Basic Integration Formula Sheet

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Basic Integration Formula Sheet Basic Integration Formulas

Integration Formulas Formulas

Definite (Riemann) Integral

Fundamental Theorem of Calculus I (FTC I) b

  0 du = du  = 0   du = du  = u  u +  + C   C    k du = du = ku  ku +  + C   C    u u du = du  =   (for n (for  n   = −1) n + 1   cos u cos  u du = du = sin u sin  u +  + C   C    sin u sin  u du = du =  −cos u cos  u +  + C   C    sec u du = du = tan u tan  u  + C   + C    sec u sec  u tan u tan  u du = du = sec u + C  + C    csc u du = du  =  −cot u cot  u +  + C   C    csc u csc  u cot u cot  u du = du =  −csc u + C  +  C    e du = du  = e  e + C  1   a du = du  = a + C,a > 0 >  0 ln a ln  a   1  du  = ln |u| + C  u   2

2

u

u

u

tan u du = du  =

 − ln | cos x| + C  = ln | sec u| + C 

     

2

n

lim

2

2

||∆ ||∆→0||

Fundamental Theorem of Calculus II (FTC II)

 || ∆|| = ∆x Partitions of unequal width:  || ∆||  = max(∆x max(∆x )

d dx

a

Continuity   Integrability The Converse is NOT True (see example below)

dx

a

Example: 2

 

n

f ( f (x) dx = dx  = f   f ((c)(b )(b

a

i=1

i=1 n

Area of the region bounded by the graph of  f ,  f , x-axis, and x and  x  = a  =  a  and  x  = b  =  b

i=1 n

n

i=1 n

3

2

where r where r i

i=1 n

2

4

30

i=1

Upper and Lower Sums n

I.R. I.R. =

i

i=1

f ( f (mi ) is the min. value of  f  of  f  on the subinterval.

n→∞

 f (f (M  )∆x )∆x i

i=1

f ( f (M i ) is the max. value of  f  of  f  on the subinterval.

b

− 1)∆x 1)∆x,

 

| | = − ln | sec u − tan u| + C 

csc u du = du  =

 − ln | csc u + cot u| + C 

Right Endpoints: a + i  + i∆ ∆x, for i for  i  = 1, . . . , n ∆x  = (b a)/n

−

du =  g  (x) dx

 ⇒ du =  du  = g  g  (x)dx, dx,

 ⇒

 

g (b)

b

f ( f (u) du

 ⇒

 

a

g(a)

a

a

b

n

b

c

a

b

a

b

i−1

i

a

i=1

n

b

c

b

i

a

a

0

a

b

i−1

i=1

b

b

2

b

a

a

b

a

b

sec u du = du = ln sec u + tan u + C 

f ( f (x) dx

f ( f (g(x))g ))g (x) dx

a

a

a

Left Endpoints: a + (i (i for i for  i  = 1, . . . , n

b

a

f ( f (u) du   1. f ( f (x) dx = dx = 0 Numerical Integration      2. f ( f (x) dx = dx  = − f ( f (x) dx  f   x  + x  +  x Midpoint Rule: Rule: f ( f (x) dx ≈ ∆x 2       3. f ( f (x) dx = dx  = f ( f (x) + f ( f (x) dx      f (f (x ) + f  +  f ((x ) Trapezoidal Rule: Rule : f ( f  ( x ) dx  ≈ ∆x     2 4. kf ( kf (x) dx = dx  = k  k f ( f (x) dx    b − a   Simpson’s Rule: Rule: f ( f (x) dx ≈ [f ( f (x ) + 4f  4 f ((x )+ 3n 5. [f ( f (x)dx ± g(x) ± · · · ± k (x)] dx )]  dx  =       2f ( f (x ) + 4f  4 f ((x ) + · · · + 4f  4f ((x ) + f  +  f ((x )] f ( f (x)  ± g(x) dx ± · · · ± k(x) dx   6. 0 ≤ f ( f (x) dx for dx  for  f  non-negative  f  non-negative     7. If  f (  f (x) ≤ g(  g (x) ⇒ f ( f (x) dx ≤ g (x) dx     8. If  f    f   is an even function, then f ( f (x) dx = dx = 2 f ( f (x) dx   a

n

C.R. C.R. = lim lim

−a

Let u Let  u  = g  =  g((x), then

i

b

 f (f (m )∆x lim lim )∆x

n→∞

i=1

i

Properties of Integration

− 1)

b

 

a

 ∈ [x  [ x −1 , x ]

+ 3n 3n

1

i

n→∞

2

f ( f (c) =

Substitution Method for Integration

b

   f (f (r )∆x )∆x  = f ( f (x) dx

Area Area = lim lim

2

− a), where c where  c  ∈ [a,  [a, b]

Average Value of a Function on [ on  [a, a, b]

Area of a Region in a Plane

n

 c = cn  =  cn  i =  n(  n (n + 1) 2  i = n(n + 1)(2n 1)(2n + 1) 6  i = n (n + 1) 4  i = n(2n (2n + 1)(n 1)(n + 1)(3n 1)(3n

f ( f (t) dt  = f   =  f ((u)u (with chain rule)

b

x dx = dx = 1

0

 

f ( f (t) dt  = f   =  f ((x)

 Mean Value Theorem for Integration

2

1

x

    d

u

 = a  + a  +  a  + .  +  . . . + a  +  a

− F ( F (a)

a

i=1

Summation Formulas Formulas n

f ( f (x) dx = dx  = F   F ((b)

a

i

 ⇒

cot u du = du = ln sin u + C 

| | = − ln | csc u| + C 

i

i

+ C 

a

b

   f (f (c )∆x )∆x  = f ( f (x) dx

Partitions of equal width:

2

√  = arcsec a u u2 − a2

n+1

n

u

  du √ a − u   = arcsin ua  +  C    du 1 u = arctan  +  C  a + u a a   du 1  | u|

 

b

a

a

a

a

−a a

9. If  f  is  f  is an odd function, then

f ( f (x) dx = dx = 0

−a

0

3

n−1

n

1