Substitution in a definite integral

[mcastillo356]

TL;DR

Need some help understanding the so-called Theorem.

Hi, PF
I am going to reproduce the introduction of the textbook; then the Theorem:

The method of substitution cannot be forced to work. There is no substitution that will do much good with the integral $\int{x(2+x^7)^{1/5}}\,dx$, for instance. However, the integral $\int{x^6(2+x^7)^{1/5}}\,dx$ will yield to the substitution $u=2+x^7$. The substitution $u=g(x)$ is more likely to work if $g'(x)$ is a factor of the integrand.

The following theorem simplifies the use of the method of susbstitution in definite integrals.

THEOREM 6 Substitution in a definite integral

Suppose that $g$ is a differentiable function on $[a,b]$ that satisfies $g(a)=A$ and $g(b)=B$. Also suppose that $f$ is continous on the range of $g$. Then

$$\displaystyle\int_a^b{\,f(g(x))g'(x)dx}=\displaystyle\int_A^B{\,f(u)du}$$

PROOF Let $F$ be an antiderivative of $f: F'(u)=f(u)$. Then

$\displaystyle\frac{d}{dx}\,F(g(x))=F'(g(x))g'(x)$.
Thus,
$$\displaystyle\int_a^b{f(g(x))g'(x)}=F(g(x))\Bigg |_a^b=F(g(b))-F(g(a))
=F(B)-F(A)=F(u)\Bigg |_A^B=\displaystyle\int_A^B{\,f(u)du}$$

EXAMPLE 5 Evaluate the integral $I=\displaystyle\int_0^8{\displaystyle\frac{\cos{\sqrt{x+1}}}{\sqrt{x+1}}\,dx}$.

Solution Let $u=\sqrt{x+1}$. Then $du=\displaystyle\frac{dx}{2\sqrt{x+1}}$. If $x=0$, then $u=1$; if $x=8$, then $u=3$. Thus

$$I=2\displaystyle\int_1^3{\cos{u}\,du}=2\sin{u\Bigg |_1^3}=2\sin{3}-2\sin{1}$$

Paradoxically, I understand the example 5, but can't deal with the theory, ie, the theorem 6. Try to focus the doubt about the theorem will be my attempt: why in the equality $\displaystyle\int_a^b{\,f(g(x))g'(x)dx}=\displaystyle\int_A^B{\,f(u)du}$, at the left, the integrand is a composite function?.

Greetings

PD: Post without preview

 

[PeroK]

Try to focus the doubt about the theorem will be my attempt: why in the equality $\displaystyle\int_a^b{\,f(g(x))g'(x)dx}=\displaystyle\int_A^B{\,f(u)du}$, at the left, the integrand is a composite function?.

What do you mean "why is it a composite function"? Because the theorem deals with the integral of a composite function. Note that "Integration by substitution" is like the inverse of the chain rule. And "integration by parts" is like the inverse of the product rule.

 

[fresh_42]

Substitute $g(x)=u.$ Then $\dfrac{du}{dx}=\dfrac{g(x)}{dx}=g'(x)$ and the integrand becomes
\begin{align*}
\int_a^b f(g(x)) g'(x)\,dx= \int_{x=a}^{x=b}f(u) \dfrac{du}{dx}\,dx=\int_{g(a)}^{g(b)}f(u) \,du=\int_A^B f(u)\,du
\end{align*}
which is the standard substitution method.

The crucial point is that we have $f(\;g\;)\cdot g'$ which allows to not only substitute $u=g(x)$ but simultaneously get rid of $dx$ by substituting it with $du.$ $g'$ is the correction factor that cancels out.

Here is the general n-dimensional transformation theorem of integrals:
https://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables

 

[wrobel]

1) $\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$
2) use the Fundamental theorem of calculus twice: $\int_a^bf'(g(x))g'(x)dx = f(g(b))-f(g(a))=\int_{g(a)}^{g(b)}f'(x)dx$

Remark: in the multidimensional case $g$ must be a diffeomorphism (if only we have no intention to measure the topological degree of $g$).

 

[PeroK]

1) $\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$
2) use the Fundamental theorem of calculus twice: $\int_a^bf'(g(x))g'(x)dx = f(g(b))-f(g(a))=\int_{g(a)}^{g(b)}f'(x)dx$

The inverse chain rule!

 

[wrobel]

Spoiler: offtop about Fundamental theo. of calc.

By the way, there are two pretty different theorems which are usually confused. The proofs are very different as well.

Theorem 1. Assume that $f\in C[a,b].$ Then
$$\frac{d}{dx}\int_a^xf(s)ds=f(x),\quad x\in[a,b].$$

Theorem 2. Assume that a function $g\in C[a,b]$ is differentiable in [a,b] and such that $g'$ is Riemann integrable in [a,b]. Then
$$\int_a^x g'(s)ds=g(x)-g(a).$$

 

[mcastillo356]

Hi, PF

1) $\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$

Nice introduction.

2) use the Fundamental theorem of calculus twice: $\int_a^bf'(g(x))g'(x)dx = f(g(b))-f(g(a))=\int_{g(a)}^{g(b)}f'(x)dx$

Let $g:[a,b]\rightarrow{I}$ be a differentiable function with a continous derivative, where $\mbox{I}\subset{\mathbb{R}}$ is an interval. Suppose that $f\,I\,\rightarrow{\mathbb{R}}$ is a continous function. Then
$$\displaystyle\int_a^b{\,f(g(x))g'(x)dx}=\displaystyle\int_A^B{\,f(u)du}$$
In Leibniz notation, the substitution $u=g(x)$ yields
$\displaystyle\frac{du}{dx}=g'(x)$.
Working heuristically (investigating or discovering? Wikipedia doesn't support nor refuse infinitesimals?) with infinitesimals yields the equation
$du=g'(x)dx$,
which suggests the substitution formula above.
Proof
Integration by substitution can be derived from the fundamental theorem of calculus as follows. Let $f$ and $g$ be two functions satisfying the above hypothesis that $f$ is continous on $I$ and $g'$ is integrable on $[a,b]$. Hence the integrals
$\displaystyle\int_a^b{\,f(g(x))g'(x)dx}$
and
$\displaystyle\int_{g(a)}^{g(b)}{\,f(u)du}$
in fact exist, and it remains to show that they are equal.

Applying the fundamental theorem of calculus twice gives (here I say: the Fundamental Theorem of Calculus is, in my personal opinion, is not the issue of this thread; it is ment to be part of the comprehended background; however, a doubt has raised: the term "twice"):

By the way, there are two pretty different theorems which are usually confused. The proofs are very different as well.

Theorem 1. Assume that $f\in C[a,b].$ Then
$$\frac{d}{dx}\int_a^xf(s)ds=f(x),\quad x\in[a,b].$$

Theorem 2. Assume that a function $g\in C[a,b]$ is differentiable in [a,b] and such that $g'$ is Riemann integrable in [a,b]. Then
$$\int_a^x g'(s)ds=g(x)-g(a).$$

Now what I think is that when we mention to use twice, we talk about the left and the right sides of the theorem of substitution in a definite integral, this is, demonstrate the following:

2) use the Fundamental theorem of calculus twice: $\int_a^bf'(g(x))g'(x)dx = f(g(b))-f(g(a))=\int_{g(a)}^{g(b)}f'(x)dx$

Since $f$ is continous, it has an antiderivative $F$. The composite function $F\circ{g}$ is then defined (why is it then defined?) Since $g$ is differentiable, combining the chain rule and the definition of an antiderivative gives

$(F\circ{g})'(x)=F'(g(x))\cdot{g'(x)}=f(g(x))\cdot{g'(x)}$

Applying the fundamental theorem of calculus twice gives

$\displaystyle\int_a^b{f(g(x)\cdot{g'(x)\,dx}}=\displaystyle\int_a^b{(F\circ{g})'(x)\,dx}$
$=(F\circ{g})(b)-(F\circ{g})(a)$
$=F(g(b))-F(g(a))$
$=\displaystyle\int_{g(a)}^{g(b)}\,f(u)du$

which is the substitution rule
Source, Wikipedia. Thanks, @fresh_42
Greetings.