1549901369-Elements_of_Real_Analysis__Denlinger_

7.6 The Fundamental Theorem of Calculus 411

From this we would write

rv:rr

Jo xsin(x^2 +~)dx = ~ [-cosuJ!/~

2

=~[-cos^3 ; +cos~]= ~[0-0] = O. D

Theorem 7.6.15 (Integration by Parts) Suppose J and g are differen-

tiable on [a, b], and their derivatives f' and g' are integrable on [a, b]. Then

both Jg' and gf' are integrable on [a, b], and

1b Jg'= J(b)g(b) - J(a)g(a) -1b gf'.

Proof. Suppose J and g are differentiable on [a, b], and their derivatives
f' and g' are integrable on [a, b]. By the product rule for derivatives, gJ is
differentiable on [a, b] and

(Jg)'= Jg'+ gf'. (25)

Since the product of integrable functions is integrable, Jg' and gJ' are
integrable on [a, b], and consequently, so is (Jg)'. Moreover, Jg is continuous
on [a, b]. Thus, by FTC-I,

1b (Jg)'= (f g)(b) - (f g)(a) (26)

However, by Theorem 7.5.l applied to (25),

1b (Jg)'= 1b Jg'+ 1b gf'. (27)

Putting (26) and (27) together,

J(b)g(b) - J(a)g(a) = 1b Jg'+ 1b gf'.

Therefore, J: Jg'= J(b)g(b) - J(a)g(a) - J: gf'. •

We can use integration by parts to develop an alternative formula for the
remainder in Taylor's theorem.

Theorem 7.6.16 (Taylor's Theorem,^11 with Remainder in Integral
Form) Suppose J is n times differentiable on an open interval containing

11. Taylor polynomials are defined in 6.5.1, R,,,(x) is defined in 6 .5.9, and Taylor's theorem
    is stated in 6.5.11.