Integration by parts r e^r/2 dr

Click For Summary
SUMMARY

The discussion centers on the application of integration by parts to the integral of r er/2 dr. The correct choices for u and dv are crucial for simplifying the integration process. The participants clarify that for the integral, one should set v = er/2 and du = r dr, leading to dv = (1/2)er/2 dr and u = (1/2)r2. The discussion emphasizes the importance of selecting appropriate functions to reduce the complexity of the integral.

PREREQUISITES
  • Understanding of integration techniques, specifically integration by parts.
  • Familiarity with exponential functions and their integrals.
  • Basic knowledge of differential calculus and the chain rule.
  • Ability to perform variable substitutions in integrals.
NEXT STEPS
  • Learn the integration by parts formula in detail, including common pitfalls.
  • Study the integration of exponential functions, particularly eax.
  • Practice variable substitution techniques for simplifying integrals.
  • Explore advanced integration techniques, such as reduction formulas.
USEFUL FOR

Students and educators in calculus, mathematicians focusing on integration techniques, and anyone seeking to enhance their understanding of integration by parts and exponential integrals.

lemurs
Messages
30
Reaction score
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...
 
Physics news on Phys.org
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
u=F(x)
then
du=\frac{dF(x)}{dx}dx

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

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

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

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

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

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

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

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.
 
Last edited:
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?
 
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.
 
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, v=\int dv = \int e^{r/2} dr.

I am leaving the last part of this up to you.
 
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
 
What is so hard about this?

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

Similar threads

  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K