1549901369-Elements_of_Real_Analysis__Denlinger_

7.6 The Fundamental Theorem of Calculus 413

J(k+l)( )

= - (k +

1

~ (x - a)k+I + [f(x) - Tk(x)]

(fill in details)

= f(x) -Tk+1(x)

= Rk+1(x).

Therefore, P(k + 1) is true, so P(k) =:> P(k + 1). •

Theorem 7.6.17 (First Mean Value Theorem for Integrals) If

f: [a,b]---+ ~is continuous on [a,b], then 3 c E (a,b) 3 l: f = f(c)(b-a).

y

y =fix)

fie)

a c b x

Figure 7.10

Proof. Exercise 16. •

Theorem 7.6.18 (Second Mean Value Theorem for Integrals) Suppose
f, g : [a, b] ---+ ~ where a < b, f is continuous on [a, b], and g is integrable and

does not change sign on [a, b]. Then 3 c E (a, b) 3 l: Jg= f(c) l: g.

Proof. Suppose f, g : [a, b] ---+ ~ where a < b, f is continuous on [a, b],
and g is integrable and does not change sign on [a, b]. Then f g is integrable on
[a, b], by Corollary 7.5.7. Since g does not change sign on [a, b], either g 2'. 0
throughout [a, b], or g :S: 0 throughout [a, b].

Case 1 (g 2'. 0 throughout [a, b]): Then J: g 2'. 0. By the extreme value
theorem (5.3.7),

3m=minf([a,b]) and 3M=maxf([a,b]).