Calculus Tables

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
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
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
1◦ =
π
180
s = rθ
s
r
q
180◦
r
π
hyp
opp
sin θ =
hyp
cos θ =
hyp
sin θ
opp
tan θ =
=
cos θ
cos θ
cot θ =
=
sin θ
opp
1
hyp
=
sec θ =
cos θ
1
hyp
csc θ =
=
sin θ
opp
opp
q
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
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.
```

Plus en détail

Plus en détail

Plus en détail

MATH 141 Fundamental Theorem of Calculus (FTC) Let f : [a, b] → R

MATH 141 Fundamental Theorem of Calculus (FTC)

Plus en détail

Plus en détail

FORMULAS

secn2 u tan u n  2

Plus en détail

Plus en détail

Integral Formulas 1. ∫ u du = 1 n + 1 u + c if n

(10) 1. Carbon extracted from a tooth found at an archaeological dig contains 15 as much radioactive as is found in teeth of living creatures. The half-life of 14C is 5700 years. Even though you do...

Plus en détail

DERIVATIVES AND INTEGRALS

DERIVATIVES AND INTEGRALS Basic Differentiation Rules

Plus en détail

Plus en détail

Plus en détail

Algebra Geometry - Stewart Calculus

y csch u du  ln  tanh u   C 109. y sech u du  tanh u  C 110. y csch u du  coth u  C 111. y sech u tanh u du  sech u  C 112. y csch u coth u du  csch u  C

Plus en détail

OFFRES !!! ANNÉE 2016 PNEUS D`ETÉ MARCA DIMENSIÓ

BRAVURIS 3 RSP3 CSC5 PRIMACY3 BRAVURIS 3 RSP3 CPC5 PRIMACY3/SAVER+ BRAVURIS 3 RSP3 CPC5 PRIMACY3 BRAVURIS 2

Plus en détail

Plus en détail

Plus en détail

Plus en détail

Plus en détail

Plus en détail

Plus en détail

Plus en détail

Plus en détail

Plus en détail