Font size
WorksheetsBC Calculus Memorization
Total questions: 115
Worksheet time: 29hrs 45mins
ln1
0
1
e
DNE
ln e
1
0
e
DNE
e≈
3.142
2.718
1.414
1.732
Area of a Trapezoid
A=2bh
A=2(b1+b2)h
A=h(b1+b2)2
A=2(b1+b2)
0k
Undefined
Indeterminate form
0
k0
Undefined
Indeterminate form
0
Rewrite:
bxa
xba
xab
x−ba
x−ab
cos 0
0
1
−1
22
cos π
0
1
−1
22
sin π
0
1
−1
22
cos 2π
0
1
−1
22
cos 23π
0
1
−1
22
sin 2π
0
1
−1
22
sin 23π
0
1
−1
22
sin 0
0
1
−1
22
sin 2π
0
1
−1
22
cos 2π
0
1
−1
22
cos 6π
22
23
21
1
sin 6π
22
23
21
1
sin 4π
22
23
21
1
cos 4π
22
23
21
1
cos 3π
22
23
21
1
sin 3π
22
23
21
1
sec θ
sin θ1
cos θ1
cos θsin θ
sin θcos θ
csc θ
sin θ1
cos θ1
cos θsin θ
sin θcos θ
tan θ
sin θ1
cos θ1
cos θsin θ
sin θcos θ
x→alimf(x) exists
Plug in a
x→a+ limf(x)=x→a− limf(x)
x→alimf(x)=f(a)
f(a) exists
Condition for IVT
f(x) is continuous on [a, b]
f(x) is differentiable on (a, b)
x→alimf(x) exists
x→a+ limf(x)=x→a− limf(x)
Analysis for IVT
There is a c, a < c < b, such that
f(a)<f(c)<f(b)
There is a c, a < c < b, such that
f′(c)=b−af(b)−f(a)
x→alimf(x) exists
f(a) exists
x→alimf(x)=f(a)
x→a+ limf(x)=x→a− limf(x)
Definition of Continuity
There is a c, a < c < b, such that
f(a)<f(c)<f(b)
There is a c, a < c < b, such that
f′(c)=b−af(b)−f(a)
x→alimf(x) exists
f(a) exists
x→alimf(x)=f(a)
x→a+ limf(x)=x→a− limf(x)
Average Rate of Change
b−af(b)−f(a)
f(b)−f(a)b−a
2f(b)+f(a)
b−a1∫abf(x)dx
Instantaneous Rate of Change
b−af(b)−f(a)
f(b)−f(a)b−a
Integral
Derivative
Derivative Fails to Exist
cusp, sharp turn
discontinuity
vertical tangent line
smooth curve
continuous
horizontal tangent line
cusp, sharp turn
discontinuity
horizontal tangent line
smooth curve
continuous
vertical tangent line
Vertical Tangent Line
Derivative is undefined
f′(a)=0
Derivative exists
Limit exists
Horizontal Tangent Line
Derivative is undefined
f′(a)=0
Derivative exists
Limit exists
Critical Numbers
Numbers that tell you what you are doing wrong
f′(a)=0
or undefined
f(a)=0
or undefined
f′(a)=0
f is increasing
f′>0
f′<0
f′′>0
f′′<0
f is decreasing
f′>0
f′<0
f′′>0
f′′<0
f is concave up
f′ increasing
f′′<0
f′ decreasing
f′′>0
f′ increasing
f′′>0
f′ decreasing
f′′<0
f is concave down
f′ increasing
f′′<0
f′ decreasing
f′′>0
f′ increasing
f′′>0
f′ decreasing
f′′<0
f has a relative maximum
f′=0 or undef.
and changes
+ to −
f′=0 or undef.
and changes
− to +
f′′=0 or undef.
and changes
+ to −
f′′=0 or undef.
and changes
− to +
f has a relative minimum
f′=0 or undef.
and changes
+ to −
f′=0 or undef.
and changes
− to +
f′′=0 or undef.
and changes
+ to −
f′′=0 or undef.
and changes
− to +
f has an inflection point
f′ has a
relative max or min
f′′ has a
relative max or min
f′′=0 or undef.
and changes
signs
f′=0 or undef.
and changes
signs
dxdtan x
sec2x
−sec2x
sec x ⋅tan x
csc x ⋅tan x
dxdcot x
csc2x
−csc2x
sec x ⋅cot x
−csc x ⋅cot x
dxdsec x
tan2x
sec x⋅cot x
sec x ⋅tan x
csc x ⋅tan x
dxdcsc x
cot2x
−csc x⋅cot x
csc x ⋅tan x
−csc x ⋅tan x
dxdcsc x
cot2x
−csc x⋅cot x
csc x ⋅tan x
−csc x ⋅tan x
∫abf′(x)dx
f(b)−f(a)
f′(b)−f′(a)
f(a)−f(b)
f′(a)−f′(b)
Average Value
b−a1∫abf(x)dx
b−af(b)−f(a)
2f(b)+f(a)
a−b1∫abf(x)dx
Displacement
∫abv(t)dt
∫aba(t)dt≈
∫ab∣v(t)∣dt
x(a)+∫abv(t)dt
Total Distance Traveled
∫v(t)dt
∫a(t)dt
∫ab∣v(t)∣dt
x(a)+∫abv(t)dt
∫ x1 dx
x1+C
ln x+C
ln∣x∣+C
∫ secxtanx dx
sec x+C
tan x+C
−sec x+C
tan2x+C
∫ sec2x dx
sec x+C
tan x+C
sec xtanx+C
tan2x+C
∫ cscxcotx dx
−csc x+C
−cot x+C
csc x+C
cot2x+C
∫ csc2x dx
−csc x+C
−cot x+C
csc xcotx+C
cot2x+C
dxd∫axf(t)dt
f(x)
f′(x)
f(a)
0
dxd∫ag(x)f(t)dt
f(g(x))g′(x)
f′(g(x))
f(g(x))
f′(g(x))g′(x)
∫axdx
∫lnaax+C
lnaax+C
lnaax
ax+C
∫a2−u21du
sin−1au+C
a1tan−1au+C
a1sec−1a∣u∣+C
∫u2+a21du
sin−1au+C
a1tan−1au+C
a1sec−1a∣u∣+C
∫uu2−a21du
sin−1au+C
a1tan−1au+C
a1sec−1a∣u∣+C
Exponential Growth:
E(x)=Cekt
E(x)=Cerkt
E(x)=ekt
E(x)=ekt+C
Area between two curves:
∫abf(x)−g(x)dx
∫abf(x)+g(x)dx
∫abf(x)g(x)dx
∫abf(x)dx
∫aaf(x)dx
0
1
-1
f(a)
Which of these is the definition of a derivative?
h→0lim hf(x+h)−f(x)
h→0lim hf(h)−f(x)
h→0lim hf(x+h)+f(x)
h→0lim hf(h)+f(x)
The intermediate value theorem (IVT) is primarily concerned with which of the following?
y-values
first derivative values
second derivative values
x-values
Which of these sums up the Mean Value Theorem (MVT)?
f′(c)=b−af(b)−f(a)
f(c)=b−af(b)−f(a)
f(c)=b−af′(b)−f′(a)
f′(c)=b−af′(b)−f′(a)
Volume using discs revolving around horizontal line.
π∫x=ax=b(top −bottom)2dx
π∫x=ax=b(top −bottom)dx
∫x=ax=b(top −bottom)2dx
∫x=ax=b(top −bottom)dx
Volume using discs revolving around vertical line.
π∫y=ay=b(right −left)2dy
π∫y=ay=b(right −left)dy
∫y=ay=b(right −left)2dy
∫y=ay=b(right −left)dy
Volume using washers revolving around horizontal line.
π∫x=ax=bR2−r2 dx
π∫x=ax=b(R−r)2 dx
∫x=ax=b(R−r)2dx
∫x=ax=bR2 −r2 dx
Volume using washers revolving around vertical line.
π∫y=ay=bR2−r2 dy
π∫y=ay=b(R−r)2dy
∫y=ay=b(R−r)2dy
∫y=ay=bR2−r2 dy
What does the Extreme Value Theorem remind us to do?
Check the endpoints! They might be the max/min.
Check the inflection points! They might be the max/min.
Check the endpoints! They give the secant slope.
Check the inflection points! They are when the derivative is 0.
What is the formula for areaof a circle given the radius?
A = πr
A = 2πr
A = πd
A = πr²
According to the quotient rule: dxdg(x)f(x) =
(g(x))2f′(x)g(x)−g′(x)f(x)
(g(x))2g′(x)f(x)−f′(x)g(x)
(g(x))2f′(x)g(x)+g′(x)f(x)
g′(x)f′(x)
In general, what is the formula for the derivative of an exponential function?
dxdax =
x⋅ax−1
axlna
ax
axlne
In general, what is the derivative for a logarithmic function?
dxdlogax=
xlna1
xlne1
x1
alnx1
V=∫ab(top−bottom)2dx
Square cross section volume
Equilateral triangle cross section volume
Semi circle cross section volume
Isosceles right triangle cross section volume
V=8π∫ab(top−bottom)2dx
Square cross section volume
Equilateral triangle cross section volume
Semi circle cross section volume
Isosceles right triangle cross section volume
V=43∫ab(top−bottom)2dx
Square cross section volume
Equilateral triangle cross section volume
Semi circle cross section volume
Isosceles right triangle cross section volume
V=21∫ab(top−bottom)2dx
Isosceles right triangle with hypotenuse as base cross section volume
Equilateral triangle cross section volume
Semi circle cross section volume
Isosceles right triangle with leg as base cross section volume
V=41∫ab(top−bottom)2dx
Isosceles right triangle with hypotenuse as base cross section volume
Equilateral triangle cross section volume
Semi circle cross section volume
Isosceles right triangle with leg as base cross section volume
cos 2 x + sin 2 x=
Move the equations to the correct spots.
Arc length from x=a to x=b
∫ab(1+(f′(x))2)dx
∫ab(1+f′(x))dx
∫ab(f′(x))2dx
∫ab(1+(f(x))2)dx
Arc length for parametric and vector equations
∫t1t2(dtdx)2+(dtdy)2dt
∫t1t2(dtdx)+(dtdy)dt
∫t1t2(dtdx)2−(dtdy)2dt
∫t1t2(dtdx)−(dtdy)dt
Total distance for vectors
∫t1t2(dtdx)2+(dtdy)2dt
∫t1t2(dtdx)+(dtdy)dt
∫t1t2(dtdx)2−(dtdy)2dt
∫t1t2(dtdx)−(dtdy)dt
Speed
∣v(t)∣
(dtdx)2+(dtdy)2
(dtdx)+(dtdy)
∫t1t2(dtdx)2+(dtdy)2dt
(dtdx)2−(dtdy)2
Match the following
+ velocity & + acc.
- velocity & + acc.
- velocity & - acc.
+ velocity & - acc.
Polar area
21∫θ1θ2r2dθ
∫θ1θ2r2dθ
21∫θ1θ2rdθ
41∫θ1θ2r2dθ
Parametric 1st Derivative
dxdy=
dtdxdtdy
dtdydtdx
Parametric 2nd Derivative
dx2d2y=
dtdxdtd(dxdy)
dtdydtd(dxdy)
dtd(dxdy)
dtdx⋅dtdydtd(dxdy)
Polar conversion:
x=
rcosθ
rsinθ
rtanθ
Polar conversion:
y=
rcosθ
rsinθ
rtanθ
Using the nth term test
If n→∞liman=0 then the series
converges
diverges
the test is inconclusive - pick a different one
Using the nth term test
If n→∞liman=0 then the series
converges
diverges
the test is inconclusive - pick a different one
A p-series where p>1 then the series will...
converge
diverge
the test is inconclusive - pick a different one
A p-series where 0<p≤1 then the series will...
converge
diverge
the test is inconclusive - pick a different one
The harmonic series
n=1∑∞n1
always...
diverges
converges
In a geometric series, if ∣r∣≥1 then the series will...
diverge
converge
test is inconclusive, pick a different one
In a geometric series, if ∣r∣<1 then the series will...
diverge
converge
test is inconclusive, pick a different one
A convergent geometric series, will converge to...
1−r1st term
r−11st term
1+r1st term
What are the requirements for the integral test to be applied:
(a) (b) (c)
PUT your answers in ALPHABETICAL ORDER
If an=f(n) is continuous, positive, and decreasing, and ∫1∞f(n)dn converges then n=1∑∞an will...
converge
diverge
test is inconclusive - pick a different one!
If an=f(n) is continuous, positive, and decreasing, and ∫1∞f(n)dn=±∞ then n=1∑∞an will...
converge
diverge
test is inconclusive - pick a different one!
Comparison test:
Let 0<an≤bn for all n.
If n=1∑∞bn converges, then n=1∑∞an (a)
If n=1∑∞an diverges, then n=1∑∞bn (b)
If n=1∑∞bn diverges, then (c)
If n=1∑∞an converges, then (d)
n=1∑∞an is unknown
n=1∑∞bn is unknown
In the limit comparison test, if n→∞lim(bnan)=a finite, positive number then....
an and bn either both converge or both diverge
an and bn both diverge
an and bn both converge
the test is inconclusive - pick a different oen!
For an alternating series, select the TWO requirements in order to prove convergence by the Alternating Series Test. (assume the series is alternating)
n→∞liman=0
n→∞liman=0
ana(n+1)<1
(series is decreasing)
ana(n+1)>1
(series is increasing)
n!(n+1)!=
1
n
n+1
n!
(n+1)!=
1
n!+1!
(n+1)n!
n!
Based on the Ratio Test for Convergence, if n=1∑∞an has positive terms and n→∞lim(ana(n+1))<1 then the series...
converges
diverges
test is inconclusive - pick a different one!
Based on the Ratio Test for Convergence, if n=1∑∞an has positive terms and n→∞lim(ana(n+1))>1 then the series...
converges
diverges
test is inconclusive - pick a different one!
Based on the Ratio Test for Convergence, if n=1∑∞an has positive terms and n→∞lim(ana(n+1))=1 then the series...
converges
diverges
test is inconclusive - pick a different one!
