Calculus Tables

Commentaires

Transcription

Calculus Tables
ALGEBRA
Lines
Special Factorizations
Slope of the line through P1 = (x 1 , y1 ) and P2 = (x 2 , y2 ):
y − y1
m= 2
x2 − x1
Slope-intercept equation of line with slope m and y-intercept b:
x 2 − y 2 = (x + y)(x − y)
x 3 + y 3 = (x + y)(x 2 − x y + y 2 )
x 3 − y 3 = (x − y)(x 2 + x y + y 2 )
y = mx + b
Binomial Theorem
Point-slope equation of line through P1 = (x 1 , y1 ) with slope m:
(x + y)2 = x 2 + 2x y + y 2
y − y1 = m(x − x 1 )
Point-point equation of line through P1 = (x 1 , y1 ) and P2 = (x 2 , y2 ):
y − y1
y − y1 = m(x − x 1 ) where m = 2
x2 − x1
Lines of slope m 1 and m 2 are parallel if and only if m 1 = m 2 .
Lines of slope m 1 and m 2 are perpendicular if and only if m 1 = − m12 .
(x − y)2 = x 2 − 2x y + y 2
(x + y)3 = x 3 + 3x 2 y + 3x y 2 + y 3
(x − y)3 = x 3 − 3x 2 y + 3x y 2 − y 3
n(n − 1) n−2 2
x
(x + y)n = x n + nx n−1 y +
y
2
n n−k k
+ ··· +
x
y + · · · + nx y n−1 + y n
k
n
n(n − 1) · · · (n − k + 1)
where
=
k
1 · 2 · 3 · ··· · k
Circles
Equation of the circle with center (a, b) and radius r :
(x − a)2 + (y − b)2 = r 2
Distance and Midpoint Formulas
Quadratic Formula
Distance between P1 = (x 1 , y1 ) and P2 = (x 2 , y2 ):
d = (x 2 − x 1 )2 + (y2 − y1 )2
x 1 + x 2 y1 + y2
,
Midpoint of P1 P2 :
2
2
If ax 2 + bx + c = 0, then x =
−b ±
b2 − 4ac
.
2a
Inequalities and Absolute Value
If a < b and b < c, then a < c.
Laws of Exponents
If a < b, then a + c < b + c.
xm
x m x n = x m+n
xn
= x m−n
1
x −n = n
x
(x y)n = x n y n
√
x 1/n = n x
√
n
x m/n =
√
n
xm =
If a < b and c > 0, then ca < cb.
(x m )n = x mn
n
x
xn
= n
y
y
√
n
x
x
n
= √
n y
y
√ √
xy = n x n y
√
m
n
x
If a < b and c < 0, then ca > cb.
|x| = x
|x| = −x
if x ≥ 0
if x ≤ 0
−a
0
a
| x | < a means
−a < x < a.
c−a c c+a
| x − c | < a means
c − a < x < c + a.
GEOMETRY
Formulas for area A, circumference C, and volume V
Cone with
arbitrary base
Triangle
Circle
Sector of Circle
Sphere
Cylinder
Cone
A = 12 bh
A = πr 2
C = 2π r
A = 12 r 2 θ
V = 43 π r 3
V = πr 2 h
s = rθ
(θ in radians)
A = 4π r 2
V = 13 π r 2 h
A = πr r 2 + h 2
= 12 ab sin θ
a
h
q
b
r
r
s
q
r
r
h
h
r
Pythagorean Theorem: For a right triangle with hypotenuse of length c and legs of lengths a and b, c2 = a 2 + b2 .
V = 13 Ah
where A is the
area of the base
h
TRIGONOMETRY
Angle Measurement
π
Fundamental Identities
radians = 180◦
1◦ =
π
rad
180
s = rθ
s
r
q
180◦
1 rad =
r
π
(θ in radians)
hyp
opp
sin θ =
hyp
adj
cos θ =
hyp
sin θ
opp
tan θ =
=
cos θ
adj
cos θ
adj
cot θ =
=
sin θ
opp
1
hyp
=
sec θ =
cos θ
adj
1
hyp
csc θ =
=
sin θ
opp
opp
q
adj
csc θ =
r
y
r
sec θ =
x
x
cot θ =
y
sin θ
=1
θ →0 θ
1 − cos θ
=0
θ
θ →0
(−
(− 23 , 12 )
5p
6
(−1, 0) p
7p
6
− θ = cos θ
2
π
− θ = sin θ
cos
2
π
− θ = cot θ
tan
2
sin(θ + 2π ) = sin θ
cos(θ + 2π ) = cos θ
tan(θ + π ) = tan θ
B
sin B
sin C
sin A
=
=
a
b
c
P = (r cosq, r sinq )
r
a
c
C
q
y
Addition and Subtraction Formulas
x
sin(x + y) = sin x cos y + cos x sin y
sin(x − y) = sin x cos y − cos x sin y
cos(x + y) = cos x cos y − sin x sin y
cos(x − y) = cos x cos y + sin x sin y
( 12 , 23 )
( 22 , 22 )
3
p
60˚ 4
3 1
45˚
p ( 2 , 2)
tan(x + y) =
tan x + tan y
1 − tan x tan y
tan(x − y) =
tan x − tan y
1 + tan x tan y
(0, 1)
p
2
120˚
135˚
150˚
4
180˚
210˚
225˚
5p
240˚
b
A
a 2 = b2 + c2 − 2bc cos A
lim
(− 12 , 23 )
) 3p 2p3
tan(−θ ) = − tan θ
π
The Law of Cosines
y
r
x
cos θ =
r
y
tan θ =
x
2 2
,
2 2
cos(−θ ) = cos θ
1 + cot2 θ = csc2 θ
The Law of Sines
Trigonometric Functions
lim
sin(−θ ) = − sin θ
1 + tan2 θ = sec2 θ
sin
Right Triangle Definitions
sin θ =
sin2 θ + cos2 θ = 1
90˚ p
Double-Angle Formulas
30˚ 6
0˚ 0
360˚ 2p (1, 0)
330˚ 11p
3
1
315˚ 6
,−
2
2
300˚ 7p
sin 2x = 2 sin x cos x
cos 2x = cos2 x − sin2 x = 2 cos2 x − 1 = 1 − 2 sin2 x
(− 23 , − 12 )
(
)
4 4p
5p 4
2
2
2
2
−
−
−
( 2 , 2 ) 3 270˚ 3p2 3 ( 2 , 2 )
1
3
(0, −1) ( 2 , − 2 )
(− 12 , − 23 )
tan 2x =
2 tan x
1 − tan2 x
sin2 x =
1 − cos 2x
2
cos2 x =
Graphs of Trigonometric Functions
y
1
y
y
y = sin x
y = cos x
1
y = tan x
2
x
p
p
2p
−1
2p
p
−2
2p
−1
y
y
y = csc x
1
−1
x
x
p
x
2p
y
y = sec x
2
1
−1
y = cot x
p
x
2p
−2
p
x
2p
1 + cos 2x
2
ELEMENTARY FUNCTIONS
Power Functions f(x) = xa
f (x) = x n , n a positive integer
y
y
y = x4
y = x3
y = x2
(−1, 1)
1
(1, 1)
(−1, −1)
−1
y = x5
2
−1
x
1
−1
−2
x
1
(1, 1)
n odd
n even
Asymptotic behavior of a polynomial function
of even degree and positive leading coefficient
Asymptotic behavior of a polynomial function
of odd degree and positive leading coefficient
y
y
x
x
n even
f (x) = x −n =
1
xn
n odd
y
y
y=
1
x3
y=
y=
1
−1
1
x
x
−1
y=
1
−1
1
1
x4
1
x2
x
−1
1
Inverse Trigonometric Functions
arcsin x = sin−1 x = θ
⇔
sin θ = x,
−
arccos x = cos−1 x = θ
π
π
≤θ≤
2
2
⇔
cos θ = x,
⇔
p
2
−
q = cos−1 x
−1
−1
π
π
<θ<
2
2
q = tan−1 x
p
2
1
−p
tan θ = x,
q
q = sin−1 x
x
−1
0≤θ≤π
q
q
p
2
arctan x = tan−1 x = θ
x
1
−p
2
x
1
Exponential and Logarithmic Functions
loga x = y
loga
(a x )
⇔
ay = x
=x
=x
a loga x
loga 1 = 0
ln x = y
ln(e x )
loga a = 1
⇔
=x
loga (x y) = loga x + loga y
=x
x
loga
= loga x − loga y
y
eln x
ln 1 = 0
y
ey = x
ln e = 1
loga (x r ) = r loga x
y
y=
( 12 )x
ex
4
e−x
( 14 )x
y
10 x 4 x
ex
y = log2 x
2x
y = log x
y = log5 x
y = log10 x
2
y=x
1.5 x
3
1
2
x
1
−1
−1
y = ln x
1
2
3
1
1
2
3
4
5
x
4
−2
lim a x = ∞, a > 1
−1
1
2
x
lim a x = 0, a > 1
x→∞
lim loga x = −∞
x→0+
x→−∞
lim a x = 0, 0 < a < 1
x→∞
lim a x = ∞,
0<a<1
x→−∞
lim loga x = ∞
x→∞
Hyperbolic Functions
e x − e−x
2
x
e + e−x
cosh x =
2
sinh x
tanh x =
cosh x
sinh x =
1
sinh x
1
sech x =
cosh x
cosh x
coth x =
sinh x
y
csch x =
y
y = coth x
3
y = cosh x
1
2
y = tanh x
1
−2
y = sinh x
−1
x
−1
1
2
−2
2
−2
−1
−3
Inverse Hyperbolic Functions
y = sinh−1 x
⇔
sinh y = x
y = cosh−1 x
⇔
cosh y = x and y ≥ 0
y = tanh−1 x
⇔
tanh y = x
x
sinh(x + y) = sinh x cosh y + cosh x sinh y
sinh 2x = 2 sinh x cosh x
cosh(x + y) = cosh x cosh y + sinh x sinh y
cosh 2x = cosh2 x + sinh2 x
sinh−1 x = ln x + x 2 + 1
cosh−1 x = ln x + x 2 − 1 x > 1
1+x
1
tanh−1 x = ln
−1 < x < 1
2
1−x
y
y = sinh−1 x
1
−2
−1
y = cosh−1 x
1
−1
2
x
DIFFERENTIATION
Differentiation Rules
1.
2.
d
(c) = 0
dx
d
x=1
dx
21.
d
1
(tan−1 x) =
dx
1 + x2
22.
d
1
(csc−1 x) = − dx
x x2 − 1
d
1
(sec−1 x) = dx
x x2 − 1
d
1
(cot−1 x) = −
24.
dx
1 + x2
23.
3.
d n
(x ) = nx n−1
dx
4.
d
[c f (x)] = c f (x)
dx
(Power Rule)
Exponential and Logarithmic Functions
d
5.
[ f (x) + g(x)] = f (x) + g (x)
dx
d
[ f (x)g(x)] = f (x)g (x) + g(x) f (x) (Product Rule)
dx
f (x)
d
g(x) f (x) − f (x)g (x)
(Quotient Rule)
7.
=
d x g(x)
[g(x)]2
6.
25.
d x
(e ) = e x
dx
26.
d x
(a ) = (ln a)a x
dx
27.
1
d
ln |x| =
dx
x
1
d
(loga x) =
dx
(ln a)x
8.
d
f (g(x)) = f (g(x))g (x)
dx
(Chain Rule)
28.
9.
d
f (x)n = n f (x)n−1 f (x)
dx
(General Power Rule)
Hyperbolic Functions
10.
d
f (kx + b) = k f (kx + b)
dx
1
where g(x) is the inverse f −1 (x)
f (g(x))
f (x)
d
ln f (x) =
12.
dx
f (x)
11. g (x) =
Trigonometric Functions
29.
d
(sinh x) = cosh x
dx
30.
d
(cosh x) = sinh x
dx
31.
d
(tanh x) = sech2 x
dx
32.
d
(csch x) = − csch x coth x
dx
13.
d
sin x = cos x
dx
33.
d
(sech x) = − sech x tanh x
dx
14.
d
cos x = − sin x
dx
34.
d
(coth x) = − csch2 x
dx
15.
d
tan x = sec2 x
dx
Inverse Hyperbolic Functions
16.
d
csc x = − csc x cot x
dx
35.
17.
d
sec x = sec x tan x
dx
18.
d
cot x = − csc2 x
dx
Inverse Trigonometric Functions
d
1
(sin−1 x) = dx
1 − x2
d
1
(cos−1 x) = − 20.
dx
1 − x2
19.
d
1
(sinh−1 x) = dx
1 + x2
d
1
(cosh−1 x) = 36.
dx
x2 − 1
37.
1
d
(tanh−1 x) =
dx
1 − x2
38.
d
1
(csch−1 x) = − dx
|x| x 2 + 1
39.
d
1
(sech−1 x) = − dx
x 1 − x2
40.
d
1
(coth−1 x) =
dx
1 − x2
INTEGRATION
Substitution
Integration by Parts Formula
If an integrand has the form f (u(x))u (x), then rewrite the entire
integral in terms of u and its differential du = u (x) d x:
f (u(x))u (x) d x =
u(x)v (x) d x = u(x)v(x) −
u (x)v(x) d x
f (u) du
TABLE OF INTEGRALS
Basic Forms
u n du =
1.
2.
u n+1
+ C,
n+1
n = −1
du
= ln |u| + C
u
eu du = eu + C
3.
4.
au
a u du =
+C
ln a
sin u du = − cos u + C
5.
cos u du = sin u + C
6.
sec2 u du = tan u + C
7.
csc2 u du = − cot u + C
8.
sec u tan u du = sec u + C
9.
csc u cot u du = − csc u + C
10.
tan u du = ln |sec u| + C
11.
cot u du = ln |sin u| + C
12.
sec u du = ln |sec u + tan u| + C
13.
csc u du = ln |csc u − cot u| + C
14.
du
u
= sin−1 + C
a
a2 − u 2
du
1
u
16.
= tan−1 + C
a
a
a2 + u 2
15.
Exponential and Logarithmic Forms
1
ueau du = 2 (au − 1)eau + C
a
1
n
18.
u n eau du = u n eau −
u n−1 eau du
a
a
eau
(a sin bu − b cos bu) + C
19.
eau sin bu du = 2
a + b2
eau
20.
eau cos bu du = 2
(a cos bu + b sin bu) + C
a + b2
21.
ln u du = u ln u − u + C
17.
u n ln u du =
22.
23.
u n+1
[(n + 1) ln u − 1] + C
(n + 1)2
1
du = ln |ln u| + C
u ln u
Hyperbolic Forms
sinh u du = cosh u + C
24.
cosh u du = sinh u + C
25.
tanh u du = ln cosh u + C
26.
coth u du = ln |sinh u| + C
27.
28.
29.
sech u du = tan−1 |sinh u| + C
1 csch u du = ln tanh u + C
2
sech2 u du = tanh u + C
30.
csch2 u du = − coth u + C
31.
sech u tanh u du = − sech u + C
32.
csch u coth u du = − csch u + C
33.
Trigonometric Forms
1
1
u − sin 2u + C
2
4
1
1
35.
cos2 u du = u + sin 2u + C
2
4
36.
tan2 u du = tan u − u + C
37.
cot2 u du = − cot u − u + C
1
38.
sin3 u du = − (2 + sin2 u) cos u + C
3
1
3
39.
cos u du = (2 + cos2 u) sin u + C
3
1
40.
tan3 u du = tan2 u + ln |cos u| + C
2
1
3
41.
cot u du = − cot2 u − ln |sin u| + C
2
1
1
42.
sec3 u du = sec u tan u + ln |sec u + tan u| + C
2
2
34.
sin2 u du =
43.
44.
45.
46.
47.
48.
49.
50.
51.
52.
1
1
csc3 u du = − csc u cot u + ln |csc u − cot u| + C
2
2
1
n−1
n
n−1
u cos u +
sinn−2 u du
sin u du = − sin
n
n
1
n−1
cosn−2 u du
cosn u du = cosn−1 u sin u +
2
n
1
tann u du =
tann−1 u − tann−2 u du
n−1
−1
n
cotn−1 u − cotn−2 u du
cot u du =
n−1
1
n−2
tan u secn−2 u +
secn−2 u du
secn u du =
n−1
n−1
−1
n−2
cscn u du =
cot u cscn−2 u +
cscn−2 u du
n−1
n−1
sin(a − b)u
sin(a + b)u
sin au sin bu du =
−
+C
2(a − b)
2(a + b)
sin(a − b)u
sin(a + b)u
cos au cos bu du =
+
+C
2(a − b)
2(a + b)
cos(a − b)u
cos(a + b)u
sin au cos bu du = −
−
+C
2(a − b)
2(a + b)
u sin u du = sin u − u cos u + C
53.
Forms Involving
67.
68.
69.
70.
71.
72.
73.
74.
75.
u cos u du = cos u + u sin u + C
55.
u n sin u du = −u n cos u + n u n−1 cos u du
56.
u n cos u du = u n sin u − n u n−1 sin u du
57.
sinn u cosm u du
sinn−1 u cosm+1 u
n−1
=−
+
sinn−2 u cosm u du
n+m
n+m
sinn+1 u cosm−1 u
m −1
=
+
sinn u cosm−2 u du
n+m
n+m
54.
Inverse Trigonometric Forms
58.
59.
sin−1 u du = u sin−1 u +
1 − u2 + C
cos−1 u du = u cos−1 u −
1 − u2 + C
1
ln(1 + u 2 ) + C
2
2u 2 − 1 −1
u 1 − u2
61.
u sin−1 u du =
sin u +
+C
4
4
2u 2 − 1
u 1 − u2
62.
u cos−1 u du =
cos−1 u −
+C
4
4
u2 + 1
u
63.
u tan−1 u du =
tan−1 u − + C
2
2
n+1
1
u
du
64.
u n sin−1 u du =
u n+1 sin−1 u −
, n = −1
n+1
1 − u2
n+1
du
1
u
n
−1
n+1
−1
cos u +
u
, n = −1
65.
u cos u du =
n+1
1 − u2
n+1
du
1
u
u n+1 tan−1 u −
, n = −1
66.
u n tan−1 u du =
n+1
1 + u2
60.
tan−1 u du = u tan−1 u −
√
a2 − u2 , a > 0
u
u 2
a2
a 2 − u 2 du =
a − u2 +
sin−1 + C
2
2
a
u
u
a4
2
2
2
2
2
2
2
u a − u du = (2u − a ) a − u +
sin−1 + C
8
8
a
2
2
2
2
a −u
a + a − u du = a 2 − u 2 − a ln +C
u
u
2
1 2
a − u2
u
du = −
a − u 2 − sin−1 + C
2
u
a
u
u 2
a2
u
u 2 du
−1
2
=−
a −u +
sin
+C
2
2
a
a2 − u 2
du
1 a + a 2 − u 2 = − ln +C
a u
u a2 − u 2
du
1 2
=− 2
a − u2 + C
a u
u 2 a2 − u 2
u
3a 4
u
(a 2 − u 2 )3/2 du = − (2u 2 − 5a 2 ) a 2 − u 2 +
sin−1 + C
8
8
a
du
u
=
+
C
(a 2 − u 2 )3/2
a2 a2 − u 2
Forms Involving
√
u2 − a2 , a > 0
u 2
a 2 u 2 − a 2 du =
u − a2 −
ln u + u 2 − a 2 + C
2
2
77.
u 2 u 2 − a 2 du
u
a 4 = (2u 2 − a 2 ) u 2 − a 2 −
ln u + u 2 − a 2 + C
8
8
2
u − a2
a
du = u 2 − a 2 − a cos−1
+C
78.
u
|u|
u 2 − a2
u 2 − a2
du = −
+ ln u + u 2 − a 2 + C
79.
u
u
du
80.
= ln u + u 2 − a 2 + C
u 2 − a2
u 2 du
u 2
a 2 81.
=
u − a2 +
ln u + u 2 − a 2 + C
2
2
u 2 − a2
du
u 2 − a2
82.
+C
=
2
2
2
a2 u
u u −a
du
u
83.
=− +C
2
(u 2 − a 2 )3/2
a u 2 − a2
76.
Forms Involving
a2 + u2 , a > 0
u 2
a2 a 2 + u 2 du =
a + u2 +
ln u + a 2 + u 2 + C
2
2
85.
u 2 a 2 + u 2 du
u
a4 = (a 2 + 2u 2 ) a 2 + u 2 −
ln u + a 2 + u 2 + C
8
8
2
a + a2 + u 2 a + u2
2
2
86.
du = a + u − a ln +C
u
u
2
2
2
2
a +u
a +u
87.
du = −
+ ln u + a 2 + u 2 + C
u
u2
84.
88.
89.
du
a2 + u 2
u 2 du
a2 + u 2
= ln u + a 2 + u 2 + C
=
u
2
a2 + u 2 −
a2
2
ln u +
101.
du
1 a 2 + u 2 + a 90.
= − ln +C
a u
u a2 + u 2
du
a2 + u 2
91.
=−
+C
2
2
2
a2 u
u a +u
u
du
= +C
92.
(a 2 + u 2 )3/2
a2 a2 + u 2
93.
102.
103.
104.
Forms Involving a + bu
=
a2 + u 2 + C
105.
u du
1 = 2 a + bu − a ln |a + bu| + C
a + bu
b
u 2 du
1 = 3 (a + bu)2 − 4a(a + bu) + 2a 2 ln |a + bu| + C
a + bu
2b
du
1 u 95.
= ln +C
u(a + bu)
a
a + bu a + bu b
1
du
+C
ln
+
=
−
96.
au
u 2 (a + bu)
a2 u u du
a
1
97.
= 2
+ 2 ln |a + bu| + C
(a + bu)2
b (a + bu)
b
a + bu 1
1
du
+C
=
ln
−
98.
a(a + bu)
u(a + bu)2
a2 u u 2 du
1
a2
99.
= 3 a + bu −
− 2a ln |a + bu| + C
a + bu
(a + bu)2
b
√
2
(3bu − 2a)(a + bu)3/2 + C
100.
u a + bu du =
15b2
√
u n a + bu du
106.
94.
107.
√
2
u n (a + bu)3/2 − na u n−1 a + bu du
b(2n + 3)
√
u du
2
= 2 (bu − 2a) a + bu + C
√
3b
a + bu
√
n−1
2u n a + bu
du
u n du
2na
u
=
−
√
√
b(2n + 1)
b(2n + 1)
a + bu
a + bu
√
a + bu − √a 1
du
= √ ln √
√
√ + C, if a > 0
a a + bu + a u a + bu
a + bu
2
tan−1
+ C,
if a < 0
= √
−a
−a
√
a + bu
b(2n − 3)
du
du
=−
−
√
√
n−1
n
2a(n − 1)
a(n − 1)u
u a + bu
u n−1 a + bu
√
√
du
a + bu
du = 2 a + bu + a
√
u
u a + bu
√
√
a + bu
a + bu
b
du
du
=
−
+
√
u
2
u2
u a + bu
Forms Involving
108.
109.
√
2au − u2 , a > 0
2au − u 2 du =
u −a
a2
a−u
2au − u 2 +
cos−1
+C
2
2
a
u 2au − u 2 du
2u 2 − au − 3a 2 a3
2au − u 2 +
cos−1
6
2
a−u
du
= cos−1
+C
110.
a
2au − u 2
2au − u 2
du
=−
+C
111.
2
au
u 2au − u
=
a−u
a
+C
ESSENTIAL THEOREMS
Intermediate Value Theorem
The Fundamental Theorem of Calculus, Part I
If f (x) is continuous on a closed interval [a, b] and f (a) = f (b),
then for every value M between f (a) and f (b), there exists at least
one value c ∈ (a, b) such that f (c) = M.
Assume that f (x) is continuous on [a, b] and let F(x) be an
antiderivative of f (x) on [a, b]. Then
b
Mean Value Theorem
If f (x) is continuous on a closed interval [a, b] and differentiable on
(a, b), then there exists at least one value c ∈ (a, b) such that
f (b) − f (a)
f (c) =
b−a
Extreme Values on a Closed Interval
If f (x) is continuous on a closed interval [a, b], then f (x) attains
both a minimum and a maximum value on [a, b]. Furthermore, if
c ∈ [a, b] and f (c) is an extreme value (min or max), then c is either
a critical point or one of the endpoints a or b.
a
f (x) d x = F(b) − F(a)
Fundamental Theorem of Calculus, Part II
Assume that f (x)is a continuous function on [a, b]. Then the area
function A(x) =
x
a
A (x) = f (x)
f (t) dt is an antiderivative of f (x), that is,
or equivalently
x
d
f (t) dt = f (x)
dx a
Furthermore, A(x) satisfies the initial condition A(a) = 0.

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