Chapter 4 notes from the Text Book Section 4.4 - The Fundamental Theorem of Calculus Theorem 4.9  The Fundamental Theorem of Calculus   If a function f is continuous on the closed interval [a,b] then ∫ a b f(x) dx = F (b)  F (a) Guidelines for Using the Fundamental Theorem of Calculus Provided you can find an antiderivative of f, you now have a way to evaluate a definite integral without having to use the limit of a sum. When applying the Fundamental Theorem of Calculus, the fallowing notation is convenient.   ∫ a b f(x) dx = F (x) ] a b  = F(b)  F (a)     (the big F is the antiderivative of little f)               For instance, to evaluate ∫ 1 3 f(x^3) dx, you can write.             ∫ 1 3 f(x^3) dx = (x^4/4) ] 1 3   =  (3^4/4)  (1^4/4) =  (81/4)  (1/4) = 20   It is not necessary to include a constant of integration C in the antiderivative because ∫ a b f(x) dx = F (x) + C ] a b  = [F(b)+C]  [F (a)+C] =  F(b)  F (a) . The constant is added in then it is subtracted out so it is cancelled. Theorem 4.10 - Mean Value Theorem for Integrals   If f is continuous on the closed interval [a,b], then there exists a number c in the closed interval [a,b] such that ∫ a b f(x) dx = f(c )(b - a). Definition of the Average Value of a Function on an Interval   If f is integrable on the closed interval [a,b], then the average value of on the interval is:    ( 1/(b - a)) ∫ a b f(x) dx. Example 4  Finding the Average Value of a Function Find the average value of f(x) = 3x^2  2x on the interval [1,4], Solution  the average value given by ( 1/(b - a)) ∫ a b f(x) dx = (1/3) ∫ 1 4 (3x^2  2x) dx                  = (1/3)[x^3  x^2] 1 4                                    = (1/3)[(4^3 - 4^2) - (1^3 - 1^2)]                                   = (1/3)[64-16-(1-1)]                                   = (48/3) = 16 Theorem 4.11  The Second Fundamental Theorem of Calculus   If f is continuous on an open interval I containing a, then for every x in the interval,   ( d/dx )[∫ a x f( t ) dt] = f( x ). /////////////////////////////////////////////////////////////////////////////////////   proof   Begin by defining F (big F is antiderivative of little f) as   F(x) = lim (Δx à0) [F(x + Δx)  F(x)] / Δx             lim (Δx à0)(1/ Δx) [∫ a x+Δx f( t ) dt  -   ∫ a x f( t ) dt  ]             lim (Δx à0)(1/ Δx) [∫ a x+Δx f( t ) dt  +  ∫ x a f( t ) dt  ]             lim (Δx à0)(1/ Δx) [∫ x x+Δx f( t ) dt] Example 8 Using the Second Fundamental Theorem of Calculus   Find the derivative of   F(x) = ∫Pi/2 x^3  cos( t ) dt    Solution  Using the u = x^3, you can apply the Second Fundamental Theorem of Calculus  with the Chain Rule as follows. F  (x) = (dF/du)(du/dx)                                      Chain Rule                                         = (d/du) [ F(x)]( du/dx)                            Definition of (dF/du)             = (d/du)[ ∫Pi/2 x^3  cos( t ) dt  ] ( du/dx)             Plug in function             = (d/du)[ ∫Pi/2 u  cos( t ) dt  ] ( du/dx)              Substitute u for (b)= X^3             = (cos u) (3x^2)                                      Apply 2nd Fundamental Theorem of Calc             = (cos x^3)(3x^2)                                           Rewrite as function of x.   Because the integrand in Example 8 is easily integrated, you can verify the derivative as follows. F(x) = ∫Pi/2 x^3  cos( t ) dt  = sin(t)] Pi/2 x^3  = sin(x^3)  sin (pi/2) = (sin x^3)-1 In this form, you can apply the Power Rule to verify that the derivative is the same as that obtained in Example 8. F (x) = (cos x^3) (3x^2)