Integration by parts r e^r/2 dr

In summary, you are having difficulty with integration by parts and need to look for a better way to approach this problem.
  • #1
lemurs
30
0
i know this is integration by parts so here is the problem I am currently confused on how the get an answer
r e^r/2 dr

I know I the formula is int udv= uv - int du v

what i am not getting is what to use for this one. I figured

e^r/2 = v and du = r

so u= r^2/2 and what will that make dv? i have always had trouble with this part.
so while i try another guys please help... make me not go insane...
 
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  • #2
Post replaced.


It appears you have a basic misunderstanding of integration by parts. It appears you have some things backwards and you are entirely missing the
treatment of the differential as an operator.

This is not quite rigorous, but if
[tex] u=F(x)[/tex]
then
[tex] du=\frac{dF(x)}{dx}dx[/tex]

Suppose [tex]F(x)[/tex] is the indefinite integral of [tex]f(x)[/tex]. Reversing the above,
[tex] du=f(x) dx[/tex]
then
[tex] u=F(x)[/tex]

Now suppose
[tex] v=G(x)[/tex]
so that
[tex] dv=\frac{dG(x)}{dx}dx = g(x)dx[/tex]

Integration by parts is simply the chain rule applied to differentiation:
[tex]\frac d{dx}(F(x)G(x)) = f(x)G(x) + F(x)g(x)[/tex]
Integrating wrt x,
[tex]F(x)G(x) = \int f(x)G(x) dx + \int F(x)g(x) dx[/tex]
Applying the definitions of u,v, du, and dv,
[tex]uv = \int v\,du + \int u\,dv[/tex]
or
[tex]\int v\,du = uv - \int u\,dv[/tex]

So now, applying to
[tex]\int r e^{r/2} dr[/tex],
you need to identify v and du such that
[tex]v\,du = r e^{r/2} dr[/tex]

For example (these are your choices),
[tex] v = e^{r/2}[/tex]
[tex]du = r dr[/tex]

Solving for dv and u,
[tex] dv = 1/2e^{r/2} dr[/tex]
[tex]u = 1/2 r^2[/tex]

Substituting on the right hand side of the integration by parts equation yields
[tex]\int r e^{r/2} dr = 1/2 r^2 e^{r/2} - 1/4 \int r^2 e^{r/2} dr[/tex]

and now we are worse off than before!

The problem is an unwise choice of u and v. A better choice will reduce the order/complexity of the problem. An unwise choice make the problem more complex.

Try looking for a choice of u and v that reduces the order.
 
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  • #3
bah but like i siad i noit getting the right answere and i tried it several times...
e^r/2 ' = e^r/2 *1/2

so dv = (e^r/2)/2

right?
 
  • #4
i understand the princple if integration by parts.. the formula abnd stuff is easy to me.. the one part i am having problem is whenI choice my dv.

if dv=e^r/2 what the hell is v. that is my problem.. most often goiung back is the problem.
 
  • #5
The fact that you are saying "dv=e^r/2" indicates to me that you don't understand the principle of integration by parts. The correct expression is "dv=e^r/2 dr". Integrating, [tex]v=\int dv = \int e^{r/2} dr[/tex].

I am leaving the last part of this up to you.
 
  • #6
UGHHHH ! that the problem I not getting what the integral of that. I know i got to take the integral.. i just leave out the dr caue i know.. what the problem is that I have difficulty going in rever and thaking a integral of e^r/3
 
  • #7
What is so hard about this?

To calculate
[tex]\int e^{r/2} dr[/tex]
use a change of variables that converts this to a form you do know how to integrate. I assume you know how to calculate
[tex]\int e^t dt[/tex]
What change of variables is needed to convert the first form to this?
 

Related to Integration by parts r e^r/2 dr

1. What is integration by parts?

Integration by parts is a technique used in calculus to evaluate integrals of the form ∫u dv. It involves breaking down a complex integral into simpler integrals and using the product rule of differentiation to solve it.

2. How do you use integration by parts?

To use integration by parts, you must first identify the function u and its derivative du, as well as the function v and its derivative dv. Then, you can use the formula ∫u dv = uv - ∫v du to solve the integral.

3. What is the purpose of using integration by parts?

The purpose of using integration by parts is to simplify complex integrals that are difficult to solve using other techniques. It allows us to evaluate integrals of a wide range of functions, including exponential functions like e^r/2.

4. How do you apply integration by parts to the integral ∫r e^r/2 dr?

First, we identify u = r and dv = e^r/2 dr. This means du = dr and v = 2e^r/2. Plugging these values into the formula, we get ∫r e^r/2 dr = r(2e^r/2) - ∫2e^r/2 dr. This simplifies to ∫r e^r/2 dr = 2re^r/2 - 2e^r/2 + C, where C is the constant of integration.

5. What are the limitations of integration by parts?

Integration by parts may not always be applicable to every integral. It is most effective when one function in the integral can be easily differentiated and the other can be easily integrated. It also requires careful selection of u and v, which can be time-consuming for more complex integrals.

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