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)

∫
_{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
∫
_{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.

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)[6416(11)]
=
(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]

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 2^{nd} 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)
