- #1
mcastillo356
Gold Member
- 560
- 267
- TL;DR Summary
- The theoretical part is no problem. I also understand the following example; but at the next paragraphs, I've got a naive question I want to share with the forum.
Hi, PF
Integration by parts is pointed out this way:
Suppose that ##U(x)## and ##V(x)## are two differentiable functions. According to the Product Rule,
$$\displaystyle\frac{d}{dx}\big(U(x)V(x)\big)=U(x)\displaystyle\frac{dV}{dx}+V(x)\displaystyle\frac{dU}{dx}$$
Integrating both sides of this equation and transposing terms, we obtain
$$\displaystyle\int{\,U(x)\displaystyle\frac{dV}{dx}dx}=U(x)V(x)-\displaystyle\int{\,V(x)\displaystyle\frac{dU}{dx}dx}$$
or, more simply,
$$\displaystyle\int{\,UdV}=UV-\displaystyle\int{\,VdU}$$
(...)
EXAMPLE 1 ##\displaystyle\int{\,xe^{x}dx}## Let ##U=x##, ##dV=e^{x}dx##.
Then ##dU=dx, V=e^{x}##.
##=xe^{x}-\displaystyle\int{\,e^{x}dx} (i.e., UV-\int{\,VdU}##
##=xe^{x}-e^{x}+C##
Note also that had we included a constant of integration with V, for example, V=e^{x}+K, that constant would cancel out in the next step:
##\displaystyle\int{\,xe^{x}dx}=x(e^{x}+K)-\displaystyle\int{(e^{x}+K)dx}##
$$=xe^{x}+Kx-e^{x}-Kx+C=xe^{x}-e^{x}+C$$
In general, do not include a constant of integration with ##V## or on the right-hand side until the last integral has been evaluated
Question: it is the appearance of ##K## to express a constant of integration:
(i) Could it have been any other letter of the alphabet?
(ii) Mention ##C'## could have been misleading?
Greetings!
Attempt
(i) It wouldn't have been appropiate neither the Greek alphabet, ment for linear equationts, physics notations (for example ##\Omega## stands for electrical resistance), nor non Latin not capital letters etc,
(ii) ##C'=0##
Integration by parts is pointed out this way:
Suppose that ##U(x)## and ##V(x)## are two differentiable functions. According to the Product Rule,
$$\displaystyle\frac{d}{dx}\big(U(x)V(x)\big)=U(x)\displaystyle\frac{dV}{dx}+V(x)\displaystyle\frac{dU}{dx}$$
Integrating both sides of this equation and transposing terms, we obtain
$$\displaystyle\int{\,U(x)\displaystyle\frac{dV}{dx}dx}=U(x)V(x)-\displaystyle\int{\,V(x)\displaystyle\frac{dU}{dx}dx}$$
or, more simply,
$$\displaystyle\int{\,UdV}=UV-\displaystyle\int{\,VdU}$$
(...)
EXAMPLE 1 ##\displaystyle\int{\,xe^{x}dx}## Let ##U=x##, ##dV=e^{x}dx##.
Then ##dU=dx, V=e^{x}##.
##=xe^{x}-\displaystyle\int{\,e^{x}dx} (i.e., UV-\int{\,VdU}##
##=xe^{x}-e^{x}+C##
Note also that had we included a constant of integration with V, for example, V=e^{x}+K, that constant would cancel out in the next step:
##\displaystyle\int{\,xe^{x}dx}=x(e^{x}+K)-\displaystyle\int{(e^{x}+K)dx}##
$$=xe^{x}+Kx-e^{x}-Kx+C=xe^{x}-e^{x}+C$$
In general, do not include a constant of integration with ##V## or on the right-hand side until the last integral has been evaluated
Question: it is the appearance of ##K## to express a constant of integration:
(i) Could it have been any other letter of the alphabet?
(ii) Mention ##C'## could have been misleading?
Greetings!
Attempt
(i) It wouldn't have been appropiate neither the Greek alphabet, ment for linear equationts, physics notations (for example ##\Omega## stands for electrical resistance), nor non Latin not capital letters etc,
(ii) ##C'=0##
Last edited: