Abhisar
  • Abhisar
Integration Help! Please guide through the steps \(\huge \sf \int\limits_{0}^{4} (6t-8)(3-8t+3t^2).dt\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
imqwerty
  • imqwerty
ok 1st we need to open the brackets :)
imqwerty
  • 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
  • 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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SolomonZelman
  • SolomonZelman
u substitution
Abhisar
  • Abhisar
You mean like this \(\sf ∫u v dx = u∫v dx −∫u' (∫v dx) dx\)
SolomonZelman
  • 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
  • SolomonZelman
I meant \(du=f'(x)~dx\)
SolomonZelman
  • SolomonZelman
should i give ordinary, reg. example?
Abhisar
  • Abhisar
Yes, please c:
SolomonZelman
  • 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
  • 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
  • arindameducationusc
I did it in a different way..... (I dont understand why it should be wrong) |dw:1440428321372:dw| I multiplied the brackets
Abhisar
  • Abhisar
Please, wait....Let me digest it c:
SolomonZelman
  • SolomonZelman
|dw:1440428430105:dw|
SolomonZelman
  • SolomonZelman
take your time
arindameducationusc
  • arindameducationusc
|dw:1440428440544:dw|
ganeshie8
  • ganeshie8
why is \(u\) substitution dropped ?
SolomonZelman
  • SolomonZelman
what do you mean ganeshie?
SolomonZelman
  • SolomonZelman
dropped, as "dropped into the thread" - saying, why is it proposed? or, "dropper" as left out?
SolomonZelman
  • SolomonZelman
dropped*
ganeshie8
  • ganeshie8
im asking why are you expanding that out instead of simply using \(u\) substitution..
SolomonZelman
  • 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
  • SolomonZelman
just *see* (typos, oh my)
arindameducationusc
  • arindameducationusc
One question,
SolomonZelman
  • SolomonZelman
Abhisar, if you want another example let me know ...
arindameducationusc
  • arindameducationusc
Without u substitution... it can be solved this way right?
ganeshie8
  • ganeshie8
Ahh okay nice :) its just that i felt this problem is begging for \(u\) substitution... expanding it out is somewhat pain...
SolomonZelman
  • SolomonZelman
Yes, arinda it is possible, certainly:)
arindameducationusc
  • arindameducationusc
okay.... :D
SolomonZelman
  • SolomonZelman
not that i would abide by that approach in such case.
Abhisar
  • Abhisar
One min, let me try this method on the original problem.
SolomonZelman
  • SolomonZelman
Sure, go ahead:) \(!^\infty\)
arindameducationusc
  • arindameducationusc
I don't know about u substitution, could you guyz teach me?
SolomonZelman
  • SolomonZelman
this is not your question, apologize to remind you-;(
SolomonZelman
  • 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
  • arindameducationusc
ya @SolomonZelman but thanks to @Abhisar to raise this question, I got to know about u substitution!
Abhisar
  • Abhisar
Ok, so applying u substitution method on the original problem, I can choose any one as u. Correct?
SolomonZelman
  • 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
  • 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
  • Abhisar
ok, let's start with \(\sf 3-8t+3t^2\) as u, then \(du = -8+6t\).
SolomonZelman
  • SolomonZelman
Tell me what is the derivative (with respect to t) of \(3t^2-8t+3\) ?
SolomonZelman
  • SolomonZelman
yes, correct LOL
Abhisar
  • Abhisar
c:
Abhisar
  • Abhisar
Now, what?
SolomonZelman
  • 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
  • arindameducationusc
I am watching :D Nice.....
Abhisar
  • 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
  • SolomonZelman
yes, nice
Abhisar
  • Abhisar
ok, now? c:
SolomonZelman
  • 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
  • SolomonZelman
the `3-8t+3t²` is your \(\cancel{\rm you}\) u
SolomonZelman
  • SolomonZelman
did i confuse you by typing too much? (sorry for this habit then)
Abhisar
  • Abhisar
No it's actually helping me. Wait a sec.
SolomonZelman
  • SolomonZelman
sure
SolomonZelman
  • SolomonZelman
if you don't understand something, then don't hesitate to ask...
Abhisar
  • Abhisar
So I write, \(\large\color{black}{\displaystyle\int\limits_{3}^{19}\color{red}{(-8+6t)}}(3t-4t^2+t3)\)
SolomonZelman
  • SolomonZelman
no, not really... you mixed it up a bit.
Abhisar
  • Abhisar
oh .. !, then?
arindameducationusc
  • arindameducationusc
its without -8+6t
arindameducationusc
  • arindameducationusc
the same expression you wrote except that!
Abhisar
  • Abhisar
\[\int\limits_{3}^{13}(3t²-8t+3).du\]
arindameducationusc
  • arindameducationusc
right @SolomonZelman ? no 3 to 19 !! @Abhisar
Abhisar
  • Abhisar
\[\int\limits_{3}^{19}(3t²-8t+3).du\] Sorry , Typo
arindameducationusc
  • arindameducationusc
yup, hope so
SolomonZelman
  • 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
  • arindameducationusc
nice concept @SolomonZelman I totally got it! WOW!
SolomonZelman
  • SolomonZelman
Note that "-8+6t" and "dx" are going away, being replaced by "du".
Abhisar
  • Abhisar
Ok, so what will be the final answer O.O
SolomonZelman
  • SolomonZelman
well what is the integral of u (or, u¹) ?
arindameducationusc
  • arindameducationusc
replace t by u, dt by du and put 3 and 19 for the limit
SolomonZelman
  • 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
  • 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
  • arindameducationusc
okay @SolomonZelman
Abhisar
  • Abhisar
oh.....ok, !!!!! u=176. so now \(\sf 3t^2−8t+3=176\) right?
SolomonZelman
  • SolomonZelman
why are you doing that, Abhisar?
SolomonZelman
  • SolomonZelman
oh the result for the integral is 176 you mean?
Abhisar
  • Abhisar
Ooops..my apologies, yeah...
SolomonZelman
  • 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
  • Abhisar
ok, one final question. In what cases we can use this method?
SolomonZelman
  • SolomonZelman
In many cases. I really want to through a lot of examples rather than giving a verbal definition.
SolomonZelman
  • 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
  • SolomonZelman
mostly, the once that have the derivative of u sitting in the integral.
SolomonZelman
  • SolomonZelman
ones*
Abhisar
  • 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
  • SolomonZelman
You are always welcome!
Abhisar
  • Abhisar
I will post my answers in a new thread since this one is becoming annoying to handle c:
SolomonZelman
  • SolomonZelman
one note: By indefinite integrals, you always need to sub in the original variable back (after you integrate with the substituted variable).
SolomonZelman
  • SolomonZelman
alright, if you like to use a new thread:) Whatever you like!
SolomonZelman
  • SolomonZelman
c(u) then
Abhisar
  • Abhisar
Yeah c; and thanks a bunch again c:

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