kirbykirby
  • kirbykirby
Integral 2b(1+t) dt
Mathematics
chestercat
  • chestercat
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kirbykirby
  • kirbykirby
\[\int\limits2b(1+t)dt\] How does my prof get b(1+t)^2 + C but I got 2bt +bt^2 +C
Mimi_x3
  • Mimi_x3
\[\int\limits2b(1+t)dt \] \[\int\limits2b(1+t)dt \] \[2b\int\limits1dt +2b\int\limits td t\] bt^2+2bt+c
kirbykirby
  • kirbykirby
It seems like my prof might have done lik a usubstituion on (1+t) but how come it doesn't give the same answer?

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Mimi_x3
  • Mimi_x3
\[2b*t+2b*\frac{t^{2}}{2} +c\] \[2bt+bt^2+c\] it is the same..
anonymous
  • anonymous
\[\int\limits2b(1+t)dt= (1+t)^2 + k\] Leave the 2b alone and put it back later since it's a constant. \[=\frac{1}{2}(1+t)^2 +k\] Put back the 2b. \[=\frac{2b}{2}(1+t)^2 +k\] \[b(1+t)^2 + k\]
anonymous
  • anonymous
your professor is right . because \[\Large \int\limits_{}^{} (ax+b)=\frac{(ax+b)^2}{a}+C\]
anonymous
  • anonymous
its 2a !
anonymous
  • anonymous
k is a constant. Just replace that k with a big C.
kirbykirby
  • kirbykirby
b(1+t)^2 = b(1+2t+t^2)=b+2bt+bt^2 but this is not the same as 2bt+bt^2 o_O
Mimi_x3
  • Mimi_x3
but im still right!
kirbykirby
  • kirbykirby
am I missing something?
anonymous
  • anonymous
yes you are !
anonymous
  • anonymous
@kirbykirby You meant to differentiate it.
anonymous
  • anonymous
You expand then differentiate.
anonymous
  • anonymous
Wait sorry you differentiate by not expanding.
Mimi_x3
  • Mimi_x3
differentiate?? this is integration!
anonymous
  • anonymous
No, he wants to get the answer back to the original expression.
anonymous
  • anonymous
When you integrate it, you can differentiate it back to the expression that you originally integrated.
anonymous
  • anonymous
@kirbykirby You differentiate that expanding expression you had, to get back your original expression. Integration is the reverse of differentiation.
kirbykirby
  • kirbykirby
Ok. I'm trying to understand though how the 2 methods though give a different answer though...
anonymous
  • anonymous
differentiate b+2bt+bt^2 with respect to t and you get: 2b+2bt=2b(1+t)
anonymous
  • anonymous
\[\Large \int\limits_{}^{}2b(1+t)dt=2b \frac{(1+t)^2}{2}+C=b(1+t)^2+C\] differentiate \[\Large \frac{d}{dt}(b(1+t)^2)+C=2b(1+t)\] you get original back . means it is correct .
anonymous
  • anonymous
Which is the original expression you had in the question...
kirbykirby
  • kirbykirby
oh is it because the b + C gets absorbed into another constant "C" using the other method o_O?
kirbykirby
  • kirbykirby
isn't it a constant here with respect to t?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
If it was respect to b, then it's a different story.
kirbykirby
  • kirbykirby
b(1+t)^2 = b(1+2t+t^2)=b+2bt+bt^2 but this is not the same as 2bt+bt^2 So is this difference because of the b being absorbed??
anonymous
  • anonymous
If you want to know why there's a C at the end everytime you integrate, it's because there are many curves that can still produce the same derivative. For example. The derivative of this curve y=x^2+1 is the same as y=x^2
anonymous
  • anonymous
That's for you to understand why there's a constant attached everytime you itnegrate. If you don't attach a C or k whatever, you get it wrong.
anonymous
  • anonymous
integrate*
kirbykirby
  • kirbykirby
No I am not asking why there is a C constant... I am trying to understand why: Integrating it gives either "2bt+bt^2 + C" or "b(t+1)^2 + C"... which gives "b+2bt+bt^2+C" which is not the same as above
anonymous
  • anonymous
\[2bt+bt^2 \neq=2b(t+1)\]
anonymous
  • anonymous
You're meant to differentiate to get the original result. The word differentiate is the key word here.
anonymous
  • anonymous
Once you integrate, you must differentiate to get back the original, otherwise, it's going to be certain that it's not the same as the original.
kirbykirby
  • kirbykirby
Because One method is using \[2b( \int\limits(1)dt+\int\limits(t)dt)\] = 2bt+bt^2 +C OR \[2b \int\limits(1+t)dt=2b \frac{(1+t)^2}{2}=2b(1+t)^2+C\]
anonymous
  • anonymous
Integration is just getting the original equation of a curve or function. When you're integrating something, that "something" is the derivative.
kirbykirby
  • kirbykirby
oops forget the extra "2" in the last expression
anonymous
  • anonymous
To be frank with you kirby, you shouldn't be using formulas when you're beginning to learn how to integrate.
kirbykirby
  • kirbykirby
I am not learning how to integrate :( I know how to integrate
kirbykirby
  • kirbykirby
I just don't see how these 2 methods are giving different answers
anonymous
  • anonymous
Still, these easy indefinite integrals require you to just think about differentiating, but working backwards.
anonymous
  • anonymous
If you were to differentiate 3x, I would probably first but an indice of 2 beside x giving me x^2 and then when you integrate s^2, you get 2x. but the original equation just has an x without the two. SO then you put a half in front of the x^2. That gives you (1/2)x^2. and then once you put back the 3, you get (3/2)x^2.
anonymous
  • anonymous
first put*
anonymous
  • anonymous
x^2*
anonymous
  • anonymous
That's how you should do it. If you have that mindset, nothing can stop you from getting good marks in integration.
kirbykirby
  • kirbykirby
I just want to be clear... I know how to differentiate... Factoring 2b and integrating "1" and "t" individually gives "2bt+bt^2+C" But if you factor 2b and integrate (1+t) using a substitution, you get "integral 2bu" = 2bu^2/2 +C= bu^2+C => b(1+t)^2+C
anonymous
  • anonymous
Why would you do it that way. You rather keep integration as simple as possible. Don't go into that mucky stuff. You will end up confusing yourself in the end.
kirbykirby
  • kirbykirby
I know it is stupid to do that, but shouldn't the answers AGREE? I mean what is the point if the two methods don't agree???
anonymous
  • anonymous
Then use one method. Simple as that.
kirbykirby
  • kirbykirby
But why does one not give the same answer as the other :( THIS is what I wanna understand
anonymous
  • anonymous
No point in pondering on two methods. I'm not using any of the two but integrating through differentiation.
anonymous
  • anonymous
If you really want to know, then go to your professor and ask him why.
anonymous
  • anonymous
Your professor's there to help you understand, not bore you to death.
anonymous
  • anonymous
*ask him/her
kirbykirby
  • kirbykirby
is it because the extra "b" in the second method gets absorbed into the constant C?? (which is why they give the same derivative)... but I never saw an answer like "2x+2+C" being absorbed as 2x+C

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