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kirbykirby Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits2b(1+t)dt\] How does my prof get b(1+t)^2 + C but I got 2bt +bt^2 +C
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.3
\[\int\limits2b(1+t)dt \] \[\int\limits2b(1+t)dt \] \[2b\int\limits1dt +2b\int\limits td t\] bt^2+2bt+c
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.0
It seems like my prof might have done lik a usubstituion on (1+t) but how come it doesn't give the same answer?
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.3
\[2b*t+2b*\frac{t^{2}}{2} +c\] \[2bt+bt^2+c\] it is the same..
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits2b(1+t)dt= (1+t)^2 + k\] 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. \[=\frac{2b}{2}(1+t)^2 +k\] \[b(1+t)^2 + k\]
 one year ago

sami21 Group TitleBest ResponseYou've already chosen the best response.1
your professor is right . because \[\Large \int\limits_{}^{} (ax+b)=\frac{(ax+b)^2}{a}+C\]
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
k is a constant. Just replace that k with a big C.
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.3
but im still right!
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.0
am I missing something?
 one year ago

sami21 Group TitleBest ResponseYou've already chosen the best response.1
yes you are !
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
@kirbykirby You meant to differentiate it.
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
You expand then differentiate.
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
Wait sorry you differentiate by not expanding.
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.3
differentiate?? this is integration!
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
No, he wants to get the answer back to the original expression.
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
When you integrate it, you can differentiate it back to the expression that you originally integrated.
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
@kirbykirby You differentiate that expanding expression you had, to get back your original expression. Integration is the reverse of differentiation.
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.0
Ok. I'm trying to understand though how the 2 methods though give a different answer though...
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
differentiate b+2bt+bt^2 with respect to t and you get: 2b+2bt=2b(1+t)
 one year ago

sami21 Group TitleBest ResponseYou've already chosen the best response.1
\[\Large \int\limits_{}^{}2b(1+t)dt=2b \frac{(1+t)^2}{2}+C=b(1+t)^2+C\] differentiate \[\Large \frac{d}{dt}(b(1+t)^2)+C=2b(1+t)\] you get original back . means it is correct .
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
Which is the original expression you had in the question...
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.0
oh is it because the b + C gets absorbed into another constant "C" using the other method o_O?
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.0
isn't it a constant here with respect to t?
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
If it was respect to b, then it's a different story.
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.0
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??
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
\[2bt+bt^2 \neq=2b(t+1)\]
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
You're meant to differentiate to get the original result. The word differentiate is the key word here.
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.0
Because One method is using \[2b( \int\limits(1)dt+\int\limits(t)dt)\] = 2bt+bt^2 +C OR \[2b \int\limits(1+t)dt=2b \frac{(1+t)^2}{2}=2b(1+t)^2+C\]
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
Integration is just getting the original equation of a curve or function. When you're integrating something, that "something" is the derivative.
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.0
oops forget the extra "2" in the last expression
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
To be frank with you kirby, you shouldn't be using formulas when you're beginning to learn how to integrate.
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.0
I am not learning how to integrate :( I know how to integrate
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.0
I just don't see how these 2 methods are giving different answers
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
Still, these easy indefinite integrals require you to just think about differentiating, but working backwards.
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
That's how you should do it. If you have that mindset, nothing can stop you from getting good marks in integration.
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.0
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???
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
Then use one method. Simple as that.
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.0
But why does one not give the same answer as the other :( THIS is what I wanna understand
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
No point in pondering on two methods. I'm not using any of the two but integrating through differentiation.
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
If you really want to know, then go to your professor and ask him why.
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
Your professor's there to help you understand, not bore you to death.
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.0
*ask him/her
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago
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