
n
X
n
X
bk
k=1
k=m
n
X
i2 =
i=1
n+i
X
ak−i
k=m+i
n(n + 1)
2
i=1
#2
"
n
X
n(n
+
1)
i3 =
2
i=1
c = nc
i=
k=1
n
X
ak =
n(n + 1)(2n + 1)
6
Arithmetic Series:
n
X
[a + (k − 1)d ] = n2 [2a + (n − 1)d ]
k=1
Finite Geometric Series:
n
X
a rk−1 =
k=1
Infinite Geometric Series: If |r| < 1,
a(1 − rn )
1−r
∞
X
a rk−1 =
k=1
a
1−r
Differentiation Rules
d
(c)
dx
d
(xn )
dx
=0
d
[cf (x)]
dx
= cf 0 (x)
d
[f (x)g(x)]
dx
d
f (g(x))
dx
d
(sin x)
dx
= f 0 (x)g(x) + f (x)g 0 (x)
= n xn−1
d
[f (x)
dx
d
dx
h
f (x)
g(x)
± g(x)] = f 0 (x) ± g 0 (x)
i
=
f 0 (x)g(x)−f (x)g 0 (x)
g 2 (x)
= f 0 (g(x)) g 0 (x)
= cos x
d
(cos x)
dx
= − sin x
d
(tan x)
dx
= sec2 x
d
(cot x)
dx
= − csc2 x
d
(sec x)
dx
= sec x tan x
d
(csc x)
dx
= − csc x cot x
Fundamental Theorem of Calculus
For a continuous function f ,
Z b
a
f (x) dx = F (x)|ba = F (b) − F (a) where F 0 (x) = f (x)
For a continuous function f ,
Z x
d
dx
f (t) dt = f (x)
a
Integration Formulas
Z
[f (x) ± g(x)] dx =
Z b
f (x) dx = −
a
Z
Z
Z
Z
Z a
Z
f (x) dx ±
Z
g(x) dx
cf (x) dx = c
Z b
f (x) dx
b
xn dx =
Z
f (x) dx +
a
Z
Z c
f (x) dx
f (x) dx =
b
Z c
f (x) dx
a
xn+1
+ C , for n 6= −1
n+1
Z
sin x dx = − cos x + C
Z
sec2 x dx = tan x + C
Z
sec x tan x dx = sec x + C
cos x dx = sin x + C
csc2 x dx = − cot x + C
csc x cot x dx = − csc x + C
Strategies For Integration
Method
Example
Expanding
(x − x−1 )2 = x2 − 2 + x−2
Using a trigonometric identity
sin2 x =
Simplifying a fraction
5x6 + 3
= 5x4 + 3x−2
x2
u-substitution: For u = g(x),
For u = x2 + 1 we have x dx =
Z
0
f (g(x)) g (x) dx =
Z
f (u) du
and so
Z
1
[1 − cos(2x)]
2
√
du
2
Z
x
du
√
dx
=
2 u
x2 + 1
For application of integrals, such as finding areas and volumes, see chapter 5 of your textbook.