# Substitution in a definite integral

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• mcastillo356
In summary, the method of substitution in integrals involves using a substitution function to simplify the integrand and make it easier to integrate. This method cannot be forced to work and is more likely to be successful if the substitution function has a derivative that is a factor of the integrand. The substitution theorem simplifies the use of this method in definite integrals. It states that if the substitution function is differentiable on the interval of integration and the integrand is continuous on the range of the substitution function, then the integral can be rewritten in terms of the substitution function. This can be derived from the fundamental theorem of calculus by applying it twice.
mcastillo356
Gold Member
TL;DR Summary
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

Last edited:
mcastillo356 said:
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.

mcastillo356
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

dextercioby, mcastillo356 and DaveE
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##).

Last edited:
PeroK
wrobel said:
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!

mcastillo356
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
$$\int_a^x g'(s)ds=g(x)-g(a).$$

Last edited:
mcastillo356
Hi, PF
wrobel said:
1) ##\frac{d}{dx}f(g(x))=f'(g(x))g'(x)##
Nice introduction.
wrobel said:
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"):

wrobel said:
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:

wrobel said:
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.

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