Microsoft Word - Calculus 2 Formula Cheat Sheet

 

Calculus II Final Exam Cheat Sheet L’Hospital’s Rule: When taking a limit, if you get an indeterminate form i.e.

±∞ 0 ,  ,etc you take ±∞ 0

the derivative of the top and bottom and evaluate the limit again…  again…  Integration by Parts Trig Substitution  If the integral contains the following root udv = uv − vdu use the given substitution and formula to  

∫

∫

b

b

and = uv |a − a vdu choose u and dv from integral and compute du by differentiating u and computing v by integrating dv Trig Stuff sin 2 x = 2 sin x cos x sin 2  x + cos 2 x = 1 1 1 + tan 2  x = sec 2 x   cos 2  x = (1 + cos 2 x)   2 1 + cot 2 = csc2  x 1 sin 2  x = (1 − cos 2 x) 2

convert into an integral involving trig  functions. a a 2 − bx 2 ⇒ x = sin θ  b a bx 2 − a 2 ⇒ x = sec θ    b a a 2 + bx 2 ⇒ x = tan θ  b

∫

Product and Quotients of Trig Functions n

n

m

m

 For  tan

 For  sin  x cos xdx we have the following: 1. n odd. Strip 1 sine out and convert rest to cosines using sin 2 = 1 − cos 2  x , then use the substitution u = cos x . 2. m odd. Strip 1 cosine out and convert rest to sines using cos 2 = 1 − sin 2 x , then use the substitution u = sin x . 3. n and m both odd. Use either 1 or 2 4. n and m both even. Use double angle and/or half angle formulas to reduce the integral into a form that can be integrated.

sec  xdx we have the following: n odd. Strip 1 tangent and 1 secant out and convert the rest to secants using tan 2  x = sec 2 x − 1 , then use the substitution u = sec x   m even. Strip 2 secants out and covert rest to tangents using sec2 = 1 + tan 2 x , then use the substitution u = tan x . n odd and m even. Use either 1 or 2. n even and m odd. Each integral will

∫

∫

1.

2.

3. 4.

 be dealt with differently. differently. Centroid b b  _   _  1 1 1  x = x[ f ( x) − g ( x)]dx    y = [ f ( x)] )]2 − [ g ( x))]]2 } dx   {  A a  A a 2

∫

∫

Parabola  focus : (0, p )    x 2 = 4 py    Directrix  y = − p 2

= 4 px  

 focus : ( p, 0) directrix : x = − p

 

 x 2

y2

+ = 1  a 2 b2 Vertices and foci are always on major axis c 2 = a 2 − b2   Make a box with sides determined by the square root of the denominators. Ellipse

 x 2

y2

− = 1  or a2 b2  y 2 x 2 − = 1  a 2 b2 Foci and vertices are always on axis determined by positive squared term. Draw box and  Hyperbola

make diagonal asymptotes. c 2 = a 2 + b2  

 

∞

Taylor Series

∑

 f '( a ))(( x − a) n n!

n =0

Ratio Test lim n →∞

an +1 an

Differential Equations  P (t ) = P0e k  ⋅t   Exponential growth

 

Separable:

 converges if < 1

dy dx

=  g ( x ) f ( y ) cross multiply

dy + P ( x) y = Q ( x) use I.F.F Absolute Convergence: If the absolute value of the  Linear: dx  series converges the series is said to be absolutely convergent.  Arc Length Cartesian b

 L =

∫ a

2

b

 dy  1 +   dx if  dx 

=  f ( x), a ≤ x ≤ b  or  L = ∫ a

2

 dx  1 +   dy  if  x = f ( y ), a ≤ x ≤ b    dy 

Parametric 2

b

 L =

Polar 2

 dx   dy    +   dt    dt   dt   

∫ a

2

b

 L =

 dr   r +  d θ     d θ  

∫

2

a

 Surface Area

Cartesian & Parametric 

Area of Polar (not surface area) b 1 2  A = r d θ θ    

b

∫

∫

S = 2π r ⋅ L   a

a

2 Cartesian to  Polar:  x = r cos θ     y = r sin θ 

Midpoint Rule b

∫

 ___

___

___ 

 f ( x)dx ≈ ∆x[ f ( x1 ) + f ( x2 ) + ..... + f ( xn )] )]  

a

Trapezoid Rule b ∆ x  f ( x)dx ≈ [ f ( x0 ) + 2 f ( x1 ) + 2 f ( x2 ) + .. ... + 2 f ( xn −1 ) + f ( xn )] )]   2 a

∫

Simpson’s Rule b ∆ x  f ( x)dx ≈ [ f ( x0 ) + 4 f ( x1 ) + 2 f ( x2 ) + 4 f ( x3 ) + ... + 2 f ( xn − 2 ) + 4 f ( xn−1 ) + f ( xn )] )] 3 a

∫

 Polar to Cartesian: r 2 = x2 + y2

tan θ  =

  Common Integrals

∫ kdx = kx + c ∫ ∫ x

 x n dx = −1

1 n +1

x n +1

dx = ln | x | +c

1

1

∫ ax + b dx = a ln | ax + b | +c ∫ ln udu = u ln(u ) − u + c n

∫ e du = e

u

+c

∫ cos udu = sin u + c ∫ sin udu = − cos u + c ∫ sec udu = tan u + c   ∫ sec u tan udu = sec u + c ∫ csc u cot udu = − csc u + c ∫ csc udu = − cot u + c 2

 

2

∫ tan udu = ln | sec u | + c ∫ sec udu = ln | sec u + tan u | +c 1

  − u tan 1   + c  a a 1

∫a

2

∫

− u du = sin 1   + c a a2 − u2

+u 1

2

du =

 y