Calculus, Chapter 6

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Calculus, Chapter 6
2010-05-09 23:11:21

Flashcards for Chapter 6
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  1. General Form of an Antiderivative
    G(x) = F(x) + C

    C is the constant of integration
  2. Notation for Antiderivatives
    y = f(x)dx = F(x) + C
  3. Initial Conditions and Particlular Solutions
    Given: (3x2-1)dx = x3-x+C Passes through (2,4)

    F(x) = x3-x+C F(2) = 8-2+C F(2) = 4 when C= -2

    So, particular solution: F(x) = x3-x-2
  4. Sigma Notation
    • n
    • ai = a1 + a2 + a3 +...+an
    • i=1
  5. Important Summation Formulas
    Sigma i = [n(n+1)]/2

    Sigma i2 = [n(n+1)(2n+1)]/6

    Sigma i3 = [n2(n+1)2]/4
  6. Lower Sum
    (sum of inscribed rectangles)
    s(n) = Sigma f(m1)(change x)

    mi = 0 + (i-1)(change x)
  7. Upper Sum
    (sum of circumscribed rectangles)
    S(n) = Sigma f(M1)(change x)

    Mi = 0+i(change x)
  8. Subintervals = change x = [b-a]/n
  9. Drefinition of the Area of a Region in the Plane
    the area of a region bounded by the graph of f, the x-axis, and the vertical lines x=a and x=b is:

    Area = limSigma f(ci)(change x) where (change x)=[b-a]/n
  10. Definite Integral
    Fundamental Theorem of Calculus
    • b
    • Sigma f(x)dx = lim Sigma f(ci)(change xi) = F(b)-F(a) = F(x)
    • a

    a=lower limit; b=upper limit

    *remember the line thing...
  11. Average Value of a Function
    • b
    • [1/(b-a)]Sigma f(x)dx = f(C)
    • a
  12. 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)(b-a)
    • a
  13. 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
  14. 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
  15. 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 undu = [(un+1)/n+1]+C