Math 141 Equation Sheet
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. www.prep101.com 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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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
We’ve helped over 50,000 students get better grades since 1999!
Need help for exams?
Check out our classroom prep sessions - customized to your exact course - at www.prep101.com
(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. www.prep101.com 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
Our Course Booklets - free at prep sessions - are the “Perfect Study Guides.”
