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

  1. 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.
  2. 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

 

  1. 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 ).

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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)