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General Form of an Antiderivative
G(x) = F(x) + C
C is the constant of integration

Notation for Antiderivatives
y = f(x) dx = F(x) + C

Initial Conditions and Particlular Solutions
Given: (3x ^{2}1) dx = x ^{3}x+C Passes through (2,4)
F(x) = x ^{3}x+C F(2) = 82+C F(2) = 4 when C= 2
So, particular solution: F(x) = x ^{3}x2

Sigma Notation
 n
 a_{i} = a_{1} + a_{2} + a_{3} +...+a_{n}
 i=1

Important Summation Formulas
Sigma i = [n(n+1)]/2
Sigma i^{2} = [n(n+1)(2n+1)]/6
Sigma i^{3} = [n^{2}(n+1)^{2}]/4

Lower Sum
(sum of inscribed rectangles)
s(n) = Sigma f(m_{1})(change x)
m_{i} = 0 + (i1)(change x)

Upper Sum
(sum of circumscribed rectangles)
S(n) = Sigma f(M_{1})(change x)
M_{i} = 0+i(change x)

Subintervals = change x = [ba]/n

Drefinition of the Area of a Region in the Plane
the area of a region bounded by the graph of f, the xaxis, and the vertical lines x=a and x=b is:
Area = _{lim}Sigma f(c_{i})(change x) where (change x)=[ba]/n

Definite Integral
Fundamental Theorem of Calculus
 b
 Sigma f(x)dx = _{lim }Sigma f(c_{i})(change x_{i}) = F(b)F(a) = F(x)
 a
a=lower limit; b=upper limit
*remember the line thing...

Average Value of a Function
 b
 [1/(ba)]Sigma f(x)dx = f(C)
 a

Mean Value Therorem for Integrals
if f is continuous on the closed interval [a,b] then there exixts a number c in the closed interval [a,b] such that:
 b
 Sigma f(x)dx = f(c)(ba)
 a

Second Fundamental Theorem of Calculus
When we defined the definite integral of f on the interval [a,b] we used the constant b as the upper limit of integration and x as the variable of integration. We now look at a slightly different situation in which the variable x is used as the upper limit of integration.
(d/dx)Sigma dx

Guidelines for Integration by Substitution
 1. Choose a Substitution; choose the inner part of a composite function to sub
 2. Compute du = g'(x)dx
 3. Rewrite the integral in terms of the variable u
 4. Evaluate the resulting integral in terms of u
 5. Replace u by g(x) to obtain an antiderivative in terms of x
 6. Check your answer by differentiating

General Power Rule for Integration
Sigma [g(x)] ^{n} g'(x) dx = {[[g(x)'] ^{n+1}]/n+1} +C
 Equivalently, if u=g(x) then:
 Sigma u^{n}du = [(u^{n+1})/n+1]+C

