Integration Help!
Please guide through the steps
\(\huge \sf \int\limits_{0}^{4} (6t-8)(3-8t+3t^2).dt\)

- Abhisar

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- schrodinger

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- imqwerty

ok 1st we need to open the brackets :)

- imqwerty

wait....the method with which i did it was really long but i got the answer 176 wait lemme think of a shorter one :)

- arindameducationusc

hmmm. I can do this...
but as @imqwerty is doing.
I don't need to interfere.
If he has trouble then I will jump in @Abhisar

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## More answers

- SolomonZelman

u substitution

- Abhisar

You mean like this
\(\sf ∫u v dx = u∫v dx −∫u' (∫v dx) dx\)

- SolomonZelman

Well, we know that if you have something like an abstract case below:
\(\large\color{black}{ \displaystyle \int_{a}^{b} \color{red}{f'(x)}~{\rm G}\left(\color{red}{f(x)}\right)dx }\)
f'(x) is just some other function that is derivative in relation to the f(x) part,
then you set \(u=f(x)\)
\(du=f'(x)~du\)
\(x=a~~~\rightarrow~~u=f(a)\)
\(x=ba~~~\rightarrow~~u=f(b)\)
and you would then have:
\(\large\color{black}{ \displaystyle \int_{f(a)}^{f(b)} {\rm G}\left(\color{red}{u}\right)du }\)

- SolomonZelman

I meant \(du=f'(x)~dx\)

- SolomonZelman

should i give ordinary, reg. example?

- Abhisar

Yes, please c:

- SolomonZelman

\(\large\color{black}{ \displaystyle \int_{2}^{4} 2x\cdot (x^2-3)^{100}dx }\)
you don't want to epand, do you? O~O
you set \(u=x^2-3\)
and then
\(du=2x~dx\) (see how nice it is that you have that "2x" part in the integral)
and when x=2, u=\((\color{red}{2}^2-3)=1\) (because u in relation to x is x²-3)
and when x=4, u=\((\color{red}{4}^2-3)=13\)
So you get:
\(\large\color{black}{ \displaystyle \int_{1}^{13} (u)^{100}du }\)

- SolomonZelman

and then the rest in that example is obvious (note, that I change the limits of integration equivalently, so you don't need to sub back the x)

- arindameducationusc

I did it in a different way..... (I dont understand why it should be wrong)
|dw:1440428321372:dw|
I multiplied the brackets

- Abhisar

Please, wait....Let me digest it c:

- SolomonZelman

|dw:1440428430105:dw|

- SolomonZelman

take your time

- arindameducationusc

|dw:1440428440544:dw|

- ganeshie8

why is \(u\) substitution dropped ?

- SolomonZelman

what do you mean ganeshie?

- SolomonZelman

dropped, as "dropped into the thread" - saying, why is it proposed?
or, "dropper" as left out?

- SolomonZelman

dropped*

- ganeshie8

im asking why are you expanding that out instead of simply using \(u\) substitution..

- SolomonZelman

Well, arinda.. propsed that approach as it is also a posibility (in this case) for those that aren't familir with u sub.
I just say that derivative of 3t^2-8t+3 by so I said u sub right off

- SolomonZelman

just *see* (typos, oh my)

- arindameducationusc

One question,

- SolomonZelman

Abhisar, if you want another example let me know ...

- arindameducationusc

Without u substitution... it can be solved this way right?

- ganeshie8

Ahh okay nice :) its just that i felt this problem is begging for \(u\) substitution... expanding it out is somewhat pain...

- SolomonZelman

Yes, arinda it is possible, certainly:)

- arindameducationusc

okay.... :D

- SolomonZelman

not that i would abide by that approach in such case.

- Abhisar

One min, let me try this method on the original problem.

- SolomonZelman

Sure, go ahead:) \(!^\infty\)

- arindameducationusc

I don't know about u substitution, could you guyz teach me?

- SolomonZelman

this is not your question, apologize to remind you-;(

- SolomonZelman

You can make your own tho' there are plenty people (just on this post, certainly on the site) that can help you much better than little me

- arindameducationusc

ya @SolomonZelman
but thanks to @Abhisar to raise this question, I got to know about u substitution!

- Abhisar

Ok, so applying u substitution method on the original problem, I can choose any one as u. Correct?

- SolomonZelman

Well, you can choose anything technically, but I really don't promise it will be easier if you choose not the ideal one

- SolomonZelman

You know that when you substitute u=f(x)
in any case,
then whatever the derivative of this f(x) is (we will reasonably call it f'(x))
then the new du replaces "f'(x) and dx"

- Abhisar

ok, let's start with \(\sf 3-8t+3t^2\) as u, then \(du = -8+6t\).

- SolomonZelman

Tell me what is the derivative (with respect to t) of \(3t^2-8t+3\) ?

- SolomonZelman

yes, correct LOL

- Abhisar

c:

- Abhisar

Now, what?

- SolomonZelman

and knowing that
u=3t²-8t+3
you can find the corresponding limits of integration
when t=0, u=what?
when t=4, u=what?

- arindameducationusc

I am watching :D Nice.....

- Abhisar

and knowing that
u=3t²-8t+3
you can find the corresponding limits of integration
when t=0, u=3
when t=4, u=19

- SolomonZelman

yes, nice

- Abhisar

ok, now? c:

- SolomonZelman

So, what you do then is:
\(\large\color{black}{\displaystyle\int\limits_{0}^{4}\color{red}{(6t-8)}(3-8t+3t^2)\color{red}{dt}}\)
the red part is replaced by du
and the `3-8t+3t²` piece is your you u, so you put that in too.
Also, put in your new limits of integration that correspond to your "u".
\(\large\color{black}{\displaystyle\int\limits_{3}^{19}(u)\color{red}{du}}\)

- SolomonZelman

the `3-8t+3t²` is your \(\cancel{\rm you}\) u

- SolomonZelman

did i confuse you by typing too much? (sorry for this habit then)

- Abhisar

No it's actually helping me. Wait a sec.

- SolomonZelman

sure

- SolomonZelman

if you don't understand something, then don't hesitate to ask...

- Abhisar

So I write,
\(\large\color{black}{\displaystyle\int\limits_{3}^{19}\color{red}{(-8+6t)}}(3t-4t^2+t3)\)

- SolomonZelman

no, not really... you mixed it up a bit.

- Abhisar

oh .. !, then?

- arindameducationusc

its without -8+6t

- arindameducationusc

the same expression you wrote except that!

- Abhisar

\[\int\limits_{3}^{13}(3t²-8t+3).du\]

- arindameducationusc

right @SolomonZelman ?
no 3 to 19 !! @Abhisar

- Abhisar

\[\int\limits_{3}^{19}(3t²-8t+3).du\]
Sorry , Typo

- arindameducationusc

yup, hope so

- SolomonZelman

\(\large\color{black}{\displaystyle\int\limits_{0}^{4}(-8+6t)(3t^2-8t+3)dt}\)
You are starting from \(\color{blue}{u=3t^2-8t+3}\)
(You are sort of doing a chain rule for u)
\(\large\color{black}{\displaystyle\frac{du}{dt}=6t-8}~~~~\Longrightarrow\) \(\large\color{black}{\displaystyle \color{blue}{du= (-8+6t)dt}}\)
\(\large\color{black}{\displaystyle x=0~~~\rightarrow~~~u=3(0)^2-8(0)+3~~~\rightarrow~~~u=3}\)
\(\large\color{black}{\displaystyle x=0~~~\rightarrow~~~u=3(4)^2-8(4)+3~~~\rightarrow~~~u=19}\)
Now you put in the following into the integal:
1) "u" instead of "3t²-8t+3"
2) "du" instead of "(-8+6t)•dt"
3) Lower limit of u=3 instead of "t=0"
4) Upper limit of u=19 instead of "t=4"
YOU GET: \(\large\color{red}{\displaystyle\int\limits_{3}^{19}(u)du}\)

- arindameducationusc

nice concept @SolomonZelman
I totally got it! WOW!

- SolomonZelman

Note that "-8+6t" and "dx" are going away, being replaced by "du".

- Abhisar

Ok, so what will be the final answer O.O

- SolomonZelman

well what is the integral of u (or, u¹) ?

- arindameducationusc

replace t by u, dt by du
and put 3 and 19 for the limit

- SolomonZelman

\(\LARGE \rm \color{black}{\displaystyle\int\limits_{3}^{19}(u)du=\left.\frac{u^{1\color{royalblue}{+1}}}{1\color{royalblue}{+1}}~\right|}^{_{19}}_{^3}\)

- SolomonZelman

arinda what you said is just make everything the same but instead of using "u", use "t"
it is going to be the same exact thing, but there is no need in doing that

- arindameducationusc

okay @SolomonZelman

- Abhisar

oh.....ok, !!!!!
u=176. so now
\(\sf 3t^2−8t+3=176\) right?

- SolomonZelman

why are you doing that, Abhisar?

- SolomonZelman

oh the result for the integral is 176 you mean?

- Abhisar

Ooops..my apologies, yeah...

- SolomonZelman

yes, and just to clarify, we are NOT substituting back the t, once we found the equivalent limits of integration for u and placed THEM in.

- Abhisar

ok, one final question. In what cases we can use this method?

- SolomonZelman

In many cases. I really want to through a lot of examples rather than giving a verbal definition.

- SolomonZelman

\(\large\color{black}{ \displaystyle \int{}\sin^{80}(x)\cos(x)~dx }\)
(you can see that u=sin(x), and du=cos(x) dx)
AND NOTE: when you find the integral of u, yuo DO need to sub back the x into the result.
from now i will post just the type of problems, not their solutions (but if you will be unable to do one of them, then go ahead and ask in this or if you like - a different post)
\(\large\color{black}{ \displaystyle \int{}8x^3\sqrt{x^4+5} ~dx }\)
\(\large\color{black}{ \displaystyle \int{}x\cos(x^2)~dx}\)
(I am using x, but it can be any other letter, I am just used to x)
\(\large\color{black}{ \displaystyle \int{}x(x^4-1)~dx}\) (this is a little more advanced tho)
want more?

- SolomonZelman

mostly, the once that have the derivative of u sitting in the integral.

- SolomonZelman

ones*

- Abhisar

Let me do them first, I think they will be enough. But thank you very much for the help. I really appreciate it, you have been tutoring me since an hour, that means a lot c:

- SolomonZelman

You are always welcome!

- Abhisar

I will post my answers in a new thread since this one is becoming annoying to handle c:

- SolomonZelman

one note:
By indefinite integrals, you always need to sub in the original variable back (after you integrate with the substituted variable).

- SolomonZelman

alright, if you like to use a new thread:)
Whatever you like!

- SolomonZelman

c(u) then

- Abhisar

Yeah c; and thanks a bunch again c:

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