- #1

mcastillo356

Gold Member

- 568

- 273

- 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: