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Consequences of the Fundamental Theorem

Module by: Lawrence Baggett. E-mail the author

Summary: The first two theorems of this section constitute the basic “techniques of integration” taught in a calculus course. However, the careful formulations of these standard methods of evaluating integrals have some subtle points, i.e., some hypotheses. Calculus students are rarely told about these details.

The first two theorems of this section constitute the basic “techniques of integration” taught in a calculus course. However, the careful formulations of these standard methods of evaluating integrals have some subtle points, i.e., some hypotheses. Calculus students are rarely told about these details.

Theorem 1: Integration by Parts Formula

Let ff and gg be integrable functions on [a,b],[a,b], and as usual let FF and GG denote the functions defined by

F ( x ) = a x f , and G ( x ) = a x g . F ( x ) = a x f , and G ( x ) = a x g .
(1)

Then

a b f G = [ F ( b ) G ( b ) - F ( a ) G ( a ) ] - a b F g . a b f G = [ F ( b ) G ( b ) - F ( a ) G ( a ) ] - a b F g .
(2)

Or, recalling that f=F'f=F' and g=G',g=G',

a b F ' G = [ F ( b ) G ( b ) - F ( a ) G ( a ) ] - a b F G ' . a b F ' G = [ F ( b ) G ( b ) - F ( a ) G ( a ) ] - a b F G ' .
(3)

Exercise 1

  1. Prove the preceding theorem. HINT: Replace the upper limit bb by a variable x,x, and differentiate both sides. By the way, how do we know that the functions FgFg and fGfG are integrable?
  2. Suppose ff and gg are integrable functions on [a,b][a,b] and that both f'f' and g'g' are continuous on (a,b)(a,b) and integrable on [a,b].[a,b]. (Of course f'f' and g'g' are not even defined at the endpoints aa and b,b, but they can still be integrable on [a,b].[a,b]. See the remark following (Reference).) Prove that
    abfg'=[f(b)g(b)-f(a)g(a)]-abf'g.abfg'=[f(b)g(b)-f(a)g(a)]-abf'g.
    (4)

Theorem 2: Integration by Substitution

Let ff be a continuous function on [a,b],[a,b], and suppose gg is a continuous, one-to-one function from [c,d][c,d] onto [a,b][a,b] such that gg is continuously differentiable on (c,d),(c,d), and such that a=g(c)a=g(c) and b=g(d).b=g(d). Assume finally that g'g' is integrable on [c,d].[c,d]. Then

a b f ( t ) d t = c d f ( g ( s ) ) g ' ( s ) d s . a b f ( t ) d t = c d f ( g ( s ) ) g ' ( s ) d s .
(5)

Proof

It follows from our assumptions that the function f(g(s))g'(s)f(g(s))g'(s) is continuous on (a,b)(a,b) and integrable on [c,d].[c,d]. It also follows from our assumptions that gg maps the open interval (c,d)(c,d) onto the open interval (a,b).(a,b). As usual, let FF denote the function on [a,b][a,b] defined by F(x)=axf(t)dt.F(x)=axf(t)dt. Then, by part (2) of the Fundamental Theorem, FF is differentiable on (a,b),(a,b), and F'=f.F'=f. Then, by the chain rule, FgFg is continuous and differentiable on (c,d)(c,d) and

( F g ) ' ( s ) = F ' ( g ( s ) ) g ' ( s ) = f ( g ( s ) ) g ' ( s ) . ( F g ) ' ( s ) = F ' ( g ( s ) ) g ' ( s ) = f ( g ( s ) ) g ' ( s ) .
(6)

So, by part (3) of the Fundamental Theorem, we have that

c d f ( g ( s ) ) g ' ( s ) d s = c d ( F g ) ' ( s ) d s = ( F g ) ( d ) - ( F g ) ( c ) = F ( g ( d ) ) - F ( g ( c ) ) = F ( b ) - F ( a ) = a b f ( t ) d t , c d f ( g ( s ) ) g ' ( s ) d s = c d ( F g ) ' ( s ) d s = ( F g ) ( d ) - ( F g ) ( c ) = F ( g ( d ) ) - F ( g ( c ) ) = F ( b ) - F ( a ) = a b f ( t ) d t ,
(7)

which finishes the proof.

Exercise 2

  1. Prove the “Mean Value Theorem” for integrals: If ff is continuous on [a,b],[a,b], then there exists a c(a,b)c(a,b) such that
    abf(t)dt=f(c)(b-a).abf(t)dt=f(c)(b-a).
    (8)
  2. (Uniform limits of differentiable functions. Compare with (Reference).) Suppose {fn}{fn} is a sequence of continuous functions on a closed interval [a,b][a,b] that converges pointwise to a function f.f. Suppose that each derivative fn'fn' is continuous on the open interval (a,b),(a,b), is integrable on the closed interval [a,b],[a,b], and that the sequence {fn'}{fn'} converges uniformly to a function gg on (a,b).(a,b). Prove that ff is differentiable on (a,b),(a,b), and f'=g.f'=g. HINT: Let xx be in (a,b),(a,b), and let cc be in the interval (a,x).(a,x). Justify the following equalities, and use them together with the Fundamental Theorem to make the proof.
    f(x)-f(c)=lim(fn(x)-fn(c))=limcxfn'=cxg.f(x)-f(c)=lim(fn(x)-fn(c))=limcxfn'=cxg.
    (9)

We revisit now the Remainder Theorem of Taylor, which we first presented in (Reference). The point is that there is another form of this theorem, the integral form, and this version is more powerful in some instances than the original one, e.g., in the general Binomial Theorem below.

Theorem 3: Integral Form of Taylor's Remainder Theorem

Let cc be a real number, and let ff have n+1n+1 derivatives on (c-r,c+r),(c-r,c+r), and suppose that f(n+1)I([c-r,c+r]).f(n+1)I([c-r,c+r]). Then for each c<x<c+r,c<x<c+r,

f ( x ) - T ( f , c ) n ( x ) = c x f ( n + 1 ) ( t ) ( x - t ) n n ! d t , f ( x ) - T ( f , c ) n ( x ) = c x f ( n + 1 ) ( t ) ( x - t ) n n ! d t ,
(10)

where TfnTfn denotes the nnth Taylor polynomial for f.f.

Similarly, for c-r<x<c,c-r<x<c,

f ( x ) - T ( f , c ) n ( x ) = x c f ( n + 1 ) ( t ) ( x - t ) n n ! d t . f ( x ) - T ( f , c ) n ( x ) = x c f ( n + 1 ) ( t ) ( x - t ) n n ! d t .
(11)

Exercise 3

Prove the preceding theorem.

HINT: Argue by induction on n,n, and integrate by parts.

REMARK We return now to the general Binomial Theorem, first studied in (Reference). The proof given there used the derivative form of Taylor's remainder Theorem, but we were only able to prove the Binomial Theorem for |t|<1/2.|t|<1/2. The theorem below uses the integral form of Taylor's Remainder Theorem in its proof, and it gives the full binomial theorem, i.e., for all tt for which |t|<1.|t|<1.

Theorem 4: General Binomial Theorem

Let α=a+biα=a+bi be a fixed complex number. Then

( 1 + t ) α = k = 0 α k t k ( 1 + t ) α = k = 0 α k t k
(12)

for all t(-1,1).t(-1,1).

Proof

For clarity, we repeat some of the proof of (Reference). Given a general α=a+bi,α=a+bi, consider the function g:(-1,1)Cg:(-1,1)C defined by g(t)=(1+t)α.g(t)=(1+t)α. Observe that the nnth derivative of gg is given by

g ( n ) ( t ) = α ( α - 1 ) ... ( α - n + 1 ) ( 1 + t ) n - α . g ( n ) ( t ) = α ( α - 1 ) ... ( α - n + 1 ) ( 1 + t ) n - α .
(13)

Then gC((-1,1)).gC((-1,1)).

For each nonnegative integer kk define

a k = g ( k ) ( 0 ) / k ! = α ( α - 1 ) ... ( α - k + 1 ) k ! = α k , a k = g ( k ) ( 0 ) / k ! = α ( α - 1 ) ... ( α - k + 1 ) k ! = α k ,
(14)

and set h(t)=k=0aktk.h(t)=k=0aktk. The radius of convergence for the power series function hh is 1, as was shown in (Reference). We wish to show that g(t)=h(t)g(t)=h(t) for all -1<t<1.-1<t<1. That is, we wish to show that gg is a Taylor series function around 0. It will suffice to show that the sequence {Sn}{Sn} of partial sums of the power series function hh converges to the function g.g. We note also that the nnth partial sum is just the nnth Taylor polynomial TgnTgn for g.g.

Now, fix a tt strictly between 0 and 1.1. The argument for tt's between -1-1 and 0 is completely analogous.. Choose an ϵ>0ϵ>0 for which β=(1+ϵ)t<1.β=(1+ϵ)t<1. We let CϵCϵ be a numbers such that |αn|Cϵ(1+ϵ)n|αn|Cϵ(1+ϵ)n for all nonnegative integers n.n. See (Reference). We will also need the following estimate, which can be easily deduced as a calculus exercise (See part (d) of (Reference).). For all ss between 0 and t,t, we have (t-s)/(1+s)t.(t-s)/(1+s)t. Note also that, for any s(0,t),s(0,t), we have |(1+s)α|=(1+s)a,|(1+s)α|=(1+s)a, and this is trapped between 1 and (1+t)a.(1+t)a. Hence, there exists a number MtMt such that |(1+s)α-1|Mt|(1+s)α-1|Mt for all s(-0,t).s(-0,t). We will need this estimate in the calculation that follows.

Then, by the integral form of Taylor's Remainder Theorem, we have:

| g ( t ) - k = 0 n a k t k | = | g ( t ) - T g n ( t ) | = | 0 t g ( n + 1 ) ( s ) ( t - s ) n n ! d s | = | 0 t ( n + 1 ) × α n + 1 ( 1 + s ) α - n - 1 ( t - s ) n d s | 0 t | α n + 1 | | ( 1 + s ) α - 1 | ( n + 1 ) | ( t - s 1 + s | n d s 0 t | α n + 1 | M t ( n + 1 ) t n d s C ϵ M t ( n + 1 ) 0 t ( 1 + ϵ ) n + 1 t n d s = C ϵ M t ( n + 1 ) ( 1 + ϵ ) n + 1 t n + 1 = C ϵ M t ( n + 1 ) β n + 1 , | g ( t ) - k = 0 n a k t k | = | g ( t ) - T g n ( t ) | = | 0 t g ( n + 1 ) ( s ) ( t - s ) n n ! d s | = | 0 t ( n + 1 ) × α n + 1 ( 1 + s ) α - n - 1 ( t - s ) n d s | 0 t | α n + 1 | | ( 1 + s ) α - 1 | ( n + 1 ) | ( t - s 1 + s | n d s 0 t | α n + 1 | M t ( n + 1 ) t n d s C ϵ M t ( n + 1 ) 0 t ( 1 + ϵ ) n + 1 t n d s = C ϵ M t ( n + 1 ) ( 1 + ϵ ) n + 1 t n + 1 = C ϵ M t ( n + 1 ) β n + 1 ,
(15)

which tends to 0 as nn goes to ,, because β<1.β<1. This completes the proof for 0<t<1.0<t<1.

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