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Calculus 2 Reviews
Calculus 2 Unit 8 (Test 3)
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Cards (18)
If f' is
continuous
on [a,b] , then the
length
of the curve y =
f(x)
is given by
L =
int
from
a to b sqrt
(
1 + (f'(x)
)^
2
)dx
or
L =
int from a to b sqrt
(
1 + (dy/dx)^2
)
dx
Steps to find
arc length
Find the
derivative
of
f(x)
and
square
it
Fill it into the
arc length equation
Do a
u-sub
of the
inner part
of the square
root
,
integrate
by
parts
, or any of the
integration techniques
Evaluate
it at the
bounds
of the
problem
and
solve
Integration by parts formula
int
(
udv
) = uv -
int
(
vdu
)
∫
u
d
v
=
\int_{ }^{ }udv\ =
∫
u
d
v
=
u
v
−
∫
v
d
u
\ uv\ -\ \int_{ }^{ }vdu
uv
−
∫
v
d
u
Trigonometric Substitution, sqrt (
a^2
-
x^2
)
Side,
sin(theta) = x/a
-> x =
a sin(theta)
Use:
1
-
sin^2(theta) = cos^2(theta)
Trigonometric Substitution, sqrt (a^2 + x^2)
Hypotenuse
,
tan(theta) = x/a
-> x=
a tan(theta)
Use:
1
+
tan^2(theta) = sec^2(theta)
Trigonometric Substitution, sqrt (
x^2
- a^2)
Side
,
sec
(
theta
) =
x
/
a
-> x=
a sec(theta)
Use:
sec
^2(theta) - 1 =
tan^2(theta)
sine
=
opp
/
hypo
cosine =
adj
/
hypo
tangent
=
opp
/
adj
sec
=
hypo
/
adj
csc =
hypo
/
opp
cot
=
adj
/
opp
If a curve has the equation x = g(y); c ≤ y ≤ d, & g'(y) is
continuous
, then:
L =
int
from
c
to d
sqrt
(
1
+
(g'(y))^2
) dy
or
L =
int
from
c
to d
sqrt
(
1
+
(dx/dy)^2
) dy
* usually used when the original f(x)'s derivative is not defined at the integral
tan^2(x)
+ 1
=
sec^2(x)
sin^2(x)
=
1
- cos^2(x)
Area
of a
Surface
of
Revolution-
About the
x-axis
S
=
S\ =
S
=
∫
a
b
2
π
f
(
x
)
1
+
(
d
y
d
x
)
2
d
x
\ \int_{a}^{b}2\pi f\left(x\right)\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx
∫
a
b
2
π
f
(
x
)
1
+
(
d
x
d
y
)
2
d
x
S
=
S\ =
S
=
∫
c
d
2
π
y
1
+
(
d
x
d
y
)
2
d
y
\ \int_{c}^{d}2\pi y\sqrt{1+\left(\frac{dx}{dy}\right)^{2}}dy
∫
c
d
2
π
y
1
+
(
d
y
d
x
)
2
d
y
Area
of a Surface of Revolution- About the
y-axis
S
=
S\ =
S
=
∫
a
b
2
π
x
1
+
(
d
y
d
x
)
2
d
x
\ \int_{a}^{b}2\pi x\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx
∫
a
b
2
π
x
1
+
(
d
x
d
y
)
2
d
x
S
=
S\ =
S
=
∫
c
d
2
π
g
(
y
)
1
+
(
d
x
d
y
)
2
d
y
\ \int_{c}^{d}2\pi g\left(y\right)\sqrt{1+\left(\frac{dx}{dy}\right)^{2}}dy
∫
c
d
2
π
g
(
y
)
1
+
(
d
y
d
x
)
2
d
y
Steps to find the area of a surface of revolution
Find the
derivative
of
f(x)
and
square
it
Fill it into the
arc length
part of the equation and combine it with the
1
( use
LCD
)
Fill it into the
formula
,
simplify
, and
evaluate
the
integral
Rotate around the x-axis = use
'y'
Rotate around the y-axis = use
'x'
Arc Length Formula
( in terms of x)
A
=
A\ =
A
=
∫
a
b
1
+
(
f
′
(
x
)
)
2
d
x
\ \int_{a}^{b}\sqrt{1+\left(f'\left(x\right)\right)^{2\ }}dx
∫
a
b
1
+
(
f
′
(
x
)
)
2
d
x
Arc Length Formula
( in terms of y)
A
=
A\ =
A
=
∫
c
d
1
+
(
f
′
(
y
)
)
2
d
y
\ \int_{c}^{d}\sqrt{1+\left(f'\left(y\right)\right)^{2\ }}dy
∫
c
d
1
+
(
f
′
(
y
)
)
2
d
y
a
^
2
+
2ab
+ b^
2
= (a + b)^2
Factoring formula
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