# Why do we need somewhat lengthy proof of the “Integration by Substitution for Definite Integrals” theorem?

I see in many real analysis books, for example the one I'm going to say, the author first proved the Integration by Substitution Indefinite Integral Version $\int f(g(x))g'(x)dx=F(g(x))+C$, which the proof is easy. But the next thing let me feel curious, why did he then proved the Integration by Substitution "Definite Integral" Version(see the figure) with so lengthy manner? Having know that $\int f(g(x))g'(x)dx=F(g(x))+C$, and combine it with the Fundamental Theorem of Calculus, the definite integral $\int_a^b f(g(x))g'(x)dx$ is simply the difference of its antiderivative $F(g(x))$ at two points $a$ and $b$, $[F(g(x))]^b_a$. Why we need a relatively lengthier proof here?

Update: (Exercise used in this theorem)

• The essetial idea is same and the lengthier proof is mainly due to bookkeeping (ensuring that everything is OK). – Paramanand Singh Oct 31 '17 at 14:30
• May i ask, what book this is from? – Michal Dvořák Apr 29 '18 at 20:49
• @MichalDvoĆák A great book on one-variable analysis, The Real Numbers and Real Analysis by Ethan D. Bloch. See this – Eric May 1 '18 at 3:35
• @MichalDvoĆák Also I have just found that you had asked a question. If you want a complete and clear reference about constructing $e^x, \ln x$ etc in a throughly rigorous way, you would also glad to have a look this book. – Eric May 1 '18 at 3:37

The Fundamental Theorem of Calculus says that if $F'(t) = f(t)$ and $f$ is integrable on $[a,b]$ then

$$\int_a^b f(t) \,dt = F(b) - F(a).$$

Just knowing that the derivative of $F(g(x))$ is $f(g(x))g'(x)$ is not enough. You need to also know that $f(g(x))g'(x)$ is integrable. Well, when $g'(x)$ is continuous then $f(g(x))g'(x)$ is continuous and hence integrable, but we don't know that $g'(x)$ is continuous, only that it's integrable. So we need a different argument that shows that $f(g(x))g'(x)$ is integrable. This is easy enough to do and is what is done in the first paragraph.

The second paragraph is verifying that if $F$ is an antiderivative of $f$ then $F \circ g$ is an antiderivative of $(f \circ g)g'$. That's covered in the next 5 sentences (first say why $F$ exists, then say that $F' = f$ then say that $(F \circ g)' = (f \circ g)g'$).

Because $f$ is continuous then $f$ is integrable by Theorem 5.4.11, and we also know that $f$ has an antiderivative, which is $F$.