Math 141 Equation Sheet
MATH 141 EQUATION SHEET
•
SIMPLE INTEGRATION FORMULAS
•
∫ Kdx = Kx + C ( K and C are constants)
•
x n+1
=
+C (n
x
dx
∫
n +1
•
∫e
•
∫ x dx = ln x + C
n
x
≠ −1 )
cosines (or sines) and solve using
substitution.
If both m and n are even we use the
formulas
1 − cos 2 x
1 + cos 2 x and
sin 2 x =
cos 2 x =
2
2
Second case:
•
dx = e x + C
1
•
SUBSTITUTION
∫ f ′(u ( x)) ⋅ u ′( x)dx = f (u ( x)) + C
INTEGRATION BY PARTS
∫ udv = uv − ∫ vdu
FUNDAMENTAL THEOREM OF CALCULUS PART I
u ( x)
d
f (t )dt = f (u ( x)) ⋅ u ′( x)
dx ∫a
FUNDAMENTAL THEOREM OF CALCULUS PART II
∫ tan
m
x sec n xdx
2
If n is even we isolate sec x we then
change the remainder of the secants to
tangents using the formula
sec 2 x = 1 + tan 2 x and use substitution
( u = tan x )
If both n and m are odd we isolate
sec x tan x then change the
remainder of the tangents into secants
using the formula tan 2 x = sec 2 x − 1
and use substitution ( u = sec x )
Note that in the second case the case where
n is odd and m even is not covered, this case
is dealt with case by case.
TRIGONOMETRIC SUBSTITUTIONS
b
•
For x 2 − a 2 we use the substitution
a
•
For x 2 + a 2 we use the substitution
∫ f ( x)dx = F (b) − F (a)
where F ( x ) is an antiderivative of
f (x )
•
IMPROPER INTEGRALS
An integral
b
∫ f ( x)dx
is an improper integral if it
a
satisfies at least one of the two following
conditions:
•
a or b are − ∞ or ∞
•
f is undefined for any point in the interval [a,
b]
In the case of an improper integral we replace a
or b (or the point in between, or both) with a
letter, evaluate the integral then take the
appropriate limit. If we get a finite answer the
integral is called convergent otherwise it’s a
divergent integral.
∫ tan
∫
m
x sec n xdx .
First case:
•
∫ sin
m
x cos n xdx
If m or n is odd we isolate
cos x
or
x = a ⋅ tan θ
For
a2 − x2
x = a ⋅ sinθ
AREA BETWEEN TWO CURVES
The area between two curves is given by a
definite integral. Say the area in question is
between
to:
we use the substitution
x = b . It’s value is equal
∫ " higher curve" − " lower curve" dx
A
+ ...
( x − a)
For each factor of ( x 2 + a ) we put
Ax + B
+ ...
( x 2 + a)
For each factor
.
a
VOLUMES
The main formula for finding the volume
generated by the revolution of an area over an
b
axis is:
π ∫ R2 − r 2d ?
a
If the axis is parallel to the x-axis then it’s a dx, if
the axis is parallel to the y-axis it’s a dy.
AVERAGE VALUE OF A FUNCTION OVER AN
INTERVAL
interval
[ a, b ]
is given by 1
f (x )
b
b − a ∫a
f ( x)dx
over the
.
LENGTH OF A CURVE
f
be a function continuous on [a, b]
differentiable on (a, b). The arc length of the
curve from x = a to x = b is given by
Once we have the denominator factored (and
of course the degree of the numerator is less
than the degree of the denominator AND
substitution failed) we do the following:
•
For each factor of ( x − a ) we put
•
and
By higher curve I mean the one more positive.
Let
•
x=a
b
The average value of a function
PARTIAL FRACTIONS
TRIGONOMETRIC INTEGRALS
In this part we turn out attention to integrals of
two forms sin m x cos n xdx and
x = a ⋅ secθ
Once we have the partial fractions we proceed to
the common denominator, then to finding the
values of the constants and finally we compute
the integral of each of the partial fractions we
found.
(L) n
we put a partial
fraction for every power going from 1 to
n
Ex: For ( x 2 +3) 3 we put the partial fractions
Ax + B
Cx + D
Ex + F
+
+
( x 2 + 3) ( x 2 + 3) 2 ( x 2 + 3) 3
b
L = ∫ 1 + ( f ′( x)) 2 dx
a
If the roles of x and y are reversed we use the
formula L =
d
∫
1 + ( g ′( y )) 2 dy
c
AREA OF A SURFACE OF REVOLUTION
Let
f
be a function continuous on [a, b]
differentiable on (a, b) such that f ( x ) ≥ 0 on
[a, b]. The surface area S of the surface of
revolution generated by revolving the graph of f
on [a, b] about the x-axis is given by
b
S = 2π ∫ f ( x) 1 + ( f ′( x)) 2 dx
a
sin x , we change the remaining of the
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(If the function g ( y ) on [c, d] is revolving about
the y-axis then the surface area S of the surface
of revolution is given
by
S = 2π
d
∫ g ( y)
1 + ( g ′( y )) dy
2
)
SEQUENCES AND SERIES
•
Given a series
The Integral test
the series
c
∞
∑a
n =1
PARAMETRIC EQUATIONS
∞
FIRST DERIVATIVE
1
∫
1.
and the improper integral
n
n →∞
SECOND DERIVATIVE
The Comparison Test
∑
convergent.
If
an ≥
∑
is given by L =
∑
t2
∫
t1
2
⎛ dx ⎞ ⎛ dy ⎞
⎜ ⎟ +⎜ ⎟
⎝ dt ⎠ ⎝ dt ⎠
n →∞
∑a
•
and
∑a
bn
n
n is
∑
x = r cosθ , y = r sin θ
The Root Test
1.
n is divergent.
∑b
n
∑a
n
, If
lim n a n < 1 , the series is absolutely
n→∞
convergent.
and
2.
lim n a n > 1 , the series is divergent.
3.
lim n a n = 1 , the test is inconclusive
n→∞
n→∞
The Alternating Series Test
series of the form
∞
⎛ y⎞
r = x 2 + y 2 ,θ = tan −1 ⎜ ⎟
⎝x⎠
AREA OF THE REGION BETWEEN THE ORIGIN AND
THE CURVE
∞
∑ (−1)
n
β
1
A = ∫ r 2 dθ
α 2
∑ (−1)
n =1
n
an .
a n is convergent if:
1.
lim a n = 0 and,
2.
a n +1 < a n
n→∞
for all n. (The series is
decreasing)
•
Absolute and Conditional
Convergence
Given a series
∑a
n it is either
∑a
LENGTH OF CURVE
1.
absolutely convergent if
If r = f (θ ) has a continuous first derivative for
α ≤ θ ≤ β and if the point P (r ,θ )
2.
convergent
conditionally convergent if
∑a
∑a
n
is
n is convergent but
traces the curve exactly once as θ runs from
α to β then the length of the curve is given by
β
•
a n+1
= 1 , the test is inconclusive.
n →∞ a
n
lim
Given a series
bn is
both converge or both diverge
n =1
From rectangular to polar coordinates:
∫
α
3.
n
We apply this test to alternating series i.e.
From polar to rectangular coordinates:
r = f (θ ), α ≤ θ ≤ β
n
∑a
If lim a n = c > 0 then
3.
2
POLAR COORDINATES
∑b
divergent then
The length or the arc of the curve between two
t2
∑
convergent then
points corresponding to parameter values t1 and
a n+1
< 1 , the series is absolutely
n →∞ a
n
lim
a n+1
We apply this test to series of positive terms.
The comparison test can be applied in 3
forms:
1.
If
an ≤ bn and bn is
ARC LENGTH OF A PARAMETRIC CURVE
, If
2.
lim
> 1 , the series is divergent.
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a
•
2.
n
convergent.
f ( x)dx both converge or both diverge.
dy
dy
= dt
dx dx
dt
⎛ d ⎛ dy ⎞ ⎞
⎜ ⎜ ⎟⎟
d 2 y ⎜⎝ dt ⎝ dx ⎠ ⎟⎠
=
2
dx
dx
dt
∑a
3.
2
⎛ dr ⎞
r2 + ⎜
⎟ dθ
⎝ dθ ⎠
•
n is divergent
divergent if
∑a
n is divergent.
RATIO TEST
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