Question

I need help on integration by part. Someone helps me, please. I don't understand.

Answer 1

\[\int_0^\infty x\dfrac{1}{b}e^{-x/b}dx\]

Answer 2

To my prof, let \(u = -e^{-x/b}\rightarrow du=\dfrac{1}{b}e^{-x/b}dx\)
and \(dv =dx \rightarrow v =x\)
But it is weird to me. If you can show me the "human" way, it would be great.

Answer 3

LIATE
L-log
I-inverse
A-Algebraic
T-trigonometric
E-exponent
i learn thats how we shd select the "U" to take the derivative of

Answer 4

Can you get what his mean?

Answer 5

sorry but i havent work with such complicated problem before...

Answer 6

i understand half of it so far though

Answer 7

That's good, show me please.

Answer 8

well they let u=x and the derivative of x is 1 ...im not sure if u will understand how ill explained maybe u shd check this link out https://www.youtube.com/results?search_query=patrickjmt+integration+by+parts

Answer 9

or @UsukiDoll

Answer 10

Expected value of exponential distribution?

Answer 11

Yes, but the way my Prof take integration by part killed me. I don't get it.

Answer 12

So you want to really go over (conceptually) about the method in general?

Answer 13

I'm sure SolZel has something for you :)

Answer 14

Ok, do whatever you can to help me understand it. I do appreciate.

Answer 15

Well, let's suppose that \(\color{#000000 }{ \displaystyle f }\) and \(\color{#000000 }{ \displaystyle g }\) are some (differentiable) functions of \(\color{#000000 }{ \displaystyle x }\).

Then,
\(\color{#000000 }{ \displaystyle \frac{d}{dx} [f \cdot g]=f'\cdot g+f\cdot g' }\)

this is just a product rule, ok?

Answer 16

yes

Answer 17

So, when we integrate both sides, we get:

\(\color{#000000 }{ \displaystyle f \cdot g=\int \left(f'\cdot g+f\cdot g'\right)~dx }\)

\(\color{#000000 }{ \displaystyle f \cdot g=\int \left(f'\cdot g\right)~dx+\int\left(f\cdot g'\right)~dx }\)

\(\color{#000000 }{ \displaystyle f \cdot g-\int \left(f'\cdot g\right)~dx=\int\left(f\cdot g'\right)~dx }\)

this is just the general derivation of the integration by parts.
I'll write it with the x, in the "official" way.

\(\color{#000000 }{ \displaystyle\int f(x)g'(x)~dx= f(x)g(x)-\int f'(x)g(x)~dx}\)

Answer 18

This is just a proof of it.
And the purpose of this is to kill the f(x) by differentiating (since we will no longer have the f(x) inside the integral). (For example, if f(x) is hard to integrate.)

Answer 19

great explanation @SolomonZelman

Answer 20

Sounds more like surcasm, because I wouldn't get satisfied by such... (trying my best -:( )

Answer 21

Woah.................. It is a very brand new way to look at the integration by part.
Thank you so much. You are genius!!

Answer 22

i couldnt have done it like this thumbs up

Answer 23

The usual notations though have v, dv and u, du.
And that often causes confusion

Answer 24

Is there any way to go on usual way?

Answer 25

The way I wrote it, which is more like "from the product rule"
\(\color{#000000 }{ \displaystyle\int f(x)g'(x)~dx= f(x)g(x)-\int f'(x)g(x)~dx}\)

The way a person who integrates frequently would write:
\(\color{#000000 }{ \displaystyle\int u~dv= uv-\int v~du}\)

Answer 26

So, really,

\(\color{#000000 }{ \displaystyle u=f(x)}\) \(\tiny \\[0.7em]\)
\(\color{#000000 }{ \displaystyle \frac{du}{dx}=f'(x)\quad \Longrightarrow \quad du=f'(x)~dx}\)

AND

\(\color{#000000 }{ \displaystyle v=g(x)}\) \(\tiny \\[0.7em]\)
\(\color{#000000 }{ \displaystyle dv=g'(x)\quad \Longrightarrow \quad du=g'(x)~dx}\)

you can compare the two:

\(\color{#000000 }{ \displaystyle\int \color{red}{f(x)}\color{blue}{g'(x)~dx}= \color{red}{f(x)}\color{green}{g(x)}-\int \color{green}{g(x)}\color{darkgoldenrod}{f'(x)~dx}}\)
\(\color{#000000 }{ \displaystyle\int \color{red}{u}~\color{blue}{dv}= \color{red}{u}\color{green}{v}-\int \color{green}{v}~\color{darkgoldenrod}{du}}\)

Answer 27

This is all very abstract.
If you want, I can do an example or a couple....

Answer 28

I think I got it. That is good enough for me. Again, thank you so much.:)

Answer 29

Anyway, I will do one example ... just in case

Answer 30

Example1:

\(\color{#000000 }{ \displaystyle\int 3x^2\ln x~dx }\)

\(\color{#000000 }{ \displaystyle\int \color{red}{(\ln x)}~\color{blue}{(3x^2~dx)}= \color{red}{(\ln x)}\color{green}{(x^3)}-\int \color{green}{(x^3)}~\color{darkgoldenrod}{((1/x)~dx)}}\)

\(\color{#000000 }{ \displaystyle\int \color{red}{(\ln x)}~\color{blue}{(3x^2~dx)}= x^3\ln x-\int x^2~dx}\)

\(\color{#000000 }{ \displaystyle\int \color{red}{(\ln x)}~\color{blue}{(3x^2~dx)}= x^3\ln x-2x\color{grey}{\rm +C}}\)

Answer 31

You can also get the general idea of IBP by looking at the discrete case, appropriately named summation by parts.

Consider two sequences, \(a_k\) and \(b_k\), and the following generalized partial sum of their product:
\[S_{n,N}:=\sum_{k=n}^Na_kb_k=a_nb_n+a_{n+1}b_{n+1}+\cdots+a_{N-1}b_{N-1}+a_Nb_N\]where clearly \(n\le N\). Basically, you can think of \(a_k\) and \(b_k\) as the widths and heights of a sequences of rectangles, and so the product \(a_kb_k\) gives the area of the \(k\)th rectangle. Summing them up, you basically have a numerical approximation to some definite integral determined by the actual sequences \(a_k\) and \(b_k\).

Anyway, you take the difference of the \((n+1,N+1)\)th and the \((n,N)\)th partial sums, which gives
\[S_{n+1,N+1}-S_{n,N}=\sum_{k=n}^N(a_{k+1}b_{k+1}-a_kb_k)=a_{N+1}b_{N+1}-a_nb_n\]because the summation telescopes.

Let's add two terms that cancel each other to the sum, \(a_kb_{k+1}\) (or \(a_{k+1}b_k\) will also do - you'll notice that the choice won't matter).
\[S_{n+1,N+1}-S_{n,N}=\sum_{k=n}^N(a_{k+1}b_{k+1}-a_kb_{k+1}+a_kb_{k+1}-a_kb_k)\]
The added term is basically the area of the \(k\)th rectangle with one fixed dimension, while the other is swapped with the corresponding dimension of the \((k+1)\)th rectangle.

We can group the terms of the sum like so:
\[S_{n+1,N+1}-S_{n,N}=\sum_{k=n}^Nb_{k+1}(a_{k+1}-a_k)+\sum_{k=n}^Na_k(b_{k+1}-b_k)\]
Because we know this sum is equal to \(a_{N+1}b_{N+1}-a_nb_n\), moving one of these sums to the other side yields the final result:
\[\sum_{k=n}^Nb_{k+1}(a_{k+1}-a_k)=a_{N+1}b_{N+1}-a_nb_n-\sum_{k=n}^Na_k(b_{k+1}-b_k)\]which has a very similar structure to the IBP formula.