# Calculus, Chapter 6

 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: (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 Sigma Notation n ai = a1 + a2 + a3 +...+ani=1 Important Summation Formulas Sigma i = [n(n+1)]/2 Sigma i2 = [n(n+1)(2n+1)]/6 Sigma i3 = [n2(n+1)2]/4 Lower Sum (sum of inscribed rectangles) s(n) = Sigma f(m1)(change x) mi = 0 + (i-1)(change x) Upper Sum (sum of circumscribed rectangles) S(n) = Sigma f(M1)(change x) Mi = 0+i(change x) Subintervals = change x = [b-a]/n 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 Definite Integral Fundamental Theorem of Calculus bSigma 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... Average Value of a Function b[1/(b-a)]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: bSigma f(x)dx = f(c)(b-a)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 sub2. Compute du = g'(x)dx3. Rewrite the integral in terms of the variable u4. Evaluate the resulting integral in terms of u5. Replace u by g(x) to obtain an antiderivative in terms of x6. 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 undu = [(un+1)/n+1]+C AuthorAnonymous ID18552 Card SetCalculus, Chapter 6 DescriptionFlashcards for Chapter 6 Updated2010-05-10T03:11:21Z Show Answers